The Whole of A Level Maths | Pure | Revision for AQA, Edexcel, OCR AND WJEC

Hey guys, I'm really excited about this because this is the beginning of everything you need to know for a

level

math. This is the

pure

section and the lovely mystery will take you through each skill, one at a time, for each skill in here is a much longer video that breaks down each skill at three different

level

s, lots of questions, so if there is anything what you are not very sure about, go and watch the longer video, many examples are explained and solved. help you make sure you can develop each of these skills on my website.

There are thousands of multiple choice questions waiting to help you review and make sure you are confident in these skills so you can take the exam and apply them. when necessary the index laws normally involve a base number let's call it n and something happens to the base number for example when multiplying it now with the index laws we also have an index involved so we might have the power a in our first n and the power b in our second so the law is that if you are multiplying the base number and to simplify it you can add the powers so let's take a look at an example of this in action so you have 7 to the power a 2 multiplied by 7 to the power of 3 and that can be simplified to 7 to the power of 5.

Now there are also laws like this for subtraction and that's when you divide the base numbers so that by dividing the base numbers, then the two indices will be removed, so again, looking at an example, you could have 3 to the power of 5 divided by 3 to the power of 2 and that will simplify to 3 to the power of 3 by removing the powers, we also have a version for multiplication, we have your base number and you multiply a and b by what It looks like you actually have a base number in this situation, this will lead to a power a and then all of that will be to a power b, so if you have a parenthesis and you basically have two powers, you will multiply the powers together, so, looking at another For example, let's say we have 4 to the power of 2. and then that's all to the power of 4, then all together it will give us 4 to the power of 8.

We can also have one where you divide the powers and instead of writing this as a division, probably it will be um more. It is commonly seen as a fraction but fractions and divisions are the same thing now let's say a is the numerator and b is the denominator so what would this look like? Well, we are going to have our base number, it will be raised to the power of a and then it will be up to the root b, so if I look at an example, if we have the cube root of a number, let's say this number will be a2 and 2 is raised to let's say 6, then that would be equal to 2 to the power of 6 over 3. and then of course you can simplify that 6 divided by 3 is 2, so that would end up being 2 to the power of 2. so always look for opportunities to simplify these fractions if Now there are some additional details of the laws of indexes, for example, if you have a number raised to one, then you should stick to writing the number.

If you have a number raised to zero, then it is equal to 1. If you have a number raised to a negative power, then that is the same as 1 over that number raised to the positive power, so look at an example of this, if If we were to say 3 to the power of negative two, that will be equal to one divided by three to the power of two and actually three to the power of two is nine, so it will be equal to one divided by nine, just as you can simplify fractions, you can also simplify fractions of these cases and actually multiply things if that makes it simpler. but for now we will just focus on the basic rules.

I've used numbers for all my examples, but you can use algebra, so for example, you could have a to the power of 2 multiplied by the eighth and that will be a to the power. of five or you could have a to the power of two with brackets and then you could have said a b on the outside and that would be a to the power of a and that would be 2 times b is 2b, so we can have algebra involved. We can also have a combination of letters and numbers, for example we could have 3a squared times 4a squared, so use the normal rules to make all the numbers three times four equal twelve, then you would follow the laws of indexes for the indexes two plus two. is four so three a squared times four a squared would be twelve a squared three times four is twelve two plus two is four if this happened with the brackets, be very careful, you could have, for example, two a to the power of two, all to the power of three, now you don't just do two times three or six and write two a to the power of six, that's the wrong answer because not only does a get raised to the power of three, so you do two times three or six, but two is raised to the power of three and two to the power of three is eight, so be careful because it can also have multiple letters, so you could have a to the power of two and b to the power of three multiplied by a and then b to the power of four then you make the letters separately a two and one that is two plus one is a square three notice that there is no one in the a well as I said before nothing to the power of one is just that number then do the same with b's three more four is seven, so we can have a combination like that, the last thing we need to keep in mind is the negative numbers, you can have double negatives and those four normal rules, double negatives, which you will find particularly when you divide because it definitely takes away the powers and a times one of the powers is negative so be very careful with the double negatives there and the other is with the fraction when you have n raised to the negative power a you can turn that into a fraction or make it positive this can also happen with a positive m a to the a would be 1 over n to the negative a, so you're not actually making the fraction positive, what you're doing is reversing the sign in the power if you're converting it to a fraction.

One last thing to keep in mind is that they all have the same base number. Now let's say the base numbers are different, so let's say we have seven to the power of two multiplied by two to the power of 3. Now, because the base numbers are different from seven, another two instead of a seven and a seven are a two and a two, so in that situation you can simplify it, but what you could do is say that seven to the power of two is. forty-nine and 2 to the power of 3 is 8. which will give you 49 multiplied by 8 and that will give you a total of 392. so if you have the same base numbers, you can simplify if you don't have the same base numbers The best you can hope for is to change them to the values ​​and get a value.

The same also happens with addition and subtraction. We know we have loss from multiplication, division, brackets and roots. We don't have any rules. to add the same base number or subtract the part number and that is because we are answering the rules and again like the example did if you are adding or subtracting the best you can hope for is to convert them into the values ​​and solve that way we have some laws with certs so for example if you have a root say root a and you multiply it by say root b then you can simplify it to the square root of a multiplied by b and the same thing will work with division if you got the root of a divided by the root of b o written as a fraction, that is the same as the root of a divided by b o again written as a fraction, so let's look at some examples, let's say we have the square root of nine multiplied by the square root of seven then we could say that everything together is equal to the square root of 63 if we had the square root of one hundred over the square root of four then we could say that it is equal to the square root of one hundred divided by four and let's write them in a little more detail so that it is the square root of hundred divided by 4, which is the square root of 25.

You can also have laws for addition and subtraction now, these have the same base root, so we have to say a lot of root n plus b a lot of root n, so That would give us a plus b a lot of root n. Now the general term seems quite confusing but let's look at an example, if we had three lots of root seven plus two lots of group seven then all that means is that we would get five lots of root seven and the same would work with the substraction. Now you can also simplify sets, for example the previous one. answer we got root of 25, well, we can find the square root of root of 25. 25 is a square number 5 times 5 is 25.

Another way to simplify is that in the first question we had root of 9 times root of 7. There is another way to write that. root of 9 we can solve it root of 9 is three, so we could have written root of seven three times, which would have given us three root of seven, so it would be much simpler than writing root of 63, we are using smaller numbers. small ones and in fact what you can do is work backwards, they may give you the root of 63 as a question and you have to recognize that 63 is a square number multiplied by something else, it's 9 times 7 and then you can change the 9 times 3 and then you can write your 3. root of 7 that way, so in the first example you can also work backwards.

Also with gentlemen we have something called rationalizing the denominator, so let's say we had 33 over root 11. That looks good, but we don't like having roots at the bottom of a fraction, so what we do is use equivalent fractions, we take 33 over the root of 11 and we multiply the denominator by the root of 11 and the numerator by the root of 11. Now, we multiply by those particular numbers if you multiply a root. by itself you will get just that number, so root of 11 times root of 11 you are squaring it and the square root cancels out, which leaves you with a number greater than 11. and then in the numerator we are going to have 33 lots over 11.

So that's a better way because with a lot of math you don't want an irrational number as a denominator, so you might be asked to get rid of it and let's just make a note at the top if you have to say root of a multiplied by root of a, so that will give you another thing to look at to rationalize the denominator is that we can actually do the division here, we have 33 over 11. 33 divided by 11 is three, so that will give us three lots of root 11. Another thing to keep in mind when the denominator Rashford can have something like this and the denominator can be some brackets with a certain it and with a number with it now with this you can't just multiply by root three because you multiply the one by root three , which will leave you the root of three at the bottom, so again we'll just take our numbers, you might even see the question without the brackets when you're doing the working eight put the brackets in and you want to multiply it by a negative version of the same bracket.

Now what it does is it expands this quadratic and you can have quadratics with services and just be careful when you multiply a third by a number that you know root of three times one is one root of 3 and root of 3 times root of 3 will give you only 3. so be careful with things like that, but what this would do is for the numerator we would have 2 times root 3 and 2 times negative 1, so it's pretty simple, but with a quadratic that you're going to do, you have kind of four multiplication terms root three times root three equals three three times negative one will give negative one root three one times root three will give us a positive root three and then one multiplied by negative 1 will give us negative 1.

Now, when you get this, what you will notice is that the two middle terms will cancel to give you zero and you will probably have two number terms as well. in the denominator, so the numerator will remain 2 root 3 minus two, well the denominator, the negative root three and the positive root three cancel, so we have three taking away one is two and then we can even simplify that because we can actually divide both terms by two two lots of root three divided by two is 1 root 3 and negative 2 divided by 2 is negative 1. Now you can't always do that, but always look at the situations where you can do it when expand the brackets, we'll look at a few examples, so you could have something like a linear expansion where we have a parenthesis, for example, two lots of three x's plus one and the way to do that is to multiply the two by the three x's, which which will give us six x and then multiply the two. by one, which will give us 2.

It is very important to multiply by both because the common mistake is to multiply by only one of those, you may have a letter on the outside like x and on the inside you may have 4x plus the letter y, like this that again the same method x times four of x look at the numbers first of all we only have four then look at the letters we have two lots of x that's x squared then we do x multiplied by y and we can't do for them, so we can't do something, just leave it as is x multiplied by y, just write x y, then you can have a quadratic expansion, so we could have something like five x plus two and nine x. minus three so again multiply the terms five x times nine x you could let the numbers first five times nine be forty five then look at the letters minus 3. look at the numbers first 5 times minus 3 is minus 15 and then we have an x ​​in there too.

Next we are going to do 2 times 9x, which would be 18x and then 2 times negative 3, which isminus 6. so these are the four multiplications that you are going to do the first term in the first parenthesis is multiplied by both things since I can divide it then the second term in the first parenthesis is multiplied by both things in the second parenthesis now I'm just I'm going to Write the answers below with a little more space and what you will find in the quadratics are often the middle two terms that can simplify negative 15x and positive 18x, so 18x subtracts 15x is 3x, so our final answer would be 45 x squared plus 3x subtracts 6. and we can't combine the terms x, x squared and the x's stay separate and we can't combine things with x and normal numbers, so we have three terms for our final answer, all we need to look for in quadratics with thirds involved, so we could have x many root seven plus five, all squared. and the square symbol means that we're going to have this parenthesis twice and then we can do our multiplication, so x root of 7 times x root of 7.

So what's going to happen there is we're going to have x multiplied by x , which is x. squared and then root 7 times root 7 now, if you multiply a root by itself, you will cancel out the root symbol q by squaring it, so it will be seven, so it will be seven x squared, we will do root of x seven times five, which will give us five now again it's a quadratic because often the middle terms are simplified, you have five watts of x37 and then another five watts of x root 7, so in total we have 10 lots, so our final answer is 7 x al square plus 10 lots of x times root 7 and then o25 one last thing to keep in mind is a cubic where you have three brackets so let's say we had x plus one x plus two and x plus three now the method here is to start as quadratic and just Look at the first two, so I'm going to highlight the first two and multiply them x times x is x squared x times 2 is 2x and 1 times x is 1x that's 3x total and 1 times 2 is 2. and once You already have that, so you can multiply it by the third parenthesis and again you will use the same method.

You only have three terms this time. Now I'm not going to draw the arrows here. I'm going to do x squared times x. is x cubed and then I'll do x squared times three is three our second term done now we make the third term 2 times x is 2x and 2 times 3 is 6. then all we have to do is add all of that together to get our final answer and all those terms together we have our x cubed term just has one we have two x terms squared three and three which is six x squared we have two x terms nine x and two x which is eleven x and the number ten at the end plus six so again if you see a cubic, pretend it's a quadratic, multiply the first two once you get that answer, then write in your third parenthesis and pretend it's a quadratic again and multiply it again so it's basically a cubic, that's two quadratics in a row for quadratic equations, you might be on to something. like x squared minus 2x minus 24 and the task here is to put it in double brackets so we need to factor this now we know that x multiplied by x will give us x squared we need to think about how to get the negative 24. now a A good technique what this is for is listing the factors of 24 so you can divide it by one, two, three and four, you can't divide it by five and while you can divide it by six, we've all written it down and we're done. our list, so please note that I went through all the numbers one by one to get a complete list of factors and then jumped at random.

Now they are all two numbers that are multiplied to give 24, but they are going to be added to make a negative 2. So we have to go through our list and say well which of these, since they all add up to 24, which pair of numbers could make a two and it looks like it will be the one at the end four and six omega two and how. one six omega two you would have a negative six and a positive four to make a negative two, so now it's factored, an additional question might be something like having it as an equation equal to zero and figuring out what x would be if everything were equal. to zero you simply have to make x equal to zero or rather one of the square brackets equal to zero, so for example, in the first square bracket, if six, which is zero and that will give the answer. equals zero because it doesn't matter what the second parenthesis would be and it would be two in this situation zero times two is zero in the same trick with the second parenthesis x would have to be negative four for the second parenthesis to be zero, it would be negative four plus four, which is zero and then no matter what the first parenthesis is, I think it's going to be negative ten, because you multiply by zero and you're going to get a zero.

You can also have quadratics like this with a coefficient. in the x squared, so 3x squared minus 12x minus 50. so again we want to get double brackets, you know, two numbers that multiply to give negative 15 as the numbers between those brackets and the option is three times x finished our list, so how can we get a 12 from those numbers? Well, we can, but another trick is because we have a coefficient on the x squared of three and that works its way into one of the brackets that actually goes to multiply the second parenthesis by three and to multiply the second parenthesis by three, so We can multiply one of its factors by three, so let's look at more options, we could multiply the one by three or we could multiply the fifteen by three we can multiply the three by three or we can multiply the five by 3.

Now, which of Will these options give us a negative 12? It looks like we have a couple of answers here, we could have 3 and 15. We have 3 and 15 either way, so what's going to happen with 3 and 15? You have to think about where these numbers originally came from, so I'm not going to write 3 and 50 in parentheses, I'm going to write the original numbers, the first choice of which is 1 and 15. Now, because the 1 is multiplied by 3, that will be in the opposite parenthesis and it would look like this, so what negative 12, so it will be negative 15 plus something. to get to 12.

If we look at the other option and sometimes you can have multiple answers, then the 3 and the 15 are actually the 15 that was multiplied from five by three, so the five would have to be in the parenthesis opposite the 3x to get to 15 and the three remain as they are, then again for this the largest will be negative, so I will be negative five and plus three now, interestingly, you have multiple answers, which is solving it equal to zero, you will always have it. get the same answer, so for the first set of brackets three x plus three that means three x would have to be negative three and if three x is negative three divide three by three x has to equal one now you will also notice that the second parentheses of the first option, if option again. the first parenthesis in the first option if x were five, that would also give zero.

Now both examples have two answers for x, but you can have one answer, two or zero, and the way to find out is using something called a discriminant. the discriminant is b squared we take away for a c so let's take a look at some examples so let's take a look first at x squared minus 2x minus 24 which we start with a is the coefficient of x squared in this case it's one if there is no number b is the coefficient on x and c is the number at the end, so b squared will be negative 2 squared, subtract 4 times a, which is 1 times c, which is negative 24.

That will give 4, subtract a negative 48 which in total will give us 52. Now, what does this tell us about the number of roots? Well, we'll leave that for a moment and come back to this, let's take a look at another example, so let's take a look at x squared plus. x minus 8. is good a b and c are going to be the same a is the coefficient of x squared b is the coefficient of x and c is the coefficient of the number at the end so b squared would be 1 squared and they take away 4 for a, which is 1 times c, which is 8. all together that will give us 1, we take away 4 times 1 times 8, that will give us 32. so all together it will be negative 31 Okay, so that is a negative number, now again I am going to explain what that means.

Look at one more example and then we'll explain all three together, so we'll look at four x squared minus four x plus one. we have b squared which will be the negative four squared again the term in the x subtracted four times a which is four and c which is one which gives us 16 we subtract 16 which is zero and now we can talk about what all this means with the discriminant so if you have a positive like in the first example let's write positive as if it were greater than zero that means that therefore we are going to have two roots we have two answers for x if it is less than zero like in our second example that means that we have no roots , there is no way you can factor x squared plus x plus eight and finally if we are equal to zero, then what that means is there will be one root, there will only be one answer, so when you have your factorization If you have the two brackets, you will find that both brackets actually have exactly the same number and that means you only have one answer.

Another formula is the quadratic formula and the quadratic formula is negative b plus or minus, so 'We'll do it both ways and we'll get two answers, the square root of b squared minus 4ac and you'll notice that it's discriminant and that's why a negative discriminant means there are no answers because you have to take the square root of the negative number you can't and that's all in 2a with a, b and c coming from the same source as they were for the discriminant a, the coefficient on x squared b, the coefficient on so for our first question negative b would be negative two, so we have double negatives. quite a bit here, so we have the square root of b squared, which is negative two squared minus four lots of a and six of four times one for a and multiplied by negative 24 for c, all of that will be divided by two lots of a. so that's two lots of one is two now minus minus two is four and four plus or minus the square root of and we already worked with a discriminant for this and it was 52 and then we divided everything by 2.

And in fact, looking back, 52 was wrong because I multiplied by 2 and I should have multiplied by 4. so actually four times one times 24 would be 96 instead, when we add the four that will give us one hundred, now one hundred makes more sense actually because then that is a square number and that means we can simplify it very well, so I knew right away that this was wrong because the root of 52 wouldn't actually give me

whole

number answers. Now you solve this 4 plus root 100, which is 10 is 14 divided by 2. which is 7. which is not one of the original answers I have, so I know I made a mistake somewhere with the original question.

My work here. I'm just looking at things I've noticed. negative negative 2 ended up being a positive 4. now actually negative negative 2 will simply be a positive 2. so now it will make more sense because now I have 2 plus root of 100 2 plus 10, which is 12, which divided by 2 is going to give us a 6 which we have in our final answer and then once we try it with a plus, we try with a negative 2 taking away the root of 100, so 2 taking away 10 gives us negative 8. We'll divide that by 2 and get the negative 4 which we also had in our original answer now if you did this and then the root was not a number squared and you would get a decimal, you rooted it, then you have two options, you know the workers totalizer and you have a rounded decimal answer or let's just leave your answer as 2 plus or minus root of 100 over 2, all the equivalent of something that can't be simplify and that should also be fine unless you have been asked to provide two roots or equivalent and Maybe you have also been given attention.

We're going to go over some different ways to solve simultaneous equations and by simultaneous what that means is that we're going to have two equations, so we could have 2x minus 5y equals negative 14. and 5x minus 4y equals negative one, so that we have two equations. The reason we have two equations is that we can't calculate what x and y are unless we have two, so we have two variables. We need two equations. Now the way to solve this is that we need to make the coefficients of x or y equal and the only way to guarantee thatis to multiply them together, so let's say we are going to try to make the x's equal to each other, the two and the five, multiply the two by the five and the five by the two and when you multiply make sure you multiply all the terms, so two x times five is ten 4y times 2 is negative 8y and negative one times two is negative two so what What we have done now is we have made the x terms equal, they both have the same coefficient, so now they have the same coefficient as we need to cancel to get zero and the way this time is by removing them from each other 10x, the conclusion is 0.

Sometimes we have to add and that is when the signs are different in them, but in this case they are both positive, we can remove them, so removing 10x from each other. 0 then negative 20y subtracting negative 8y in total will give us negative 33y or a little bit like that but I've made a mistake now if you look at this more closely it's negative 25 as negative eight so we have got a double negative so in It's not actually negative 33, we're going to add eight, which will give us negative 17y, so now we're removing two terms. Now check with the third term, which is negative 70, take away negative two again. that will be double negative gives us negative 68. now negative 17y is equal to 68 which is an equation that we can solve we can divide both sides by negative 17. so negative 68 divided by negative 17 gives us four so y is equal to four now that we know that y is equal to four, we can find what x is and we do this by substituting into one of the original equations.

I'm going to choose the bottom one because I think it has smaller numbers overall, so no matter which one we choose the same answer right away, so 5x minus 4y equals negative 1. So we know that y is 4. You can see that 5x minus 16 equals negative one and then we can solve that we can add 16 to both sides. giving us that 5x equals 15 and divide both sides by 5 so that x equals 3. The last thing you need to do is record your answers very clearly so that x equals three and y equals four, so we have an In a few steps here we have equalized one of the variables, whether it's x or y, okay, we do this by multiplying them together and then we add our takeaway to cancel out that variable.

In this case, it was a takeaway and we figured it out. the equation then we substituted into one of the original equations the number that we just got and it is equal to four and then we solved it so that all together we do about five things. Other examples to consider here are things like x squared plus three y equals fifteen. so we have a square sign on this and then we have them without a square sign x plus y is equal to five, so another way to solve this is sector substitution. What we can do is rearrange this to make y the subject and we can Let's say y is equal to five minus x.

I can substitute that into our original equation, so we have x squared plus three lots of y and we know that y is now five minus x, so if we expand the brackets we get that x squared equals. to the x term the first three times minus on the left side we now have a quadratic and by calculating the quadratic we will get our answers now if we factor this quadratic because there is no number term at the end we are just going to take x8 so I divided both terms by x so to find out what is x we ​​want everything to be equal to zero, so if subtract three is zero and then multiply that x by zero, so there are two different ways to make this equal to zero, all we do now is substitute those numbers into our original equation which would give us zero plus and equals five and would give us three more y is equal to five, that means that y would have to be equal to five r and it would have to be equal to two at the end writing the final answers our first set of answers was x being equal to zero and y being equal to five and our second set of answers was that x is equal to 3 and y is equal to 2.

The last thing to keep in mind is that you can also do this with linear equations, for example if we said that y is equal to 2x minus seven and y is equal to the square root of negative two x plus nine. What we could do with these two is we could combine them, they are both equal to y, so we made them equal to each other and that will give us two x minus seven is equal to the square root of negative two x plus nine and then remember that you can rearrange them and then that will give you a quadratic because our first step will be to root both sides up to the right square of both sides to get rid of the root that would give us 2x minus 7 squared equals negative 2x plus 9.

Now I won't do the rest of this because we have already seen it. but we rearrange it and get a quadratic and it's like the last example, you're probably going to get two answers for x and once you get two answers for x and plug them into whatever original equation is to determine which one is y, we'll have a look at how to factor a cubic and which we'll see is x cubed minus x squared minus eight we know it must be x multiplied by x multiplied by for 2. There are many different ways to do it and we don't really want to use trial and error and expand lots of square brackets, so we need to use a simpler way. method now we're going to use a little bit of trial and error and the way we do it is using substitution.

I need to guess what our first factor will be. You'll see why in a moment, so let's assume our first factor. will be when total, that actually gives us four, which means that one is not a factor, so now we could try negative one or we could try uh, positive two, so that will be for two, two cubed minus two squared, minus eight lots of two. plus twelve, that will give us eight, minus four, minus sixteen plus twelve and all together that gives us zero, which means that two will be one of the answers, so if x is two, then we will have x minus two in one of the brackets because I'm going to say that it is equal to zero, if all of this were equal to zero, then x being equal to two would mean that our first bracket is zero because we take two away from two, which makes everything equal to zero, so now we know one of our factors, now we can use the method to find the other two factors and that method is long algebraic division, so we'll take our factor x minus 2 and we'll divide our expression by that's x cubed minus x squared minus 8x plus 12. so we use long algebraic division, so the way this works is we can't really divide by x minus two, that's a little complicated, we're just going to divide by the first term x I'm going to do each term one at a time look at x cubed first so let's divide x cubed divided by x is x squared now that's not really our answer, we don't divide by negative 2. so what we do now is multiply x squared by our divisor uh original x squared times x is x cubed and then x squared times negative 2 is negative 2 x squared, so what that means is that we didn't actually get the negative 2x squared. from this, we can now include it, so by doing that division that we have taken into account according to the expression x cubed, subtract 2x squared, then we remove it, you know, from the entire expression, now the first two terms will be calculated to make zero, then have negative x squared remove a negative two x squared we could remove the double negative so that would be negative x squared plus which would be a positive the same thing, so we divide by our first term and we have seen what is left over.

Now we're going to divide our new second term, not the original second term and the new one, by x, so x squared divided by x will give us x. so of course we didn't divide by negative two, so we multiplied it to account for the negative two x times x x squared x times negative two is negative two x then we took away what we found from what was left of our expression the first two The terms can't have to choose zero and negative eight x remove negative two double negative x, which doesn't show a negative six x and then we have the plus twelve that we haven't taken anything away from, so that's our second term.

We've already dealt with the third term, now it's the third term of our new expression minus six x and we're dividing it by x, so minus six x divided by x is negative six, so again we're not dividing by negative 2, so we have to Keep in mind that multiplying by negative 6 times x is negative 6x and negative 6 times negative 2 is a positive 12. If we take them away now we get 0. and if you get zero, you have no remainder, that means x minus two actually divides into x cubed minus x squared minus eight x plus twelve and therefore it's a factor, so it's a little proven what our substitution showed us before, now what it leaves us with is this, we know that x minus 2 is a factor but also the answer to the division of the factor x squared plus x minus six and now what we can do is that our first parenthesis is fine, but our second parentheses are quadratic and we can convert it to two types of linear parentheses. so now you just use your quadratic factoring skills, now minus six you want two numbers that are multiplied to give six and they're going to add to make one x well two times three would make six and then get positive one and make positive three minus two , so we're just using our factoring quadratics using inspection skills and there's a factored cubic that's x minus 2 x minus 2 and an x ​​plus 3.

We can even simplify that we have two lots of the first parenthesis. we can give it a little square sign there and then write our third bracket, then functions is something that we can use for substitution, for example, so we have the function that we call f of x a function of x and we say that is equal to three x plus five so if we want the function of let's say five, then we swap all the instances of x with five, which will be three lots of five plus five, three times five is fifteen and plus five is twenty, so that's the main use of functions.

Now there are some things with functions that you need to be able to do, the first thing you need to do is substitute functions several times, so let's say we want to find f g of 3. Well, what could that mean if we say that g of x is equal to 11x and f of x is equal to x plus three now we have two functions that we can look at so we have the function of x so to find gf of 3 what we do is start with the inside, we start with the letter closest to the number, so I choose what the f is and what I'm going to do first is calculate f of x and that will give me f of three , which is three plus three, which is six, another f, then we do g and it will be g for our new number and new answer, which is six, and g for six, which will be eleven times 6, which is 66 for our answer, so To substitute in multiple expressions we start with the expression closest to the number being substituted, first we do f and then we do g.

Now we can also find an expression that is g. of f of x we ​​can combine them without doing substitution now the way we do it is we are going to start with the outside with g and g is 11 x but what we are going to do is we are going to exchange the x with the function f which is x plus 3 Now if we substitute this and shoot 3, we should still get 66, so let's try it so that g f of 3. It's going to be 11 loops with 3 plus 3 3. plus 3 is 6 and 6 times 11 is 66, so that method really works so we look at just regular functions and substitute and look for compound functions where we combine multiple functions the next thing we need to look at are the inverse functions so let's say we have a function of x and it is x minus nine over six now we can find the opposite of this so we put a number in this and we got an answer let's say x was 10 then we would have 10 take with nine is one and one divided by six is ​​a sixth so we get a sip now the inverse function is a function where we put a sixth and we get ten out of the other side we work in the opposite direction now the method to do this is to first translate this to almost linear type of equation so that y is equal to x minus 9 over 6. and then once you've done that, flip the two symbols so that x andand be the other way around, which will give us then we can add 9 to both sides and that will give us 6x plus 9 equals y and once we get to that stage, all we do now is change y again for f of x, but this time f minus one x, which shows which is an inverse function, which is what the negative inverse function does. is six x plus nine, so remember the previous example, if we substitute a sixth into this, we should get ten on the other side, so if a

whole

. one and 1 plus 9 is 10. so check that it's actually going to work, so let's go over the kinds of things that we've done, we've done substitution with functions, we've drawn composite functions, we've combined multiple functions and we've done inverse functions, so what three different types of partial fractions are about taking an expression, say 12x minus 8, all divided by x.squared minus 5x minus 24 and changing it instead of being a fraction changing it to two fractions now how you change it to two fractions has to do with how you factor this so your first job is to factor now if you can factor the numerator 4 as a factor and factor the denominator is a quadratic so you can put it in double parentheses and we should have x plus 3 and x minus eight when you factor the quadratic using whatever method you want now once you have it we can see the fractions we need again it is still a fraction our new fractions will have x plus three and x minus eight as denominators but the numerators we don't know what they will be , so let's call them a and b now that we're at it.

I'm going to have to try to simplify that, so what I'm going to do is write everything down again so I have some space to work because I'm going to draw all of this and make There are a lot of crossed outs for all of this, so what the left side is written again we also want to make the right side very nice again and what we're going to do is try to get rid of the fractions, that's our first step, so what I want what we need to do is multiply by all the denominators, like this that we have x plus 3 as one of the denominators, so we're going to multiply the three terms that we have here by x plus three, then the other denominator is x minus eight, so we're going to multiply everything by x minus eight also now you might think what's the point with this because we haven't gotten rid of the denominators but then we can simplify three, we can say well, x plus three divided by x plus three is one and then I can't slightly multiply the numerator by one, it will stay the same and x minus a divided by x minus a that will cancel and also do the same on the right side, we have x plus 3 and x 3 cancel comes out and x minus a and what that will give us if we then write this again is 4 lots of 3x minus 2 and that's all that's left on the left side and on the right side we will have a lot of x minus a plus b a lot of x plus three now we can substitute the values ​​of x to get rid of one of the variables because I can't solve equations, we have a and ab, we need to get rid of one of them, so we substitute two things to get rid of one x would have to be equal to eight because then the parentheses would be eight, minus zero and a multiplied by zero is zero, so if we said that plus 3. and then you can wear it down, so on the left side 3 to 8 is 24, then we take away 2 to get 22 and multiply by 4 to get 88.

That's equal to 8 plus 3 is 11, so it's 11b , so b must be equal to 8 so this is when we got rid of a and got b now I need to get rid of b to get a now in the parentheses b we have a positive 3, so if we take 3 away from that negative 3 plus 3 is 0 and b multiplied by 0 is 0. so let's substitute we have 4 lots of 3 multiplied by minus 3, then I'll take away 2 and that will equal many negative 3s, minus 8 and the b bit has canceled to give us 0. solving this 3 times minus 3 is negative 9 take away 2 is 11 and multiply by 4 is negative 44. right side negative 3 take away 8 negative 11 that's negative 11a so divide by negative 11 and a is equal to four so now when we write again our fractions, let's write an answer a is four, so it would be four over x plus three and b is eight, so it would be eight over x minus eight, so I'm just going to Shade that now to show that it's our final answer, like this which is highlighted in pink.

This is our answer. Now there's some extra stuff here, so, for example, let's say b was negative eight. What we would do is have plus minus eight and minus eight. As a numerator what you would do is have negative eight so that you can reduce the negative symbols in the numerator so that they are the type of symbol between the fractions. The other thing is that this is a regular fraction, the numerator is smaller than the denominator and very small, we're looking at the powers of x, you know, 12x, we have x to the power of 1 and then with x squared, that's x to the power of 2.

Now, if that were the other way around and you had a higher kind of value of x at the top in terms of its power if the power of x was higher at the top then what you would have to do is divide it to try convert it into a mixed number and the way to do it is long algebraic division and if you divide the two things, you can also create a long division, then you will get a new fraction that you can deliver. First we'll draw the line y is equal to negative two x minus three, so there are two key parts to this, the first part is the y-intercept and the y-intercept with the line going to cross the y-axis that crosses at negative three, so it is a complete quantity zero minus three, it has zero as the x coordinate because we cross on the y axis that is where x is equal to zero the second important characteristic is the gradient which in this case is negative two so with the gradient the gradient is a change in y over the change in x now if it is an integer we need to write this as a fraction, what is y over x?

So I had your whole number. Well, you can write whole numbers as fractions, just write them over one. So what we're saying is that every time the y coordinate changes by negative two, the x coordinate will change. changes by one, so if you go down in the y coordinates and you cross one of the x coordinates, the place that was highlighted in pink there, we go down two for the y direction and cross one in the x direction and we just keep repeating that, keep repeating that pattern one and again and also do it in the other direction, the opposite, this is go up two and then cross minus one and we get a line like this.

Now I'm just going to draw this line. I will be using a ruler and a pencil. I just want to show you how to connect all the points, so what we have here is the line and it's equal to negative 2x minus 3. Now we'll look at something else here. What if we are? not plot that at all, what if we're plotting the modulus of negative 2x minus three and for modulus we have these? They're almost like square brackets, but they're just straight lines on either side, so what does this mean now with modulo it means we can only have positive values, so let's write some notes on that.

The modulus, so the lines on either side mean only positives and likewise, what we actually want are positive values ​​of um and first, so we look at our line, you'll notice that the first part and the left side all of this is positive since all the y coordinates are positive the x coordinates are negative so we are only concerned with the y coordinates since they are positive now we have got the positive side then it becomes negative and what we are saying is that modulo only has positive y values, so when we go down to the next coordinate that we have plotted here, which is negative one negative, what we really want to put in is minus one positive one, so the quadruple y is now positive with the intersection with the y axis, which is zero minus three, so it will be zero plus three.

The next one I plotted was one on the x axis and minus five on the y axis, so a negative five becomes a five and we see the same kind of gradient pattern here, but it goes in the opposite direction, so when you're plotting modulo, you're going to plot this section here and join it together, and then this is the line and it's equal to the modulus of 2x minus 3. Now the effect of the modulo is that you can probably see that it's almost like the line is bouncing off the x axis. , so with our original line it will go directly through the x axis. axis, but the modulus is now bouncing around, so that's something to keep in mind when asked to find the modulus of something.

The next thing we're going to look at is how to solve equations using a graph, so we're going to solve 3x plus 8 equals negative 7x subtract 2. and we can solve this using the graph now, the way we do it is first, we take a look at the left side, we have three x plus eight, let's draw the line y is equal to three x plus eight, then we look at the right side and we have negative seven two now the green and the pink part are equal to each other in our original equation, so we can make them both equal to y, they are still equal to each other, so let's draw the two lines so that three x plus eight, the intersection with y being at eight and the gradient the difference in y over to cross by one and then we can plot the graph this way, then we can draw a straight line, so there's this line, now look at our second line, minus 7x, take away 2. so the y intercepts at minus 2 and the gradient is it's going to be negative 7 over 1 so we go down 7 on the axis and we're going to cross one on the x axis now we're going to go down once as you can see before we go off the edge so we're going in the other opposite direction we're going to go up 7 and I'm going to end up like this and I can't really draw the next one now the important thing here is that I only draw the little segment of line that I have I can estimate what's going to continue so if you use a ruler you'll get a really accurate line.

The important thing about these two lines is that you can see that they both intersect and they both intersect, so we want to look at the point of intersection now the x coordinate is minus one and the y coordinate is at five and this is actually the solution to our equation because we have an x ​​coordinate and a y coordinate and we're looking for x and x is negative one so that's our answer separately, making them both equal to and and the place where the lines intersect. Your answer is finished, next we will draw quadratic lines, so the first one will be y is equal to x squared plus x minus two.

Now the usual method for plotting things like this is to substitute, so if you pick a coordinate and substitute it in, you'll get some values ​​that you can use, so let's say x is equal to three, when x is equal to 3, that will give us 3 at square plus 3 minus 2, which together will be 9 plus 3 minus 2, which will be 10. and what you just found is the y coordinate, so when x is 3 we would have the coordinate 3 10 and we can plot that on the graph Now we could do that for each value of x, but this is going to take a while, so let's look at some shortcuts.

The first shortcut is that the number at the end of this is negative 2, which is the y-intercept, like with straight lines, so we know that our curve will go through negative 2. Now the y-intercept is when x equals 0. This means we don't need to substitute 0 for x. We can also factor this out. I'm going to check it to factor. We have a review slide on that. When you factor this you will get x plus two and x minus one. Now, if we solve this to be equal to zero, we will get that x is equal to negative 2 and x is equal to a positive 1. the opposite sign because what we are trying to achieve here is to make one of the brackets equal to zero, so we multiply by zero and we get the answer zero, so it is equal to zero, so by solving the quadratic we know the x intercepts that we know that Let's go through x is one and x is negative two, so I have two more coordinates again without having to do any substitution.

Now, once you have them, you can think about what else I need to replace. So we can see that we have negative two zero one three, we don't have two, so you might want to substitute two and see what that gives us, so two squared plus 2 minus 2. That will give us 4. so that will be the coordinate 2 4 and we can plot 2 4. So it's almost like we don't really want to substitute each value, so we can use the x and y intercepts to get an idea of ​​which one to use. substitute because then after x is three, if we substitute x for four, it will go off the graph and it will be a value that we won't be able to use, so we don't want to substitute anything bigger than those, there's one more.

What is useful and this is that atDifferentiate this we can find the inflection point of the curve so we differentiate d and by dx we reduce the power x squared becomes 2x and then when you lower the power you use it by one so the power 2 becomes power one zero and you multiply it by zero it will disappear, so d and times d x are two x plus one, what does this help us? Well, that is the gradient of the curve, so when that is equal to zero, it is when the curve has a straight gradient that is going to be the inflection point, so we say that two x plus one is equal to zero for the point of inflection, which means we have two x equals negative one, so x equals a negative half, which gives us the coordinate for the inflection point.

We know that x is going to be minus 0.5. You'll see that we don't have the y coordinate for that, so we have to substitute it so that it gives us minus a half square plus minus a. We take two away from half, so a negative half squared will give us a quarter and a quarter plus a half. negative, so negative plus will be calculated as a negative subtraction, so we will have a quarter minus half. It will give us a negative quarter and then we take away two, so it will be negative two and a quarter in total, which gives us the coordinate minus half minus two and a quarter and that will be the inflection point of the curve, so with It's a little complicated to find approximately.

Here we can see that the inflection point is actually not very far below the y-intercept for this one. Now you will notice that some coordinates are missing and the trick here is that these graphs are symmetrical, for example on the opposite side. side the inflection point will come to the same height as the y-intercept, then we have the x-intercepts that we both have and then on the right side after the x y set goes up by four, so on the left-hand side it will also go up four, it will have the same height as with the green coordinate, then we go up to the top, you go one to the left, it will have the same height as the next coordinate so you can use a symmetry of the curve to fill in the missing coordinates and now we can draw it and don't make straight lines here try to make it curved, you have to draw it freehand, make sure it goes through all the points and there is the curve.

Let's review the techniques we use. We were able to find the y-intercept simply by inspecting the formula. We were able to find the x-intercept by factoring it and solving for zero. We were able to find the inflection point by differentiating it and then we were able to find all the coordinates. that's what we wanted using substitution and you will need to use two additional substitutions, although differentiation also requires substitution to find the y coordinate of the inflection point, this is how we plot a quadratic, let's look at another example, so we have y equals negative x squared plus three find the x intercepts now, if you factor this, we're going to eliminate x as a factor that will give us, we're going to eliminate the negative as a factor as well, so it's a negative negative 3. and if you solve that is equal to 0, that means that in the bracket x has to be equal to three and outside the bracket x has to be equal to zero, so the sets x and y are at zero and positive three.

It's interesting for this case that if you don't have that kind of number at the end and the y intercept is zero, it may be the case that the x and y intercepts will be the same, in fact, they will be one of them and it will have to be zero. Next, we can differentiate this for the inflection point, so d y times d x will make the two go down minus two x. to the power one reduce the power and it will be plus three and what we want is for that to be equal to zero for the inflection point, so we can take three from both sides and divide both sides by negative two so that x equals three divided by two will equal one and a half, so now we can substitute the y coordinate with one and a half, so we want uh minus one and a half squared plus three lots of one and a half so you can write three. over two or you can write a one and a half, it doesn't matter which way you write it, now we can substitute it to have the one and a half, so I'm going to square it.

I want the negative version of that. It's going to be negative 2.25, then we'll add three lots of 1.5, which will give us 2.25 or two and a quarter. The really important thing here is that this is a negative half at the beginning, don't square the negative half. number that you are substituting in which is 1.5 one and a half and then the negative symbol appears then remember that you are doing the disease first then use attractions, so this will give us the coordinate one and a half two and a quarter that we can plot so that is the point curve inflection, now we need more information to plot the bottom of this graph, so what we're going to have to do now is make some more substitutions, since we have values ​​for zero, one and a half and three could be useful to look at four now I have a space to write this on the page but we are going to do four squared which is sixteen or the negative of that to negative 16 then we are going to add in three lots of four, that gives us negative four in total, for what that will be the coordinate four minus four with the symmetry.

We also know that negative one will always be d in negative four, it will be the same height as we might want. try uh 5 too, so we want 5 square 25, so negative 25 plus 3 lots of 5 and that gives us negative 10. so we have 5 across, we have 10 down and then the negative 2 of the symmetry will be there like Okay, I'm going to start substituting like I did with one and a half and like I did with two and three in the previous question, so now we can draw the graph because I don't need any multiplication, I can see. That negative 2 and 5 have taken me to the bottom, so any of the substitutions will be a waste of time, so there's our final answer.

Next, let's plot a cubic. Let's plot y equals x. cubed minus three x squared minus will be zero 3. so y-intercepts always have an because I have a review slide on that, so I'm just going to give you the answer, but remember that some long algebraic division will be helpful, so the factoring will be x plus 1 x minus 1 and x minus 3. Again, that factoring will give you It will take a little time and will be worth knowing a couple of points on their own. Now, if we solve it equal to zero, we will get the x-intercepts and we will make it equal to zero. one of the brackets has to be zero, so the first bracket could be a negative one because negative one plus one is zero, the second bracket could be positive one and the third bracket could be positive three, so those are our intercepts with x, we have one negative positive one and positive three now we have this data on what the inflection points of the curves are and this can be done by differentiation, so again we start with a full cubic, differentiate it, reduce the power and reduce the power by one you know, reduce the power two and there's the number one upside down, you multiply 2 by 3 it's 6 and you reduce the power by 1 and you should get this now at the turning point, this will be equal to 0 because that's where the gradient is flat. there is no slope at an inflection point now if we factor this quadratic we will get our inflection points now again we have a different slide to review the factoring.

I think this was the user's quadratic formula and what it will give you is x is equal to 2.2 and minus 0.2 to a decimal, so we know the location of the inflection points, but we only have the x coordinates, we want the y coordinates, so to get the y coordinates we're going to substitute them. our original equation, so we're going to do 2.2 cubed, we take away 3 lots of 2.2 squared, we take away 2.2 and then we add 3. and so on for the coordinate and that will give us negative 3.1 to our decimal place for the first one, then we do the Same with the second we have - negative 0.2, we are going to cube it, we are going to take away three lots of negative 0.2, we are going to take away negative 0.2, that will be double negative, so we add 0.2 and then we are going to add three and that gives us three points one to a decimal so now we can plot the inflection points the first term point x will be 2.2 and y will be negative three point one and you can Let's not plot it exactly with the kind of level of detail, let's do the best we can.

The second inflection point curve this way, what we need to do now is see how it goes off the graph, so we need more substitutions, so I'm not going to write this, but basically you know. I have one point times 1 and then I have a little bit more than two and I have three, so let's take a look at four, so we'll have four cubes minus three lots of four squares minus four and plus three and that. gives us 15. So what this tells us, this is going to go off this graph very, very quickly, in fact, it's going to be something like this, it's not actually going to make x equal to four before it goes off, so the top will be at and is equal to 10. when I get to 15, I'll be about two-thirds of the way there, let's look at negative two, so we want negative two cubed, negative three lots of negative two, negative negative two, double negatives adding two and then We're going to add three and that gives us negative fifteen, it's going to be the same in the other direction, yeah, when you get to 10, it's going to be two-thirds of the way to negative two, because when we get to 15, It's going to be all the way and there's our cubic laid out pretty accurately, I think actually, so let's look at all the different techniques.

Look at the y-intercept for the number at the end of your equation. You will get the x. -intercept by factoring the cubic, so you might be doing a little bit of trial and error here, you know, try canceling due to one as give us zero, you might be using long algebraic division and then we use differentiation to find the turning points once you have them, then what you might want to do is try to replace them with a couple of additional points to get a full picture of what you are missing. Inequalities can be solved on a graph and you'll see some really nice advantages. in a moment, the first one we'll look at is y is greater than negative two x minus seven and y is greater than or equal to seven x plus two, so what you want to do is draw the two lines and let's start with the top line has an intersection with the y axis at negative seven, so we can plot it and the gradient is negative two, so it could go down two every time we cross one on the x axis and then in the opposite direction. up in two and we can draw the line now you'll notice I'm going to draw a dashed line for this and the reason is this is a greater than line so we make a dashed line if it's greater than or equal to you make a solid line and we'll talk of the importance of this in a moment once we have both lines, our second line is y is greater than seven x plus two, so we have our y intercept at plus two and our gradient is seven. it's going to go up seven times each time in the opposite direction, it's going to go down seven times each time, so we're going to have something a little bit like this, now you'll notice that I'm making a bold line, a solid line, this is approximately greater than or equal to, so you have the equals sign involved, so that's where you have your solid line.

Now the benefit of this is that we're not looking at the intercept point here that would solve the equation if these were equal to each other and we don't have equal signs, so we're looking at areas for our answer, so let's look at the first one and it's greater than minus two x minus seven, that means we want things that are greater than the line, we want things that are above the line and you can see it in blue. I'm just highlighting above the green line for the pink line which is also larger so we want things above the pink line so we're going to highlight it and what we want is you want to shade the part of the graph that I colored twice and which is a little bit at the top, you can see that it's part of the green area and this part of the pink area, so let's now completely shade the bits here towards the bottom of the graph.

The left side that was just above the pink line, on the right side, was just above the green line, so we don't want them, so we're going to delete the extra shading that I added to the exercise and therefore, the blue shaded area. This is my answer, there it iswhere both inequalities are true. What you can do now is choose the coordinates of that area and they will satisfy both lines, so let's say we choose the coordinate here, which is negative um 2 4. So if you substitute in negative 2 and 4, what you will find is change to negative 2 for x, you will always be greater than four, that's what it means, so you can have any coordinate here as you answer, but it will often ask for you. know exactly any coordinate now, what about dotted line and thick line?

Now with the thick line you are allowed to have coordinates that are on the line with the green line. You are not allowed coordinates on the line, so it is greater than or not equal. to your load coordinates on the line, any quadrant on the pink line is fine if it's larger than that, you're not allowed any of the coordinates directly on top of the line, so that's a really important thing to keep in mind account, let's take a look at another inequality so this time we will see y is greater than negative x squared plus two x plus two I'm going to see y is less than two draw the two lines so that y is less than two draw y is equal to two but it will be a dashed line because it is not equal if we read a it is simply less than not less than or equal to, so we are not allowed any of the coordinates actually on that line, it is just the ones below the line, so we're going to plot the quadratic, so use all the things we reviewed to plot quadratics.

I'm not going to analyze it in too much detail so that we know that it will go on the y axis, we know that if we factor it we will get the x-intercepts and I'm going to factor, but I think it will be in 1 negative 1 and 2. And you can know how to plot the quadratic mu substitutions and obtain precise coordinates. Now I'm not going to map this out too precisely. I just want you to have an idea of ​​what we're up to. doing y is greater than, so it will be a dashed line again, so it will be something like this, then we solve this inequality again, it forces you to put the appropriate effort into drawing it and you want to find more coordinates than I do. there, but y is less than 2, so we want everything below the pink line and y is greater than the quadratic, so we want everything above the green line.

I want to shade the shaded area twice, now the bit in the middle which is has only been shaded once, so let's rub the top part which is just above the green line so we can shade that, yeah, from the left sides and right, which is not only below the pink line but it's also above the green line so we can have the entire area on the left and then it's actually the same thing on the right as well, so the entire area on the right is below the pink line, it is smaller, but it is above the green line, it is more than that, so we can shade the entire right side and even below the pink line, so for this question we have two shaded areas and that is something that can happen with water sports but not all the time sometimes you have a little bit in the middle or a little bit on top so you have different options but this is a way out that we can get sometimes with quadratics we can also solve inequalities algebraically so the based example would be something that looks like an equation to solve so let's say two lots of eight x minus six is ​​greater than or equal to sixty eight just pretend you solve an equation but you keep the sign as it is instead of having an equal sign so you can divide it by 2, which gives you 8x minus 6 is greater than or equal to 34.

You can add 6 to both sides to give 8x is greater than equal to 40. and then you can divide both sides by 8 saying that look at x squared plus two 2x minus 8 is greater than 0. Now, if we were to pretend that this is equal to 0, we could factor this quadratic and then we factor it x times x. is positive the equation t gives us two x and negative two times four gives us negative 8. if we solve for e equal to 0 we would have x is equal to 2 and you might think well, okay, we can just put the greater symbol that x is greater than two and x is greater than negative four.

What's happening here is that greater than negative greater than two is redundant because anything greater than two is also greater than negative four, so that's a little clue that something is going wrong here. The key to this, even though we're solving algebraically, we're going to need a little graph to show this, so we draw. our curve now I'm not going to draw it precisely now I know that the y intercept is going to be negative 8 and all the x intercepts are two right and negative four and this is pretty much all the information we really need so I'm I'm going to draw a curve, no I'm too worried about how accurate it is, you'll see why in a moment, so it'll be something like this.

I was saying that is greater than zero, so we also want y to be equal to zero here, like when Solving them graphically we will do the same thing, so we want greater than zero, so by shading the parts of the curve are greater than zero and the parts are greater than zero, these bits here are the part of the curve that is greater than zero, so we will look at this in more detail, so the first bit that is greater than zero occurs at negative two and the second bit which is greater than 0 turns out to be minus 4.

Now just look at where the graph goes. We are going below negative 4, we are going towards negative 5. That means that x will be less than negative 4 because the answers less than that are on the line, on the other side we see both, we can see our lines. going towards three, it will be things greater than two, so we have two different answers with different symbols, different arrows playing in different directions x can be greater than two which is above zero or x could be less than negative four because then the graph es will also be above zero in terms of the y coordinates, so again we have two answers because we have two separate parts of the graph shaded.

Now let's look at another example. in negative a get something like this, so if we pretend that it is equal to zero, x would be equal to zero and x would be equal to a positive seven, then we know that seven will also be an x ​​coordinate. Now if we draw this graph, it will look like this and again I'm not too worried, it's true, I don't know how high it will actually go, but it will be something like this and we want when it is greater than or equal to zero, so what part of the curve is above zero so the y axis shades it, it kind of loops in the middle.

Now let's look at the numbers, so we have where x is zero and we have a x is seven so you can see that. we are allowed values ​​between zero and seven, so when we write this we know that x is equal to zero and x is equal to seven, we want x to be greater than zero and x to be less than seven and then because the originals greater than or equal to, we can put the equal signs in them and another thing we can do is write that in an alternative form so that x is greater than or equal to zero and x is less than or equal to seven.

They can be combined so that zero is less than or equal to x and x is less than or equal to seven, so they are both the same and represent the same information, but you can combine them to write minus our two examples here, so what we just did above zero on the axis and you can see that we have the curve type and it's all in one piece in our first example. see when y is above zero we have two separate pieces so that's just a hint how to write it in two separate pieces you want two separate answers like x is greater than two and x is less than negative four if everything is together and that's the clue you want it all combined so you want zero to be less than x which is less than seven looking at transformations of functions we have the black line on each graph and this is the line f of formula of this line, now the quadratic will be an x ​​squared, we don't know exactly what form it will take, so each one is y is equal to f of the first graph in blue we're going to make y equals f of x plus 2.

So if we do that, what will happen is the graph will shift two places in x. -axis because 2 is next to x and so if you just take important coordinates, for example the it comes out the graph at the top will shift in two places, so if you look at a graph like this now, you might think this has gone back two, not forward two, so you're always going to think when you're. transforming functions will do the opposite of what you think you're almost saying if you add one x to two you want x to appear two places before if you take before and we can say about this if you look at the coordinates, each coordinate has decreased by two and the y coordinate has remained the same, let's take a look at another transformation, so take a look at y is equal to f of x plus two now this time more two is on the outside of the brackets, so the effect can rearrange this and if you remove two from both sides but y remove two is equal to f from x and then we can see that around the last one where x has had something added to it, this time y something was added to it, you're going to do the opposite of what you think, so if y is minus 2, Actually, we're going to add two, so again let's take important coordinates like, for example, the x-intercepts, which are going to increase on two to get a new graph where we go off the graph that I got on two, now that goes off the graph, so it might not be the most useful to look at them, you could look at the y-intercept, which will also increase by two.

They see that the y-intercept is one versus the x-intercept, so it will also be one in the other direction. now we can make a little sketch of this line, so it will look like this, so everything has been shifted up two places now again, if you look at the coordinates, what's happening with the quadrants is that the x coordinates are staying the same. like Same thing, but you're adding two to the y coordinates, so what we have here are basically translations, so we have translation on the x axis when you have your number in parentheses with the x and then we have translations on the y. -axis with the numbers outside the brackets and if you rearrange it you can see that it will be next to the y so let's take a look at some more transformations so let's move on to the second graph because I want to write too many on the same graph because it will will get confusing, so the next transformation will be y is equal to f of 2 x, so we think about what will happen to the coordinates.

Here are inside the brackets the numbers next to the x, so Something is going to happen with the x coordinates, it was the opposite of what we expected, this will be multiplied by two, so the opposite will be divided by two , so the x coordinates could be divided by two and the y. the coordinates will remain the same so again find some important coordinates so again we got the x intercepts one of them is at negative one so divided by two will be negative half one is at negative three divide it by two and will be minus one. point five so we know the curve is going to be here the y intercept is when x is zero so that's not going to change if you divide zero by two you still get zero and you use symmetry to have the point on each side also So what about the graphics at the bottom of the page?

Well one is right after one so it's going to be half right after 0.5 and then the other one is just above negative 5 so divide that by 2 and a little bit above negative 2.5 and so we can draw that and what you can see is that if you divide the x coordinates by two, you actually make the curve more squished, it takes up half the space on the x axis. Let's take a look at another one, so Let's see that y is equal to 2 lots of f of x now, again, we can rearrange it to get that number in y, so that y is equal to half of y is equal to f of x, so dividing by two now, as always, does the opposite of what you would expect, so if you divide the y coordinate by two in the equation of our answer, we will multiply it by two, this will be y times two, so we multiply all the y coordinates by two now the x intercept is really important here because that's been y zero if you multiply zero by two you still get zero that's not going to move around whatwill change where it leaves the page so that's where y is 10 so that goes to end in y is 20. well, we can't plot that, but we'll see that y is 5, when we duplicate it we'll get 10, which is where it leaves the page, now all you have to do is look at the two places where y is equal to 5.

If you duplicate it to 10, you simply draw the same coordinate in the same place on the x-axis, but then you move up to 10 in the y axis and we can actually, probably draw this now, maybe a more important point is the inflection point of the curve, so when y equals one, if you double it it will be at y equals two, so that the curve will look like this and so what? what you can see here is a curve again, it looks like it was squashed, it got smaller, what actually happened is it got a lot bigger.

You can see the negative part of the graph at the bottom, twice as large in the bottom half. also twice as big as the top half, it's just that you can't see the top half because it has moved off the page. Now this is more equivalent to an enlargement. We are stretching and flattening the curve, making it bigger and smaller. It is effectively a kind of extension, but it would be better to call it an extension, whereas with the translation you moved the curve. around the next one that we are going to see is y is equal to f of negative x now, if the next one is disabled at x, all this means that the x coordinates will become negative versions of themselves, so if you look at the so the bottom of our curve will now be on the other side the y-intercept is really important because x is 0.

Negative 0 and 0 are the same thing, that's not going to change it, we can use symmetry to see in the other side what that height will be and then maybe look at where it goes off the graph, so we'll go off the graph at 1 and negative 5. so that's communicable minus one and positive five, so a curve will look like this, you'll notice that it looks exactly the same, it's just on the other side of the graph and effectively this is a reflection on the y axis, so they are reflections and it's a little strange that you know that if it's a reflection on the y axis and it's the line of symmetry of the mirror line, then it is the real x -coordinates that become negative and stay the same.

The next one to look at is y is equal to negative f of x and again let me rearrange this. We can put the negative in y by multiplying both sides by negative one, so what this means is. that the y coordinates will become negative and logically I assume that this will be an x ​​axis, reflects the reflection now again if you look at the coordinates so that the y coordinates are negative, the equals, but then the inflection point on the negative one, the negative version of that negative negative is the positive, so the inflection point of the curve will be the other way around, the mental steps on the three, does that become? negative three and then look at the place where it goes off the page at one and negative five will stay the same, instead they will go to negative 10, so we have a reflection on the x-axis and reflection on the y-axis to look at now, The last thing we need to look at is something interesting called modulus, so with modulus, the first thing we'll look at is y is equal to the modulus f of x and we use these lines to show the modulus symbol, what this means is that the x coordinates are going to be positive, so we have a positive x and y is going to stay the same, so let's take a look at our graph, where are the positive x coordinates? the positive on the right side that I'm highlighting in purple, this part will stay the same.

Now we're going to draw some lines on the negative part of the x-axis, but instead of being the line, you can see the rest. of the curve where the negative x-axis values, what we're going to do with the negative side is plot the positive x-axis values ​​again and it's actually going to look like this, so all we've done is all the negative values of x. We have put the y coordinates of the positive values ​​of x, let's look at the version for y which would be y is equal to the f modulus of x again, the lines showing that is positive, so when you look at the positive, so we look at the curve, we look at the part of the curve where y is positive, so I'm going to highlight that in purple and this is all part of our answer, it will overlap with the previous answer also now, when we get to the part of the curve where y is negative, we simply make it positive, so the inflection point of the curve at negative 1 has to be positive.

The y coordinates have to be positive now, so it's going to go up to a positive one and it's going to look like this, it's almost like you bounced off the x axis when you're drawing the line, so there are the four transformations that you notice that have. translations we have magnification, so you know, it stretches and contracts, we have reflections and we have the module. One last point with this is that they may ask you to do a combination of these, so you may get and it will be the same. to negative f of 2x and that means they're both going to do a sort of squish on the x axis by a factor of two and then mirror it on the x axis, so again be careful when you have combinations. of these, you do both transformations to the x and y coordinates, we can see different things we can do with a straight line and for all these questions we will use the same pair of coordinates, so we will see everything we can do with the coordinates minus 9 minus 6 and three seven so what can we do with a line between those two coordinates?

The first thing we can look at is the midpoint of the midpoint of the line. What we need to do is look at the x. and the y coordinates and look at the words in the middle of those coordinates so the way to do it is for the x coordinates add them together let's see what x1 and x2 are so let's say our first set of brackets is x1 and y1 our second The set of parentheses is x two and y two, add x squared and divide by two and then do the same with the y coordinates, which will give us negative nine plus three divided by two and will give us negative six plus seven. divided by two and then we calculate that negative nine plus three is negative six and we divide it by two we will get negative three, then we can do it with the coordinates and negative six plus seven will give us one and then one divided by two will give us half, so that the midpoint of the quantity in the middle of the line is negative three and a half.

Next we will look at the length of the line to find the length of the line we are on. We're going to use the Pythagorean theorem, so we need to know the type of distance at x and y to make our triangle have the distance at x, we want the difference between x1 and x2 and then we're going to square that, so that's going to be the first. side of our triangle, then we will add to that the difference between y1 and y2 and that will be the second side of our triangle and then we will take the square root of our answer to find the hypotenuse, which will be the line length, so it will be negative 9 minus 3, which will give us negative 12. then we have our and, so we have negative 6, negative 7, which will give us negative 13 and what is the square root of the answer to that will give us negative 12 squared is 144. minus 13 squared will be 169. in total that is 313 and the square root of 313 gives us 17,692 decimals, so that will be the length of the line next we can find the gradient of the line the gradient is the change in and over the change in x, so let's look at our y coordinates, they're actually going to be the same as the Pythagorean, so they're going to be y one minus y two and x. one minus x two and in total that will give us negative thirteen over negative twelve and then if we simplify that both negatives can become positive, it will be thirteen twelfths the gradient change to introduce the number 1 and 112.

That's how steep the line is. all the things we can do with the gradient we can have parallel gradients, so the gradient of parallel lines will have the same gradient, so a parallel line will have a gradient of 13 over 12, which again is 1 12. Perpendicular lines are lines at right angles to this line are going to have a negative reciprocal gradient and to do that you make the gradient negative and flip the fraction upside down so that 13 over 12 becomes 12 over 13. and then we can't simplify that yet more, the last thing I'm going to do is find the equation of the line and this is the technique you need to find the equation of all types of lines, given all kinds of tracks, find the equation line will be and is equal to mx more c for the equation of any line, so what we want to do is shoot this now that we know what the gradient is, it's already uh 1 y 1 12 and what we want to do is We're going to put the x and y coordinates from this and we'll also create and we will choose the two coordinates.

I'm going to choose x ui 2 because they are positive numbers, so y is 7 m. The gradient we had is at 13. over 12 as a fraction and x is three from a second coordinate, so we can solve this to find out what c is. First we can multiply three by three so that seven is equal to thirty-nine over twelve times that fraction times three. So we're going to subtract 37 over 12 from both sides. Now, if you do that on a calculator, you'll get that c is equal to three and three quarters, so we write the equation of the line. is going to be and is equal to let's write the gradient as a mixed number this time, so 1 and 1 12 x plus 3 and three quarters these are all the different things that you might be asked to find in a line, keep in mind that with this i'll set it up with two coordinates, but you could start with one coordinate and you know the gradient could give you a parallel gradient or a perpendicular gradient.

There could be all kinds of different starting points, but as you go through these things, we're going to get the numbers right, the equation of a circle tends to look like this, so we're going to have an x ​​inside a parenthesis with a number and that squared, we're going to have a y inside a parenthesis with a number and that square has to equal a number at the end, so what can we get out of this? Well, we know that the center of the circle will be the coordinate 3 minus 1. So the 3 minus 1 are from the numbers inside the parentheses. just note that the signs have been reversed, the radius is the square root of the number at the end, so the radius of that circle would be five, so that's what you normally see for the equation of a circle, now there is An alternative form, you can have this the equation of a circle looks like this the center and the radius from the previous way the way we do it is by completing the square so we complete the square for x so all we're doing is squaring this to a power of one x and then the coefficient on the term x is halved if we do that, although Not only are we going to get and then we complete the square for the y and squared becomes y 12 and it will become six, but again negative six times negative six because 36 we need to take away the extra 36, ​​so now we need to put all the number terms together to have a negative one , a negative 36 and a negative. 44.

So the pair of brackets looks good. If we put all the terms together, we'll get minus 81, which is equal to zero, and we want to put it on the right side, so we'll add 81 to both sides, so when you add all your numbers, you reverse the sign and move it to one side, which means that now we can calculate some things like this to know that the center is going to be equal to 1 6. again, the sign has been reversed and the radius is going to be the square root of 81, which is 9. this is how we use the equations of circles.

One more thing you may be asked is to find the tangent to a circle and a tangent is a line that just touches the circle on one edge. at right angles to the radius or diameter, as an example we are going to find the tangent of the equation in a circle and we are going to have a coordinate with it which is the tangent at the point 3 6. then 3 6 is on the edge of the circle and the tangent will touch the circle right at that point, so how do we do this? The first thing is that we want the gradient of the radius, the gradient, the change in y over x and we actually have two. x coordinates and two y coordinates, so let's take 3 6 first, the y coordinate and the x coordinate of 3 6.

So where is our other coordinate? Well, we know that the center of the circle will be at 5 3. So we can remove. 5 3 of this and this gives a change in y and find the tangent gradient the tangent gradient because it is a right angle it is a perpendicular line it will be a negative reciprocal so it would be negative x over y instead of y over x so a green sorry negative double negative positive and thenwe turn the fraction upside down so that the tangent gradient is two thirds, so now we can find the equation we want and it is equal to mx plus c, we are going to substitute in the y values, it will be a point on the edge of the circle you can't use the center here, so we have to use three-sixths, so y is six, x is three and now we know the gradient, the gradient is two-thirds, so if we expand the bracket that will give us six over 3 which is 2. so 6 is equal to 2 plus c so take 2 a from both sides c will be equal to 4. that means the equation of the tangent will be y is equal to 2 3 x plus four, let's say what you want to do, let's see a little what a tangent looks like.

Someone draws a rough circle here so that the tangent goes through a circle and touches it at a point and from the center to that point on the edge. we have a radius, it will be a right angle, so we have the tangent, we have the radius and it is a right angle. The parametric equation is an alternative way of writing equations, so let's say we have x equals t minus one and y equals. at two t squared this is a parametric equation now what does this mean? Well, we're going to go through the process of converting this into the type of equation that you're used to and once we've done that, we can then look back and think about why this might be used, so what I want to do is make both equations are equal to t so we can set them equal to each other and merge them so that for x it is quite simple, we can add one to both sides and we get that t is equal to a different color so we can color code, we have y is equal to two t squared, so what do we do?

What we're going to do is divide both sides by 2, which will give us that half of y is equal to t squared and then we can take the square root of both sides to get the square root of half y is equal to t. and now they are both equal to t we will make them equal to each other so that and be the subject so we can square both sides, that will give us x minus 1, all squared and a half and on the other side, now we can multiply both sides by 2 and that will give us y z equals two lots of x plus one squared, you might want to multiply the brackets, so if you multiply the brackets you will get y equals and remember to multiply by two at the end we will also get two x squared plus 4x plus 4.

Now the point of this is substitution, like this Let's look at substitution, let's say we know that y is equal to 1 or we want to find the x coordinate when y is equal to one, so how would we do it with a parametric equation? we'll just say, well, if y is equal to one, y is going to be equal to two t squared, but we say y is equal to one, so we can solve this, we divide both sides by two to get half and then we can square root , so y will be the square root of a half, then we can substitute that for x, so we can say that x is equal to the square root of a half minus one, so very quickly we get the x and y coordinates.

Even better, we have this third variable t and it might sound something like time and if you have values ​​for t then your substitution will be even easier. Now let's take a look at y equals one with our formula at the bottom, so we can say that one is equal to two x squared plus four x plus four, so to get what x is going to be here, maybe we'll have to To rearrange this, let's remove one from both sides to get two x squared plus four x. so it's going to be negative three equals zero and then you want some kind of method to factor this, you might have to use a quadratic formula for example, and that's going to take a little bit of time and then once you do that, you're going to get its value for x comes out, so hopefully we can see here that it's a regular type of equation for a line.

It is very good to find the y coordinates. You usually go out and get a y-coordinate pretty easy. We're trying to find an x-coordinate and substituting y, it's not very useful. so again it's just an alternative way of writing it so you can make substitutions more easily in some circumstances now let's look at a more difficult version of this this was a quadratic equation let's look at a circular equation as parametric so let's have x equals sine theta plus three and we're going to have y is equal to cos theta plus seven, let's play that minus seven actually so we can see how the positive and negative work, now the way we're going to combine them is our third variable here is theta and we can't really get rid of the cos sign to get theta on its own.

What we can do is use a trigonometric identity sin squared theta plus cos squared theta equals one, so we can use this to combine the two terms making them sine theta and cos theta. The subjects we'll come back to that in a moment, let's rearrange them so that for x we're going to remove 3 from both sides, which will leave us. with sine theta is equal to x minus 3. well, cos is going to add 7 to both sides, so it's going to be cos theta is equal to y plus seven, so if we get our trigonometric identity back, we want sine theta squared to replace it now. we have sine theta sine theta is x minus three so we want sine theta squared we just need to square it then with squash theta squared we have cos theta is y plus seven so all we have to do is square it and now we combine the two parametric equations together in a single form and you can notice that it is actually the equation of a circle.

Now okay, a little bit of reorganization, but we have it right away, so now we can say things about the circle. We can see that the center of the circle. The circle is at three minus seven and we can say that the radius of this circle is the square root of one, which is one, and again there are different benefits to both shapes that you see combined, it is very easy to find the center on the radius, while that when the Repair it will be easier to substitute theta values ​​or even x and y values ​​and rearrange, but rearranging with a complete equation is going to be much more complicated.

Before looking at the binomial expansion, we have to look at Pascal's triangle. The triangle has an infinite number of rows, but let's take a look at the first eight rows. Now start with one and for the next row we will have one and one. Now it is clear what will happen with the next film. we have one, two, and one, and what's happening with this triangle is that each number is the top two numbers added together, so the two are the top two added together at the edge, you only have access to a one and a white space, so they stay as one, so for our next row the edges of Pascal's triangle will be ones, but then we have one plus two is three and two plus one is three, so we can generate the triangle like this every time adding the two numbers. above, so one plus three is four, three plus three is six and three plus one is four, so let's continue, so we have each time and the numbers above and you can see alternate rows, you have two identical middle numbers, so It is a type of even number.

From the number of terms you have, then the two middle numbers will be the same. If it's an odd number like this, then the middle two terms will be different, so let's output the last few rows. You will also see that this is symmetrical. the numbers on the left are the same as the numbers on the right now what it's used for is to find the patterns when you expand the brackets so let's say we had a quadratic, we had x plus one squared, if you expand this quadratic you get x squared plus two x plus one and compare that answer to the second row of Pascal's triangle, we have a one, two and a one in the triangle and then you look at the coefficients here, we have a one , so it matches that pattern, so when we look at x plus one cubed, that should give us x cubed plus three x squared plus three x plus one again using that pattern and then we can do it right, it's the power of four So we can use this pattern, we're going to have x to the power of four plus four x cubed plus six x squared plus four x plus one now, as well as the coefficients on the x that match the rows of Pascal's triangle, for example, the fourth row we have I have one four six four one in an expansion we have one four six four one also take a look at the powers now the powers at x are decreasing every time so we have power 4 power 3 power 2 power 1 and then We don't have a pair at the end, which means it's going to be a power 0.

We can use this pattern to write a formula, so if you have a plus b raised to a power, that's what we just did. a b four that we just did, let's say we just want the term in x squared, what would we do? What we would do by looking at this formula is first, we have n over k, now we know what is going to happen. to give us n is the power, so this is a powerful expansion and k is going to be the pearl looking, so we're looking for the power 2. so a, that's going to be our first term for x, so it's going to be x to the power of n minus k so 4 minus 2 . and then b, which is a 1 here, will be a power of k, will be a power of 2. we do this multiplication, we should get six x squared, now the key thing here is what is this parenthesis about?

It looks like a vector, we have four over two. Now what it does is it generates the terms in Pascal's triangle, so if you have n over k, this uses a formula. where you take the factorial of n, which is m multiplied by all the numbers below it, all the integers below it, now divide by k multiplied by n minus k and generate the part of the triangle you're looking for. Now n is the row and k is the column and remember that columns always start with zero so we label this for the bottom row. We'll have column zero one, two, three, four, five, six, seven and eight, but obviously that's going to be different for each row again, each row starts. with zero, then you just count, so when we say four, two, we say you want the number in row four and column two, so find row and four, then we look for column two, so zero one and two and we can see. that will give us the six, so the type of vector here, the row and the column, will give us 6. x to the power of 4 minus 2 is x to the power of 2. then we multiply it by 1 squared and 1 squared is 1, so which 6 times 1 will give us 6.

And you can see that by applying the formula we have 6x squared right away, so it's useful for finding particular terms. You could ask for examples to find the last two. terms about the first two terms, for example, or perhaps a term about a particular power or like we just did here. The last thing we need to think about is that you don't need to draw a triangle for every question, so we're talking about the vector type notation with n and k is actually a button for that on your calculator and the corresponding button is the ncr button, so type n then there is a button with n c air and then press k when you type that into the calculator you should see n you should see a c and then you should see the k and again that is the method for your calculator to know which row and column of Pascal's triangle you want.

Arithmetic progressions are some of those basic sequence types that you would use. search on gcse but it might have slightly different terminology, you could say a, which just means the nth term, so let's say the nth term is half and plus a third, so you would have to be able to generate the sequence for generate the sequence. uh I would start with one in the first term, so you would have half plus one third and you could do it without using your fraction skills, but half plus one third is five sixths and then for each pair of sequences, the way which will be expanded Is it going to increase by half each time, so for the second part of the sequence n will be two, which will be two halves, which is the full plus one third which will be one and a third in the next part of the sequence n? would be three, so you have three halves plus one third and what you might find here is that instead of working with mixed numbers, it might be better to work on improper fractions, now one half will be the same as three sixths, so just add three sixths . each time, eight sixes is the same as one and a third, but now it is easier to add the three to the numerator each time to generate a sequence and then the sequence would continue forever.

You may be prompted for specific parts of the sequence, so you may be asked to say u10 for the tenth sequence of the pattern which will be one half times ten plus one third. Now, if you were to do this with mixed numbers, a half times ten would be five and five plus a third would be five. and one third, if we can stay consistent with what we're seeing before, then we could say, well, this is 16 thirds, which would be uh 32 6. so be flexible in theway you can write your fractions depending on the context because if you have the same denominators for all parts of the question, it's easier to compare, so you see that 32 over 6 is much larger than 17 over six the fifth term now I'm looking at five and a third and I can't You have to guess how it relates to the other parts.

If you are asked to generate the nth term of the sequence, then look at this sequence and you should be able to see that it increases by 3 6 each time, which is the same as another half each time. time and then if it does the same thing every time and it does it, then you know it will be half of n, so the number at the end, the coefficient on n, will always be increasing each time. Now, if that were true, so it would be. half in the first term would be half times one, which is half and not five six, so you think of half to get the first term five six, now five six and a half, the denominator is different, what is the half? three six starting from three six to five six we have to add two sixes, which is the same as a third, that's where the third comes from in this, so this is kind of a summary of how it works in ggs, these are more complicated, Hey. fractions involved, which is quite common now, another thing you can be asked to do is add a certain number of terms in the sequence and we have a formula for this, so we have this symbol, this is the sigma symbol and the sigma symbol. it just means addition and it's going to add all the things that we'll see here now above this symbol we have n this is a number of terms that we're going to add and then below the symbol we have a number and the number underneath that's where the addition begins very will often be equal to one next to this you have the nth term so we have the previous question, we could have half n plus one third and this is the type of terminology that you will see in the exam and it asks you for the sumThe formula for this is s is equal to n over t times 2a plus n minus 1 times d, so we have to think about where all these different letters come from, so n is a number of terms to add and a is the first term. of sequence, so we know from our previous work here that the first term is 5 6 well, that's something you can generate from the nth term by saying that m is equal to one d is what we call the common difference the difference common is what changes each time so in the sequence it increases by half each time, so the common difference is half, so let's say we were given a question about this, let's say we said okay, let's summarize, we want the first five terms, which I simply turn here.

I'm going to start adding from the first term, but in parentheses you actually have the nth term, so we would substitute into the formula s equals the first five terms divided by two, two lots of the first term. a so the first term is five six two lots of five six plus n minus one so five minus one multiplied by the common difference d which is half so we can start to solve this, so five divided by two will give us uh two and half, two times five-sixths is ten-sixths, I'm going to add to that five and subtract one is four, so four times one-half we can continue working on this to have two and a half times ten-sixths.

Also, now half of four is two, so if you do ten six plus two and then multiply it by two and a half, the answer I get is six point one six recurring, which we can write as a fraction in which a place occurs . a six, then it will be nine and a six, so if you want to spend some time on this, you can continue working with fractions or you can do what I just did there and just work it into the calculator as a decimal and convert it. Going back to a fraction at the end, you're probably better off keeping it in fractions the whole time as you work, but I'm going to speed this up because it's just a review slide, so now what I've done here is we've added the first five terms, now let's write these down let's see this actually works so 5 6 plus 8 6 plus 11 6 plus 14 6 plus 17 6 ​​should give us 9.16 and on my calculator it does so we can see the formula should work one last detail is that for this formula we have k is equal to one the starting point from where we are adding is the first term now if the starting point is not one let's say there are two then you have to take the first term discard your answer because this formula it doesn't take the first term, the kind of starting point in two counts, if you were to start counting at the third term, they would make your formula, which will be the answer for all the terms up to where you are.

We're looking for and and then if you want to start counting the third term, you need to remove the first and second term, so keep in mind your starting point if you need to remove something if k was quite large, let's say k was equal. to five, you could actually use the formula to find the sum of the first four terms and use any one as four and then you would remove that from your answer, which has all the terms added up to wherever your end point was geometric regressions again, you could see some notation for the nth term you could have an is equal to and let's say for example you have 2 times 3 to the power n so this will describe a sequence where the first term of the sequence n will be 1 so it would be 2 times 3 to the power of 1. well, times 1 does not change a number, so 2 times 3 is 6. when n is 2, this would be two times three to the power of two, which is two times nine, which will give us eighteen so we would have two times three to the three power, which is fifty-four, then we would have four to the power of 162 and the power of five, which is 486.

These numbers are starting to grow quite quickly, as usual. for sequence, the number of numbers and sequence could continue infinitely. In fact, they will give me the limit for the sequence. Now, what happens in the sequence every time we multiply by three? This is a sequence multiplied by three, so geometric progressions are about sequences in which multiplying or devising arithmetic for sequences are sequences where you add or subtract each time and each one uses different formulas now for this one if you can see this multiplied by 3 every time you write 3n that's where it comes from, think well if for the first the term 3n is 3 to the power of 1, which is 3, how do I get to six?

That's where we have to add multiplication by two to get to our first term. You may be asked for specific terms, say ten. will be the tenth term, so do two times three to the power of ten, so we would take the calculator three to the power of 10 multiplied by 2, which will give 118,000 and 98, so again, when you do these types of sequences, become very , very large very quickly now again it's a special notation to look at, so we have the sigma symbol for addition and at the top it tells us how many parts they want to add and the k below it will tell us the starting point commonly you will see that k is equal to one, after this you would have the nth term, so you would have whatever you are multiplying, you know what you are multiplying and you would have the power, the formula for the sum is the the sum of the first n terms is a multiplied by 1 minus r to the power of n divided by one minus uh, we need to know the meaning of all the different letters, so m is the number of terms to add to is the first term, in this case it's six for our previous example uh is what which we call the common ratio, this is basically doing the sequence each time, so in our last example we are multiplying by three each time, so the common difference will be three, it is the same number in the x. term that is raised to the power, let's see what you got k is just the starting point of where you are, so we will usually start with the first term and if k is equal to 1. we will talk about other numbers later, so Try adding the first five terms from our original example.

Now we can do it because we already have the first five written, but what we are going to do is test if the formula works for the sum of the first five terms. the first term is six, we are multiplying by one minus the common difference three to the power of the number of terms, which is five, so let's work on the numerator and the denominators, so three to the power of five will give us 243, so which one subtracting that will give us a negative 242. then one minus, that is, one minus three, which will give us negative two, so we calculate this: minus 242 divided by negative 2 gives a positive 121. then we multiply that by 6 for our answer and we get 726.

Now let's go back to our last example, add an open and see what we get: 6 plus 18 plus 54 plus 162 plus 486, all of that together gives us 726, so we can see that the formula works. Another important example here is when the nth term will give us some kind of fraction, so we could have something like 3 times 3 to the power of negative n, an alternative way of writing that will be 3 times 1 over 3 to the power of n now if we generate this sequence , will give us the first energy term one, so three times one over three that will give us one, then when n is two, three to the power of two is nine, so it will be three times one over nine, which is three over nine, which is a third, so when n is 3, 3 to the power of 3 will be 27, that will give us 3 over 27, which will give us 1 over 9.

What you can notice is that the denominator is getting 3 times larger each. time, so what happens with this sequence is that each time we divide by three and another way to divide by three is to write it as a fraction, so the denominator actually gets three times larger each time, now with this you have another type of formula and this formula is infinite sum because what's going to happen is you know it's with the sequence one third and negative one over 27. These numbers are getting smaller and smaller so that's the limit of how much big it is.

The sequence can be even if you add all these numbers to infinity because the numbers will start to get infinitely small, so this formula is multiplied by 1 over 1 minus r, so we are missing the nth part of this sequence, so If we try to say with this example that the sum to infinity a is still the first term, which is one, we are multiplying by one over one minus r, which is the common difference now because it is divided by three every time we write it as a fraction, so we're We'll write it as a third, so now we can start solving this.

We have one multiplied by the bracket and that will be one over one minus one third, that's two thirds. Now, if you get a three-story bracket like this. it's really bad, we don't want to do this right now, you can try it on a cartilage and see what happens, but what you can basically do in this situation is put the fraction backwards and you get three over two so the card clicks, let's say. write in one divided by two thirds let me write it as a decimal 0.6 occurring will give you 1.5 as an answer and 3 over 2 is also 1.5 so what the infinity solution tells us is that with the sequence 3 times 3 raised to negative n it doesn't matter how many elements in that sequence you add, you will never get more than one and a half because every time the numbers get smaller and smaller, and smaller and smaller, the last thing we should mention is that for all these questions our starting point k is equal to one if the set point k was something else, let's say it was two, three or four, we would use these formulas and we would subtract the terms we don't want, so let's say k was equal to two once it started adding.

Starting with the second term of the sequence, you would follow this normally and then remove the first number. The first part of the sequence. Radians are an alternative way of measuring circles, so normally with degrees a circle measures 360 degrees now that we look in radians. a circle measures 2 pi radians, the reason for this is that 360 was a really good number that people chose because it has many factors that can be divided in many different ways, so 2 pi is a way to relate the measurement of a circle with some real math, there's actually a reason we used pi, while 360 ​​was used because it was a good enough number, now you can convert between the two, so to go from 360 to 2pi you need 2 divided by 180 360 divided by 180 is 2 and you need to multiply by pi to get the pi at the end and that will convert any number in degrees to radians.

You can also go in the opposite direction and convert from radians to degrees to go from 2 to 360. you need to multiply by 180 and then we'll have to get rid of the pi, so we divide by pi, so do those two things in any order. Now you can convert anything in radians to degrees, besides being able to convert them maybe using the calculator for this it also has exact values ​​that you need to know, so the exact values ​​are 0 degrees 30 degrees 45 degrees 60 degrees and 90 degrees now, yes you convert them to radians you will get zero radians pi over six radians pi over four radians pi over three radians and pi over two radians and the reason we have pi is not a decimal so for example pi divided by 2 is 1 .570796327 that number will continue with more decimals, it is a rational number, it will notfinish, there is not going to be a pattern, so instead of writing everything or as far as we can go or even rounding it to make it shorter, if we write the pi there it means that we don't need to round it, so we have our degrees and radians. now, in addition to memorizing, we must memorize what is the cause of sin and tan, these are sine theta cos theta and tan theta, so at zero the sine is zero cos is one and the tail is zero at 30 the sine is the half cos is root three divided by two and tan is one divided by root three at 45 degrees sine is one divided by root two cos is one divided by root two and tan is one at six degrees sine is root three divided for two cos is half and tan is the root of three they re 90 degrees the sine is one because it is zero and tan is not defined there is no solution for it, I suppose in some way you could say that it is infinite, you look at the sign and the calls you will see that the silent chords have the same values, but in the opposite order, so if you can remember the sign values, if you flip them over, you will have the cost values.

Knowing these exact answers is useful for several reasons, so it may not or could be useful if you are working with certificates. useful if you are working without a calculator and want to work with a full level of precision without rounding trigonometry in a right angle triangle will involve labeling a second angle, so we have the hypotenuse, the longest side, the adjacent one. is next to both labeled angles and the opposite one which is the opposite of the labeled angle which is not the correct angle now if you have some values ​​for these then you will be able to use trigonometry to find the other missing value you should also remember that the angle that we can label as theta we have the acronym so car tower that helps the lex method and each of these refers to a triangle, so we have the solar triangle formula and from left to right you simply go so h we' You have the triangle of car formula and again you just go from left to right c a h and the tip of the triangle from left to right a a now the o the h and the a correspond to which of the o h and here you have in the original diagram the s c and t refer to the angle if you want to find a missing angle theta then look at the previous triangles, you cover the s c or t you look at what is left and you should have two letters on top of each other, for example theta could be the inverse sine of o over h, that is what you get for the first formula triangle o could be the inverse cosine of a over h, or the angle could be the inverse tan of o over a, that's what it means when you have the type of letters on top of each other if you're looking for the opposite then we look at the form of triangles you cover the opposite and what is left is the mathematics you do it from the first triangle form the opposite if You have s and h next to each other, which means the sine of the angle theta multiplied by the hypotenuse and you also have to in the last triangle shape you cover the opposite of what you have. what's left over is the t for tan, so you have tan of the angle multiplied by the adjacent for side a, you cover the a in the shape of a triangle, so the first, if you cover the a, you're going to get that the adjacent is equal to the cosine of theta multiplied by h because next to each other, but in the third triangle shape, a is in a different position and everything is finished.

See that the adjacent will be equal to the opposite divided by time of the angle, the last one to think of is the hypotenuse, so again you cover the hypotenuse to find it and look at what is left in the triangle formula, so we have h in the first form of triangle that you have or about s, that means that the hypotenuse will be equal to the opposite divided by the sine of the angle. You also have h in the second triangle shape, if you cover, you have over c, so it will be the adjacent divided by the cosine of the angle, now you are only going to use one of the formal triangles and it depends on what they have given you. that has been labeled, so you may not be given the hypotenuse at all.

You are only given the opposite of the adjacent, so you can only use the third triangle shape in that example and you will only be able to find the opposite of the adjacent or the angle depending on which of the three. As a clue, now you can also use trigonometry in triangles that are not right angles and we have some formulas for those now. The way we label a triangle that is not right angle is that we would give a letter for each of the angles, so we could call them b and c are another letter for the sides what we do is make the size lowercase of the opposite angle, so we have side a, side b and side c, so lower case for a side and a capital letter for an angle, that brings us to the sine rule, the sine of an angle divided by the opposite side is equal to the sine of any of the angles divided by the opposite side, so they are all in the same proportion and we can use this to find missing sides and angles, you can also do this the other way around so you can divide the side by the sine of the opposite angle .

Another rule related to this is the cosine rule. The cosine rule is related to the Pythagorean theorem, so in the Pythagorean theorem c squared. is equal to a squared plus b squared, but this only works for a right angle triangle now, if you don't have a right angle triangle, you need to make an adjustment so that the angle is different and that adjustment is taken away 2 multiplied by a multiplied by b multiplied by the cosine of the angle c is also independent of the variable names, so we could know that we need letters in another way and we can write them in different ways, we can write a version that is b squared equals and a squared equals square. is the same, so the way it works is what you're looking for, it's going to be the other two, let's add, so you're looking for b, it's going to be a squared plus c squared, you're looking for a, it's going to be b squared plus c squared, then we make the adjustment and take away two of the squared signs, so two a c or two b c and then the cos is always the angle related to what you're looking for, so you look at the b it side.

Because the angle b you are looking for is side a, it will be the angle cos a. The last thing you need to know is how to find the area of ​​a triangle using trigonometry. Now, like the cosine rule, like the Pythagorean theorem. Related to the area of ​​the triangle, the area of ​​a triangle is equal to half the base times the height. Now we don't have a base and a height in a non-right angle triangle because the base and height have to be at right angles to each other, so let's call a a on b, so that's the formula for a normal triangle inside of the right angle, if it does not have a right angle, you have to make an adjustment, the adjustments are very easy, you multiply it by the sine. of the angle of the side you are not using, are you using a and b as your sides what is the sine of c and again, just like the cosine rule, you can do this with the other letters so that a can be half of ac sine b ? or you could have half of b c sine a plus its two sides and then the unrelated angle, so here you have all the g to c type of trigonometric formulas on one page and also don't forget the Pythagorean theorem that you will need to use. that sometimes c squared is equal to a squared plus b squared trigonometric equations that combine equation solving skills with trigonometry now you can have something like sine theta is equal to 0.4 this is something you can solve using equation solving skills now we want to get a theta is proper for this question, so the first thing we need to do is we can do the inverse of the sine on both sides when you generally form a triangle in trigonometry, sign to negative one or cos to negative one, this is pretty much what what are you doing.

In doing so, if we do the inverse sine of 0.4, we get 23.6 to one decimal place and you might think that's cool, that's the answer. We do trigonometry with triangles. This is as far as you go, however, in reality, there are multiple answers to this, now how you do it. To find multiple answers, you need to draw a sine curve so that the sine curve goes from negative one to zero to one. We need to remember the way our understanding curve is going. Now what could I do to calculate, press sine zero and that. I will give you a coordination point sine zero zero then you could say well what is sine and ninety the sine of ninety is one so you think well what sine of we get up in ninety what sine of 180 sine of 180 is zero until another sine of 90 of 270 is negative one and the sine of 360 is zero so that gives you an idea of ​​where the sine curve is going, then we can stitch it over the curve and here we have a rough sketch of a sine curve, now we might want to label the uh.

On the x axis, I'll label the y axis, which you might think of as having zero degrees 90 180 270 and 360. Now our answer is at 23.6 degrees and that corresponds to 0.4, so 0.4 is just below As we can see, it is between 0 and 360. 0.4 cuts the curve in two different places, which means that there are actually two answers. The first answer we have is right at the beginning, so it's between 0 and 90. So this is our 23.6 that we can think straight from. At first we went at 23.6 degrees, now the symmetry with these curves, if we look between 0 degrees and 180 degrees, there is an axis of symmetry through this curve, so about 23.6 on the left side using symmetry , the second value is also It will be 23.6 inches, but this time going from the right side, which is at 180, so 180 and subtracting 23.6 to go to 23.6 before it gives us the second answer, which is 156.4 degrees, so it's all about yes, using your solvent equation skills. but also recognizing that we do inverse sin because our tan there will be multiple answers, so let's look at the version with cos, so let's look at cos theta of let's do negative 0.7, so we would do the inverse of cos on both sides and that gives us what theta is equal to then we will do the inverse cos negative 0.7 and I will give the first answer 134.4 degrees.

Now again there will be a second answer between 0 and 360. and we are going to have to draw something to find it so again you have to do it yourself you have to draw the curve between negative zero and positive one then with the angles it is good to go with increments of 90 degrees eventually in the first stage we remember how the curve goes if I don't remember the curve use a calculator write cos zero which will give you one write cos 90 which will give you zero cos 180 will give you a negative one and simply putting them in will give you an idea of ​​towards where the curve goes and then You can draw it to look like this now because 0.7 will be down at the bottom grouped minus one and zero and we can see again that it will cut the curve in two different places.

Now we have the answer between ninety and one eighty. this is one hundred thirty-four point four think about how we did it the last time we entered from the left side 134.4 degrees now the axis of symmetry in a curve cos is different between 0 degrees and 360 degrees the axis of symmetry passes through the middle at 180. so we say good 134.4 on the left side. Let's do the same from the right side this time. I'm going to subtract 360. And we subtract 134.4 from 360. We have our second answer which is 225.6 degrees. We can also do this with tan, so let's take a look at the tan of 0.5, so we do the inverse of 10 on both sides and that gives us that theta equals the inverse times. 0.5 which is 26.6 degrees again, this is just our first answer, so we need to draw the graph to determine what the second answer is.

The calculator only gives one answer. She will set up a v curve on the y axis between negative one, zero and one. Set up an a math error meaning there is an asymptote here the line does not go through 90 degrees so we can make the time 180 which is zero degrees again turning 270 again gives us a math error and so 360 will also give us zero . Now this hasn't really helped us draw a curve. You could look between values. Let's try to go up in 45 degree increments. So 45 will give us one, so we know the curve will increase like this.

Add another 45. get 90 and another 45 to 90 and we will get 135, so one three five will give us a negative one, then between 180 and 270 and 45 we will get 225, the tan of 225 is positive and this gives an idea. of where the curve is going, so it will be something like this. Now I want 10 to be 0.5, so to go through 0.5 we can see that we will cut the curve in two separate places now that we have the answer right at the beginning. is 26.6, there is one special thing with a tank curve and that is that it repeats every 180 degrees, so all you need to do is take the 26.6, add 180 to it and you will get another answer and that is 206.6 degrees, now the next thing we need To speak it is a range, all these questions have a range that is between 0 degrees and 360 degrees.

Now they won't always give you this, sometimes you can have between 0 degrees and 720, so for 720 you would make these curves go twice. As long as you draw them up to 720, you could have it between negative 180 and positive 180. If you draw the part drawn here, you would also enter thenegative section, regardless of how large it is. look between 0 and 360, you can accept because these curves always repeat also every 360 degrees and those three repeat every 360 degrees, so with the two answers you have, as long as you add and subtract 360 degrees from them, you will get other answers because the important thing with these is that they go on forever, so since the curve is gone forever we need to limit it somehow again, zero 360 being the most common, but you might have others like I mentioned the other thing is that all of these are in degrees.

You may also be asked this question in radians. There are two ways to do it in radians. You can do what I've done here and then we'll get your final answers. You can multiply by pi. and divide by 180 and now your answers are in radians. The other way is that you can use the mode button on your calculator and set the radian mode of your calculator. You will be in a small r on the calculation screen and it will give you the answers. in radians right away, just make sure that once you're done with that, switch it back to degree mode, a little d display; otherwise you will be confused the next time you use trigonometry and sound happy giving you wrong answers, don't wait because it is in the wrong way with trigonometric identities we can change some types of trigonometric expressions to others, so for example if you take the sine and divide it by cos you will get tan, so you can convert tan to silent cards or vice versa using this method, then We have more complicated trigonometry words, so we have cosec, cosec is equal to one over sine, then there is sec, sec is equal to one over cos and is cot, which is equal to one over tan, and these work the other way around like well so sine is equal to one over cosec cos is equal to one over sec and tan is equal to one over cut now with tan is equal to one over cut we can write another identity which is the next cut is equal to one over tan therefore, it is equal to cos over sine and all this when you have one over something, what you have done is take the reciprocal of the number and then we take the reciprocal of a kind of normal number on its own so you'll get one on top but and it becomes a fraction but it's already a fraction like sine of cos then we take the reciprocal just flip a fraction upside down this is our first set of identities the next step is that sine theta squared plus cos theta squared equals one and by rearranging this we can discover a few things, for example sine theta squared equals one minus cos theta squared cos theta al squared is equal to one minus theta squared, then we can get versions with the set cosec and cot, so tan theta squared is equal to sec theta squared minus one, then we can say things like sec theta squared is equal to tan theta squared plus a quad theta squared is equal to cosec theta squared minus one and cosec theta squared is equal to cos theta squared plus one so this is another method where you can take the same cause of time and you can convert them each other depending on what you need.

We also have identities on how to add angles, so sines. we have angle a plus angle b equals the sine of a multiplied by the cos of b plus the cos of a multiplied by the sine of b and that works if you take away there is also a version for cos due to a plus b equals due a a multiplied by the cosine of b, remove if it is the opposite sign the sine of a multiplied by the sine of b and then if you are taking away the angle, that would be adding the angles, so there is a version of the tan if you have the angle a plus the angle b which will be equal to the tan of a plus the tan in b and then everything will be divided by one to remove the sign opposite to the tan of a multiplied by tan b then you remove the angles, instead the numerator will be removed but will be an ad for the denominator, then we have the double angle formulas, so the sine of 2 theta is equal to 2 sine theta multiplied by cos theta is also equal to, so we have another alternative, two lots of so theta, all divided by one plus tan theta squared, then we have options for this for cos and tan, so cos of two theta is equal to cos theta squared remove the sine theta squared is also equal to two lots of cos theta squared takes away one and one minus the sine theta squared and it's also equal to one minus tan theta all divided by one plus tan theta so we have a lot of different options to change the shapes of these there's also one to ten, let's put that in at the top, we have some space tens two theta equals two tan theta divided by one minus tan theta squared, so we have a big List here different genomic identities, some of which you'll need to be able to use to replace some trigonometric terms with others.

The best thing to do here is to look at the list to write it down, try to find the formula booklet for the exam. you are doing and then make sure which of these you need to remember for the exam and which ones will be given to you in booklet form. What you will find is that, although you don't need to remember the entire list because some of these are rearranged to form others so that you can remember a smaller list of these to do the rest with differentiation from first principles, you may have a curve that is something like y equals x, but actually it would be useful to see what That graph will look like, so I'm going to draw some axes y equals x squared.

It will look like this and what do we mean by differentiation if you want to know what the gradient of this line is, now the gradients how steep. the line is and you'll notice you know at the bottom it's going to be less steep so it's highlighted in green and at the edges it's going to be steeper so let's highlight it in pink so the steepness of the line depends on where it is so how can we give it a gradient? How can we tell how steep it is? So what we can do is find the gradient of just a couple of the graph.

Now it is not enough to make a single coordinate. Let's look at the gradient between two different coordinates, it's not exactly an alignment, it's going to be a straight line, it's not going to be a curve, it's going to give you a rough approximation, so what are these coordinates going to be? It's called y is equal to x squared, so if you look at where the x coordinate is of the first one you call that x puts on the we know that y is equal to x squared so our first coordinate is going to be x x squared so we look at the next coordinate now if the next coordinate is going to go down a little bit further along the this x plus h.

We have a distance of h between x and an x ​​coordinate on the x axis and we'll just say that's an unknown number because I don't really know how far we've gone, so the y is equal to x squared so it's x plus h we need to square it so we take our which we can expand later to run it, so now that we have our two coordinates, let's think about how we find the gradient now in general to find the gradient. of anything it's the change in y from the change in whatever you choose first for y also has to be the first choice for x and you do the same for x, so what we can do is substitute into this formula to find the gradient between these two coordinates, so our second y coordinate was a in parentheses x plus h integer squared and then we remove the first y coordinate which was x squared, we divide it by the second x coordinate which is expand the brackets and clean this up so you know how to expand a quadratic x times x is x squared x times h will give us x h h times x is also x h we have two of those and h multiplied by h is h squared and then we subtract x squared, We already have an x ​​squared, so it will cancel to give us zero, then for the denominator x, we subtract there is an h in the denominator, so we take an h out of both terms 2 x h becomes 2x and h squared becomes h and then we have an expression that represents the gradient between the two coordinates.

Now what we're going to do next is a little complicated, so if you look at the graph, the distance to the two coordinates on the x -h axis was h and that was just a random number chosen, we don't even know what the number is , that's why we label it with a letter. Now imagine that you make the distance between x and x plus h smaller, you make it half as big. that's going to change our calculation here and it doesn't, it just means that h is going to be more of a smaller value. What we can do is reduce the distance between the two quad bits more and more and the more we reduce the distance. between them, the more accurate our gradients will be because right now it's a straight line on a curve, but the closer they get, the less inaccuracy there will be in making that straight line live on the curve and what we do is we say that.

There will be a limit as h approaches zero so what will happen is that eventually these two quads can get so close together that the value of h is completely irrelevant and if h approaches 0 then we get just 2x on its own . we get rid of h and what you'll find is that as h approaches 0, what's left over 2x disappears is actually the gradient of the curve. Now it's not a number like you would have before with a straight line. the gradient is going to be a whole number, it's going to be a slope of two two to five statements of five zepa or you have a slope of more than a quarter and that wouldn't be very steep at all you get a number that has an x ​​in it and what that means is that if If you substitute the x then you'll leave the gradient no matter where you are on the curve, so an example at the bottom where the , so let's say, hey, let's say x is 1, then 2 times 1 is 2, it's a 2 green, which you vaguely know isn't too steep if you're right in the middle, the flattest part, that's when x is zero and any zero gradient that is not very steep now let's look at the edges, if we imagine these edges are when x is like 10 or minus 10, we substitute 10 for x and you get 2 times 10 is 20. well, that's really very steep and so you can see it working, you substitute in x and you get the gradient at that point, that's how you differ from first principles.

Now we've done this so y is equal to x squared and that's why our second coordinates were squared. Now you can do this for other things too, you can even do it for trigonometry, you know it's more complicated, curves, for cubics, you know y is equal to x cubed plus two x plus eight, whatever and all that happens is that the coordinates principles y equals x squared differentiates into two x's and I said that when you differentiate you get the gradient of the line, so the gradient of y equals 2 our differentiation notation difference in y over difference in x now you don't actually have to go through differentiation from first principles to each question, there is a shortcut and if you look at the question and the answer you might be able to see it.

See the power of the x has decreased by one, we have gone from x squared to x to the power of one, it is not shown to anyone, but there is no power in an hidden from the other. The thing is that it was a power of two and now we have a two in front and there is a connection there, so let's look at the general rule for differentiation, so if we say that y is equal to a x b then the power is b a is a coefficient in x, then the difference in y or the difference in x to the gradient will be equal, so you multiply a and b together so we can write a b for that and then what we just did we can see that the power decreases by one now with y is equal to x squared there was no a there was no coefficient in the that you have to deal with hidden information in these questions, we have the hidden in one x squared and we have the hidden power one in two 2x so again two things the power has been reduced by one and we have multiplied the coefficient on x by the power previous you can also differentiate a second time we call itd2y over dx squared then if y is equal it is a line and we differentiate it once dy times dx gives you the gradient d2y times dx squared the second derivative which is the rate of change of the gradient is how the gradient changes all the time alone you follow the same process so we're going to reduce the power by one, so x has gone down from two to one, now it's going to go down to zero and then we multiply the coefficient in front, the two times the old power one and two times one is two, so which will actually stay as two now, when you have zero power, any value raised to zero is one, so effectively this is saying two times one, so all together it's actually just two.

Now with this we have only differentiated one term, we can differentiate several terms together as long as I separate by plus and minus so we can have y is equal to x cubed plus two x squared plus three x plus four just as a random example, so We differentiate once to get the gradient, this is what we do every time the power of x decreases by one, then x to the power of three becomes x to the power of two x to the power of two becomes x to the power of one x to the power of one becomes x raised to the power of 0.

So that's the first bit, tell it with the second bit. Is the coefficient of x multiplied by the power? If there is no coefficient, it will be 1. So one times three is three two times two is four three times one is three and then with this this was x raised to the power of zero, so four times zero will actually give us zero. Now, that's not how we write algebra, so there's going to be hidden information, so it's like it's one x cubed and three x to the power of one, the hidden information here is. We are not going to show the power one over four x to the power of one.

We're not going to show zero, so we didn't write it, so now you can see the final answer. here it's going to be three x squared plus four integers have x to the power of zero in them effectively, although we don't write it down, so you multiply them by zero so they disappear, next is if you differentiate like three x's, so a number with just one x to the power of one, then you just remove the x, three x has become three, it just squares and up, where you can use the full method and show all the information, we multiply the coefficient. by the power and you can reduce the power by one just keep in mind that when you differentiate some of the squares it will become the power one and we don't write the power one so just write x so there are a few things to pay attention to now , you wouldn't write all the hip information every time, so if we differentiate the second type for the rate of change of the gradient, you'll be able to do this in your head without any extra work, so let's do it 3 times, 2 is 6, reduce the power by 1. 4x will become 4 and the 3 will disappear and it's actually that fast that you don't need to write the rest, so again the integers disappear, everything with an x ​​disappears, the x will disappear and then with pose two or higher, you go through method four.

Now there are some special cases in differentiation because it's not just some kind of normal algebra that you're going to differentiate. There is also another way to prove it. Instead of writing y equals, you can write f of. x function of x is equal to then the differentiated version would be f dash x so you can see it written both ways so first you can differentiate trigonometry so if you have sine of the angle now let's say we have a more complicated angle let's say look at the sine of 2x we differentiate the angle will remain the same but what will happen is that the coefficient in x goes to go outside as well so you have sine of 2x differentiate it you get 2 cos 2 x the same thing happens with cos so let's say we have cos of 3x when we differentiate it when we differentiate cos you get sine The angle does not change, it will still be 3x, but when we go from sine to cos, the 3 will come out, so it will be 3 sine 3x .

Another thing is that it is not as simple as differentiation being the opposite of itself. It's not like that when you differentiate because you actually get a negative sign, you can also differentiate tan, let's say we have the tan of 4x, when you differentiate it you get something called sec squared, the angle is not going to change, but the coefficient on x is going to appear, so let's multiply by that, if there is already a coefficient in this, then it will also be multiplied, so let's say we have 2 tan or 4x and then when the 4 comes out, we will multiply the 2 to give us 8. sec squared It's something you'll go through when you get to more advanced trigonometry.

By the way, another thing is that you can see logarithms, so if our function is ln x, when you differentiate it, you actually get 1 over x. you can also have all the numbers involved in this, so let's say a function of x is 2 on x cubed, if you have a power on x that's going to come out as well, then we differentiate it, you would get 6 over equal to 6 and it was still going to be greater than x if you have an exponential so let's say we have 4 to the power of x when you differentiate you get some non-response so before the power of 4.

The last special case we will look at is that there is a number called e and e is an irrational number like pi. I like the pi screen. A little easier 2 at a time, but the decimal places are not repeated. there is no pattern, but now they continue forever if you raise it to the power of x and differentiate it, something really interesting happens when differentiating e does not change it, it stays the same, so it effectively represents its own gradient, now you can have all the numbers that involve it, for example we could have 2 e to the power of 4x and when you integrate it what will happen is the coefficient on x, just like just trigonometry will come out 4 times 2 is 8, but then the e part doesn't change, it will continue being e raised to 4x, so these are some special cases that you will see to differentiate, some you will see more than others, the trigonometry ones will be the most common. you'll see if we have a curve, let's say y is equal to x squared and we take a look at what that would look like, so we have an x ​​and y axis, we have the curve, you might want to find something called tangents. and normals, so what do you mean by tangents and normals?

So a tangent is a line that descends and touches the curve at exactly one point. Now a normal is a similar concept, but a normal is at right angles to the tangent it also crossed. the curve at that point could cross it at another point further back as well, but again the point of this is that it is at right angles to the tangent, so this is the normal and you may be asked to find the equation of these lines . We will also likely be given the coordinate of the point where this happens, so let's save this question.

This was not the coordinate five two, so let's see, I would find the equations of the tangent and the normal, so the first thing we need to think about is that We have our curve y is equal to x squared and if we differentiate it we will get the gradient of the curve or something that represents a gradient of the curve, so let's reduce the power by 1 x squared becomes x to the power 1 I'm going to multiply the coefficient by the power, so if there is no coefficient it's 1 1x al square 1 times 2 is 2. so we have something that represents the gradient now because we have a coordinate that we can find the gradient at the point 5 2 so we can substitute in the x coordinate 5 and that will give us the gradient at that point 2 times 5 is 10. so we know that at that point only at that point there is a gradient and there is a slope of 10. now this tells us something about the tangent, the tangent has the same gradient as the curve at the point where the tangent is, so that the tangent will be y equal to mx plus c and we know what the gradient is and it is equal to 10x plus c all we want now is a value of c and to do that again I'm going to substitute the coordinates in point five and two so that y be equal to 2, so we have the 10 lots of x and x is 5. and we have more c, which is what we're looking for.

This means that 2 is equal to 50 plus c, so we subtract 50 from both sides, which means that minus 4k is equal to c, so Look, the tangent is y is equal to 10x minus 48. We can use a method similar for the normal, so again for the normal we know it's going to be y is equal to mx plus c, it's a straight line and we know what the gradient is. The gradient is y. is equal to 1 over 10x or our negative 1 over 10x plus c, you might be thinking where the 1 over 10x comes from and it is the same as parallel and perpendicular lines, so the normal is perpendicular to the tangent, it is at an angle straight, so take the negative reciprocal, we can follow the same process to get c, so we are substituting 5 2 so that y is 2, the gradient is negative 10.

We multiply that by x, which is 5 and then we have the plus c which means that 2 is equal to negative one half plus c because 1 over 10 times 5 is 5 over 10, which counts down to half if we add half to both sides, that means two and a half is equal to c, so Now we can write the equation of the normal y is equal to minus one tenth of x plus 2 and a half, so the important thing to remember here is that the tangent has the same gradient as the curve; the normal, which forms a right angle to it, is a negative reciprocal. negative if it's a whole number, put it over one if it's a fracture, then you put the fraction backwards for the reciprocal, we can use differentiation says things about the inflection point of a curve, so let's say we have the curve and it's equal a x cubed minus two and this is what we're going to try to figure out now, before we even start, there's one thing we're going to know and that is that with a plus 4 at the end it means that the y intersection where the curve is goes through the y axis will be at 4.

So we can find that right away, if we differentiate twice, then d-y times dx we get some information and we differentiate again d2y times dx squared, we get another set of information. so let's make it so that three times one is three, which is a power of one, two times two is four, reduce the power by one and any integer will disappear, again differentiating three times two is six, which used to be powered by one and it's just the next term we'll just keep the coefficient, so what information can we get from these? Well, the first thing is that we'll probably want to know what the inflection points of the upward curve are now to find what they are when the gradient is equal to zero, so what we can do is say: well, if the gradient is three x at squared minus four x, we can make it equal to zero and then we can factor it out to find those points.

Now you may be able to factor. You can do this by inspecting it or completing the square, but you can always do it using the quadratic formula. I'm not going to go over that here because we've already gone over it, but if you do the quantity for this, you're going to understand that x is equal to zero and x is equal to one and a third, so those are going to be our inflection points, so we know that this one will be one and a third, now the other one is amazing, it's zero, so we really should have had that. a little bit higher up we're not really drawing this, you can see that it's not at x is equal to zero at all it should be a little bit higher up and what that means is that the y-intercept and one of the inflection points are actually the same coordinate at 0 4.

If you have the y coordinate of the one and a third, then what you can do is substitute it back into y equal and that would also give you the y coordinate if you wanted another What you might want is something we call the inflection point. The inflection point is between two peaks, so it's kind of where we start to be the first curve and we start to be the second curve. That's where the second one is. the differentiation is equal to zero, the rate of change is equal to zero because it is in the middle of going up and down, so we are going to say that six x minus four is equal to zero and we can solve the equation, we can add 4. a both sides we can divide by 6 you want to know. is that we have the kind of top and bottom of the two curves zero and one and a third.

We only know which one is on the top and which one is on the bottom because I actually sketched the graph correctly, but we may not actually know which direction this cubic is, so how do we know that zero is on the top and a third at the bottom? So what we do tothis is again substituting it into six x minus four, we have six lots of zero minus four. and we have six lots of one and a third minus four, so substituting the x coordinates of the top and bottom into the second differentiation, this is the rate of change, so what we want to do is see what the rate is exchange. these points now at zero six times zero zero take away quarter minus four we have a negative exchange rate up there which means it's at zero the exchange rate is minus four that means the curve is starting to go down after it so a We call this maximum and the next one we have six times one and a third, which gives us eight and then we take away four and we get four, it's a positive four, which means that at one and a third the rate of change is four, that means the curve is going to start going up so we call it the trough it's these numbers about what happens next when you read the graph from left to right so if on the right side there's minus four you should be on top. most of what's going to go down next is going the other way or something like that, if the exchange rate is going to be four next then you have to be at the bottom for it to go up later.

One last thing is you know the top is zero. that curve is not actually the top of the graph because we can see on the right side that the cubic actually goes up and also goes down on the other side, so these are not actually the maximum minimum points, we call them local maximum. and the local minimum because of the highest lows around where they are, but later on there will be even higher points and let's label this on the chart, so one and a third was the low point and zero was the high point. one other thing you can do and it's not part of differentiation, but if you take the original x cubed minus two x squared plus four, if you make it equal to zero and you use the quadratic formula or again whatever other method you need to factor, this is going to give you the x intercepts now for this one there's only one so it's going to give you x is equal to y is approximately negative 1.1 to a decimal but often with cubics you'll get three of these and it crosses the entire three times, so that's something else that also goes a long way, so there are many different methods from different areas of mathematics, but the key is that the differentiation will give you the turning points and tell you which one is local. maximum and what is a local minimum, it will also tell you the inflection point, which is the bit between the two inflection points.

Differentiation usually works when you have single terms those terms will be separated by a plus or a minus if you have single terms or multiple terms then you can differentiate them now sometimes you may have terms that are separated by other things so let's say that They are multiplying together now for that you will use the product rule so let's go First take a look at an example. you could have something like y equals and then you could have 2 x squared plus x multiplied by x cubed plus three x square, so you can see here that you have two terms, but they must be multiplied together and this is where you use product rule so the product rule is if you are differentiating u multiplied by v then your answer will be equal to v d u times dx plus u dv times dx now I like to write that in a simpler way, so I'm going to show it my way of writing it, so we say that if we want to differentiate u multiplied by v, then what we are going to get is v multiplied by the differential of u plus u multiplied by the differential of v, so what I want to do is we want to find these different bits when we find what u is we want to find what v is so u will be our first parentheses 2x squared plus x and v will be our second parentheses x cubed plus 3x squared and we can Look with a little dash, all it means is that you differentiate it, so the dash two times two is four, reduce the power by one and x is going to differentiate to the same thing with v three times it was not a number so the addition of three reduces the power by one three times two is six and we reduce the power by one and so we have our letters and we have our differentiated version of the letters so now we just substitute so we want v multiplied times u dash, so v is x cubed plus three x squared and we're going to multiply that by u dash which is four x plus one, we're going to add to that u times v dash u is 2x squared plus x and v dash is three x squared plus six x now we just need to expand the brackets, so we have quadratics here where the configuration is x cubed times four x will be four x the power of four x is will be 12 x cubed and 3x squared by 1 will be 3 squared is three x cubed and x multiplied by six x is six , so that's thirteen, another twelve, so that will be twenty-five and another three will be 28. so 28 x cubed, then we will see the squares that we have 3x squared and 6x squared, that is 9x squared and that will be our final answer 10x to the power of 4 plus 28 x cubed plus 9 x squared.

This is probably one of the most complicated answers you can get with just x and powers, it will usually be A little bit of plastic cancels out more when you do your differentiation. You can also get this with things like trigonometry and when you do things like that, just take a look at the special cases for differentiating trigonometry, but it will work the same way as the quotient. The rule is fun when I differentiate, but the numbers are added or subtracted like normal differentiation, they don't multiply like the product rule, they divide, so you could have something like y is equal to 2 x cubed plus 3 x al square, all divided by eight. x plus five now to use the quotient rule, what you need to do is differentiate, you have u divided by v or something b divided by something else, so the answer will be v multiplied by d u times dx take away u multiplied by dv times dx and everything will be divided by v squared now again a little complicated notation that's why I like to write it and a little bit simpler so let's see my way so I would say if you have u over v and you want to differentiate them, then you are going to do v multiplied by the differential from u, take u multiply the differential of v, all divided by v squared, so first let's figure out what u and v are. u will be the numerator 2 x cubed plus 3 x squared and v is the denominator 8x plus 5. so u script we are going to differentiate the numerator 2 times 3 is 6, which is the power of 1. 3 times 2 is 6 because of the power for 1 the same for v 8 x becomes 8 and the 5 is going to disappear and now we can substitute it, so we want to do v times u dash so v will be eight x plus five and u dash is six x squared plus six x so that is what has been multiplied in parentheses, we take away from u which is two x cubed plus three x squared multiplied by v dash which is a and everything will be divided by v squared which is 8x plus 5 multiplied by itself is squared, so now we need to expand all the brackets, so 8x times 6x squared will be 48 x cubed eight x times six x will be rtax squared five times six x squared will be 30x squared and five times six x will be 30x, we are taking away 2x cubed times 8 it will be 16x cubed and 3x squared times 8 will be 24 x squared remember we are taking away the second parentheses so these will end up being negative everything will be divided by we have the quadratic to expand 8x times 8x is 64 now, it's just kind of an order, let's look at the x cubes first, we have 48 x cubed and then we have a negative. 16 x cubed, so 48, subtracting 16 will give us 32.

Then we look at one 30x then the denominator we have 64 denominator, then you can divide by and make it simpler. Maybe you can see everything and go through each term to get a result. x here I can divide 3 by x or each coefficient is even, so we can divide 3 by 2. So there are things you can do at this point with words, but that's outside the quotient rule, it's just simplifying algebra , so I will do it. leave it here, the chain rule is when we differentiate and different parts of your expression and they are separated by more or less, for normal differentiation they are not separated by multiplications for the product rule or divisions for the quotient rule, the rule of the string was going to be some kind of another function, so it could be a power or it could be a little bit of trigonometry or something, for example, you could have y equals and you could have something like 4 x cubed plus let's say 3x al square and this in itself will be raised to a power, so this could be to the power of let's say five, so to use a chain rule, what you say is d y times d x is equal to d y times d u multiplied by d u by d x, you may need to know what this is about. refers to some substitution now it might be easier to write this in a different way so if you have a function of x and x is another function of x then we can change that to b the differentiated version of that function which is a function of g of x multiplied by the differentiated version of g of x and again that still seems a little complicated so let's look at it in practice and see if it makes more sense for our function of five, so now it's going to be something to the power of five, instead of saying oh, it's something to the power of five, we're actually going to use a letter for this and we're going to use u. that's where the u came from before and we put it in the x, so our function four x cubed plus three x squared is raised to the power of five now what's inside the brackets, what has it been, you know what the function has been applied to is our g of x and that's 4x cubed plus 3x squared, so we've separated what's inside the parentheses and what's outside the parentheses.

We have done this with a power. In this situation, you can also have this with trigonometry, so when you differentiate the angle in trigonometry, then the f something could be sine of cos instead of a power five with the x bit being the bit inside the trigonometry brackets, like this which now we need to differentiate, differentiating f from u to know that we are going to reduce the power. times one, then five will become four and we multiply it by the old power, so we get five lots of u to the power of four, then we differentiate g from x, so four times three is twelve, we reduce the power by one, three times two is six, we reduce the power by one I have the differentiated version of g so now we just need to substitute it so what they say is that it is the differentiated version of f which was five of our function the power of four let's put g of x again into what was four x cubed plus three This above we have the integer 5 in front, we have one set of this bracket and we have four sets of this bracket, so we have applied the chain rule here, we still have a rather complicated expression in the end, there may be more steps here, you may want to remove more things like factors, we may want to expand the brackets, it all depends on the question you've been asked, but this is the chain rule, part of the integration of the question is to reverse of differentiation so let's say we have y is equal to 2x squared plus three x plus four if we were to differentiate this to get the gradient then we would multiply the coefficient by the power two times two is four and we would reduce the power by one anything in x We only have the coefficient left and all the numbers will disappear, so we get 4x plus 3 is the gradient of 2x squared plus 3x plus 4.

So we have differentiated, so how do we integrate? How can we go in the opposite direction? direction, so we think about differentiation, there are a couple of characteristics, the first thing is that the power decreases by one and then it is multiplied by the previous power, so the integration will be the opposite: it increases the power by 1 and divides by the new power and we can write a general version of this, so if we integrate, this is the integral symbol, if we integrate, say, ax to the power of b, then we also put dx at the end, which is part of the notation for this, so the answer increases the power by one, so b becomesb plus one and then divide by the new power set, a will divide by b plus one, so yeah, differentiation, take one away from the power integration, add one to the power differentiation, multiply by the previous power. integration is divided by the new power, so let's try it, let's do this, see what it does, so if we're going to integrate 4x plus 3, then what we do is we increase the power by 1 so that x becomes x squared and then we need to divide by the new power, then 4 is divided by 2 to get 2x squared, then for 3 we increase the power if there is no x there then it is x 0, we increase it by 1 and we get x to the power of 1. and then we divide by the new power, while the new powers are one, three divided by one is three and that's it, and now if you compare this to differentiation, we're actually missing the plus four, so when we work backwards en There's actually no way to get that plus four back, so what we do is we write plus c plus a constant.

There may have been a number, but we can't actually work backwards to find what it is, so now there is an error with the integration. One way to compensate for this is something called a definite integral. What we just did is an indefinite integral, so if we're going to integrate or 4x plus 3, what they might give you are numbers in the integral and let's take a look. This means that we are going to do the integration that we already know is 2x squared plus 3x more. So what that means is that you're actually substituting those numbers so that the first term reaches 10 and is 3.

That means we have two lots. of three squared plus three lots of three plus c then we subtract by substituting two lots of one plus three lots of one plus c and then we can solve this so that three squared is nine times two is eighteen three times three is nine plus eight then we subtract the second parenthesis one squared is one times two is two, we remove the second parenthesis, so we remove it three times one is three, but again we remove and then we have the plus c and Again, because we are removing the second parenthesis, we have to remove the plus c and if you do that, we'll notice that we have a positive plus c and a negative plus c and they will cancel, so In fact, I can solve a definite integral even if you don't know what c is, so we can add 18 plus nine, Subtract two, subtract three and all together we get 22, so we get a complete answer without needing to know every bit.

From what we integrate now, you might be wondering what this represents, so let's try it on a graph and see what the x and y coordinates actually have. We're going to have the curve that is our 2x squared plus 3x plus 4. So what does the definite integral mean? Well, we saw two things, three and one, so what this means is that you have one on the x-axis and you will have three on the x-axis, so this is what we are looking for in the middle and what happens is that when you integrate the What you actually do is find the area under the curve between the curve and the x axis so all I'm highlighting in blue this is our answer that's what we're looking for and that's what we found out was 22 so if we label it has an area of ​​22.

This requires another integration function in addition to being the opposite differentiation, so you can convert a gradient back into the line like we did with our example. Also use it to find the area between the curve and the which is cos of x so when we integrate it to get f of x when you integrate cos you get sine it works the same way it does with differentiation now you can also have numbers in when you integrate it, you will start x again, the angle will not change, but with differentiation you will be multiplying by the two, well integration is the opposite, you must divide by the two to finish. up with half the sine 2x if you do that you can integrate the sine so let's look at the sine of 3x so when you do this you will remember the differentiation when you differentiated because you got a negative sign so this is the opposite of that so if you integrate sine you will get a negative because again the angle is not going to change, we could divide by the number, it's 3, so we're going to divide by three, then we do the same thing with tan now with tan because the opposite will start. with sec squared and let's say we already have a four on the outside, so when you integrate it when you integrate sec squared you get tan, the angle doesn't change but you divide it by the angle and we already have a 4 on the outside, so 4 divided by 4 is 1.

So we get tan of 4x, we also have the converse of differentiating logs, so f of x is one over x and then when you integrate that, you're going to get ln x if f dash x was something like five over x then f of x will be ln x to the power of five we also have e if f dash x is equal to e to the power of has special numbers involved, they can be involved, for example, e at 3x, when you integrate it, you will actually get a third of e at 3x. 3x so the pair e hasn't changed the expert hasn't changed but you get the three if it's differentiating you multiply by it it's integration you divide by it another useful thing to show is that you can have things like you can have 2 plus the square root of x now there are special rules for this, it's just a matter of rewriting this into a more reasonable form so we can say well, this is equal to 2 over x raised to half the square root is the same as half x times half y then we can take that up until it's 2 to the negative half of so that x at negative half becomes x at positive half, we add one and then divide by our new power, so it will be two divided by the required half as a fraction for For example, you write it as a decimal two divided by half how many halves are there in two there are four halves and two so you can understand some complicated things in that situation another way to divide in half is to multiply by its reciprocal invert the fraction backwards you get 2 over 1, which is 2, so 2 times 2 it also gives you 4, so you're trying to integrate, you have really free notation, you have fractions, you have roots, try to change them. in a single line like I did here and then you should be able to integrate Integration by parts is when the different terms that you are integrating and separating by more or less are maybe multiplying together, so for example, let's say we have and equals a I would have something like x minus 3 and we could have x squared plus 5x so how would we integrate this?

So using integration by parts is a simple way to write your integral if you have u multiplied by p then that will be equal to u times v we take away the integral of v times u so it seems quite complicated let's try to identify what u and v are so what u will be one of our brackets, let's say it's the first bracket x minus three then we can differentiate that to get u dash because we need that to be part of our formula, which would be a v v dash, that's how it is in the formula, our second parenthesis is going to be x squared plus five answer we will have u multiplied by v, that will be x minus multiplied by a third. of x cubed plus 5 halves of x squared, take away the integral of v multiplied by u dash, so v is a third x cubed plus five over two x squared and then u dash is once we multiply it by one is not going to change and so now we need to integrate that for our final answer, so the first pair will not change here, but in this second part we need to increase the powers by one again so that x to the power of three becomes x to the power to four. we need to divide by four we are already dividing by three divided by three and four times together we get twelve and then we remove it so remove the sign we sort the second term x squared will become x cubed 3 divided by 3 we are all dividing by 2 , so 2 times 3 is 6 and we're removing it and now we have our complete expression for integration by paths.

Now there are things we can do here so you can expand the parentheses, you could try factoring things, there are other options from this point on, but what we have here is the kind of essence that I integrate by parts and yes, there can be some manipulated algebra later. The last thing to remember is that when you do this we're going to get the plus c because we're integrating, so we have to include the plus c. We also have integration through a substitution. This is when normal integration separates the terms by plus or minus, if there is multiplication involved, maybe brackets next to integration with each other by parts, if you have powers it will be integration by substitution, so you have something like and equals, so we could have three x plus two cubed and we want to integrate this, so this is how we do it, if the integral if If we have a function of g x times g dash function of u, so the f part I'm going to call f of u because I'm going to take out what's inside the parenthesis as a substitution, so it's going to be a parenthesis and it's cubed, so we have our g of x, that's what it is inside the parentheses, it's three x plus two and this is also the same as u.

Now we need the form of g dash x. so if we differentiate it, we'll get a 3. Now, what this formula means is that you can just use a shortcut and change that kind of complicated bit on the left to the simpler bit f of u on the right. that if you have g by substitution if we're multiplying this by g hyphen x which is three so we're going to put it in to get our integral we have three do it because obviously if you multiply by three we have something three times as big and it's going to mess up the answer if we also put a division by three here now it's going to work because we have to cancel each other out and now we can use the integral of f of u, so that's what we're going to do next so the integral of the function of u we've already done the function of u so we can substitute that and it was u cubed so if we differentiate u cubed we need to increase the power so that the 3 is make it a 4 and then we're going to divide by the new power and of course we get more c because we integrate, so we can substitute u back into u was equal to g of x, which was greater than three x plus two.

We also have to remember that we had the third part of this, let's put the third, that means we have got a third multiplied by a fourth, if you multiply them together, you actually get 1 12. So our final answer would be 12 of 3x plus 2 to the power of 4 plus c and again, at this point, there may be brackets to expand. There are factors to eliminate, there may be more steps, but we have done the integration using substitution. Integration is finding the area under a curve, particularly between a curve or line and the x-axis. Now look at the diagram here, it's not between. a curve or a line and the x-axis is between a curve and a line, so how can we integrate this now?

It's as simple as combining the two lines because both lines are equal to y, so x squared plus 12x plus 32 equals y and x plus 8 equals y so they are both equal to and so they are both equal to each other so x squared plus 12x plus 32 equals x plus 8. now what we could do now is combine this on one side so we can remove we can integrate that so we can integrate x squared plus 11x plus 24. and if we integrate this combined line and it will give us the area between the two lines, to integrate it we will increase the power so that x squared becomes x cubed and so let's divide by the new power x becomes x squared divided by the new power and 24 there is no x, so we're going to put in an use a definite integral and it's going to be between the the definitive integral so let's substitute the largest number first minus three that will give us one third of negative three cubed plus eleven halves ofnegative three squared plus 24 many negative three now I could write plus c but the next step we're going to remove the substitution for a smaller number which is negative eight and therefore we remove we get a negative c from this so the c's will cancel out so we can safely ignore them, so let's have one third of negative 8 cubed plus 11 halves of negative 8 squared plus 24 lots of negative 8. and if we calculate all of this, we'll get an answer for the area , so negative three cubed is negative 27 and then we divide we multiply it by three because a third will give us negative nine minus three squared is nine which eleven halves multiplied by eleven divided by two will give us 49 and a half, they want 24 times minus 3, that will give us 49 and a half. give us minus 72. moving on to the second parenthesis, let's remove this parenthesis. minus 8 cubed will give us minus 512.

I want a third of that, so we divide it by three, which will give us minus 170 and two thirds say that 0.6 is recurring now we have to take this away, it's already negative, so that gives us will give a double negative, will be positive, then negative 8 squared is 64. What is 11 halves multiplied by 11 divided by 2? to give us a positive 352. We are removing this so we are going to remove it, then we have 24 times negative 8, it will give us negative 192 and again removing this support will be double negative, so we are adding it so we can add all of this for our final answer minus nine more 49.5 minus 72 plus 170.6 recurring subtract 352 and add 192.

In total, that will give us a negative 20.8 with one decimal, now how can we have? a negative area, well we can do that, so when you get a negative area, you can assume it's positive, so you can ignore the negative sign. Returning to the graph, we are saying that the area highlighted in green has 20.8 squares shaded with one decimal. First, you can also use this method to find the area under a curve for when you have a quadratic and a cubic and the intersection creates two separate areas that I will label area number one and area number two.

There is a similar method, but there is an important one. difference halfway now I need a lot of space for this method so I'm going to try to compress by combining the two things so I'm going to use the cubic as a base now we have x cubed with the negative square 10x squared we also have a positive x squared in the other so we need to take x squared to combine them so it's negative 11x squared then we have 29x and subtract 5x on the other side we need to add 5x to both sides to get rid of that so it would be 34x then with the numbers that we have minus 20 plus 4 on each side remove 4 from both sides to give us minus 24. so this is what we are going to integrate, so if we integrate it we increase all the powers by 1 x to the power of 3 becomes x to the power of 4 and we divide by the new power x to the power of 2 because x to the power of 3 divided by the new power x becomes x squared now dividing 34 by 2 that should give us 17 and 24 to become 24 now what's different here is what is the definite integral now look at the is between one and four, so the definite integral for this will be: substituting the largest, which is four, then we are going to have a quarter of four to the power of four minus eleven thirds of four cubed plus 17 lots of four to the squared minus 24 lots of four that take away from that second number which was one, so we're going to have a quarter of one to the power of four, minus eleven thirds of one to the power of three plus 17 lots of one squared minus 24 lots of one, like this there's a lot to do there, but we need to get a number for the first area, so we have four to the power of four and then a quarter divided by four is going to give us 64. four to the power of three eleven thirds times eleven divided by three is going to give us 234.6 recurring four squared is sixty times multiplied by 17 will give us 272. then we're going to subtract 24 times 4, which is 96. we come out of the second parenthesis and remember we removed this one, so we're substituting one and we make a power of one, we will still have one, so a quarter of one was when it was negative a quarter 11 thirds is already negative, we are going to remove it, it will be double negative, so we have the positive 17 we have to remove them, they will be negative and then in the end we will go to 24, so it's already negative, we're taking it away and it's going to become positive so we can add all of this together to get the first one. area, so 64. subtract 234.6 recurring plus 272. subtract 96, subtract a quarter 1.25, then we're going to add 11 thirds, so add 11 over 3, subtract 17 and add 24 all together for this and get y is 15 15.75 or 15 and three quarters now what we need to do now is what we found, although that's just the first area, we need to find the second area, so we're going to do another substitution so that the second area is between four and six.

Look up, that's the end point, so the biggest is six. A small quarter is six to the power of four. Take the weight eleven thirds of six cubes plus seventeen times six squared and subtract 24 times 6. Then we're going to remove. the smallest one, which is four, we've already substituted it into four, so we can already get a value for this and it was the first four terms, so it was 64. subtract 234.6 recurring plus 272, subtract 96 . so it was actually five and a quarter, so we already have the value, so all we have to do now is our third substitution for every six, so we have a six to the power of four, which is a divided quarter. times four, that's 324. . six to the power of three, which is eleven thirds times eleven divided by three, which will give us seven hundred ninety-two six squares, thirty-six times seventeen, which gives us six hundred twelve and then twenty-four times. six will give us 144. let's take away five and a third and all together that should give us the second area, so 324 minus 792 plus 612 minus 104 times four minus five and a third that gives us minus five and a third, isn't that strange? that we get the same number twice?

Basically everything when we replace the six seems to have canceled out so for the areas we just take the positive so that's five and a third that's the second area so now if you want the total area we just need to add them together , so 15 and three quarters plus five and one third, that will give us, so you can write it as a decimal, it would be 21.1 with one decimal. a precise answer, you could write it as a fraction, it's 21 and 1 12 to get a precise unrounded answer to the curve here, the biggest source of error is that that second area appears as negative, but you treat it as a positive area.

The areas are always positive. so make sure any numbers you have are positive and add them at the end. Vectors give a direction and they give a length so in the first diagram we can go from a to b and we can see that it is labeled vector a. so we have two a, there are the capital letters at the point so you can see the coordinates, the positions of the vertices whatever you want to call them and a with a lowercase a is the vector, so we can say that the vector from a to b is equal to a now , the thing about vectors is that they are not limited to particular positions, so for example, from c to d you will notice that they have the same length and the same direction as from a to b, so what we can do is say well c to d is also equal to a, you will also see vector b, so from a to e we can call it vector b and what we can also do is go backwards so we can go, let's say, from m to i in the opposite. direction a b, but you'll see that in addition to going in the opposite direction, it's the same direction and the same length, so what you can do is say that the vector m is i, so the direction is really important.

We are not saying i to m, let's play m2i, we can say this is negative b, it goes in the opposite direction, another thing we can do is use multiples, for example, we could go from f to h and we will see from f to h. We have gone in the same direction as a and we have not traveled the same length, we have traveled twice as far, the length is twice the length of a, so we can say that from f to g is equal to two a, so we can do math . with the lengths we do the same with b, so let's say we go from g to o, then the vector g o would be the same.

You see the same direction as b again but it's twice as long, so we call it t b if you want to go in the opposite direction. and we go from o to g so we call it negative 2b another thing we can do is go to the middle so we can go from let's say n to the middle to o so we can see that it's the same direction as a but it's half the size so the vector from m to o is equal to half of a you can also do vector combinations let's say we wanted to go from m to j now there is no labeled vector for that direction what we could do is go from m to i which is already labeled and then we go from i to j and we can traverse the longest path and these are the charge vectors, so we are saying that the vector from m to j is equal to, so it was a negative b from m to i and a positive a from i to j, so we can combine vectors together and our second diagram has different vectors, it has a different layout, so it is still from a to b, it will still be vector a, so it is the same for this diagram just by coincidence, but if we take a look at vector b you will see that this one is different, it has a slightly different direction and we go from a to o, this time it is the vector b to o in the diagram above, you know something completely different, so just like algebra, the value of the letters can change depending on the question you have, they are not always constants.

Now you'll notice that we have another direction that we can go in and go from b to o and from b to o there is no labeled vector, so we can't label anything right away, we can't just say oh, we'll call it vector c, you can't do that, you can just use what you have been given, so what can we say about this? we should go the long way, we could go from b to a which is the wrong way down vector a to negative a and then we can go from a to o which is the right way down vector b, so we are adding b, so again with vectors you can go the long way if necessary.

You'll also notice that if, for example, we went from b to c, that's plus b and c to o is negative a, so even if we went in a different direction, we'd still get exactly the same answer, so now we know which beta is always negative a plus b , we can start using it to find other vectors, so let's say you want to go from c to e, the direct path has no labeled lines, so we can't take a direct path, but what we could do if we go from c to d and we can go from d to e and that will take us to the same end point, now c a d has the same direction and the same length as b. a or we could even write this with a little equation so we can say that c a e is equal to c a d plus d a e so we know that c is d being equal to b a o will be negative a plus b and know that d a e is the opposite direction to a so all together will give us minus 2 a plus b, so you can also start doing math with vectors.

You will also commonly see positive vectors placed first, so minus 2a plus. b you might see b minus 2a, it represents exactly the same thing, but putting a positive number in front just means you don't need to write the positive sign because if there is no symbol in front we assume it's positive, so you're just saving a space By writing it that way, you don't get any extra marks for it, but you might see it written that way on math outlines. Now there are multiple ways to write vectors, so for example, here we have the vector p and We have the vector q and so using individual letters is like what we saw in the last review slide, we can split them into vectors i and j and these they give us a little more of an idea about what the direction of the The vector is like this, so let's sketch out what these things mean, so I correspond to the x coordinate and y corresponds to j, so when I say 10 i what you mean is that we are crossing for five when we say for j we mean that' Now that you are going up four, it doesn't mean crossing and then going up, what it means is that you will go directly between the two points, but you will describe it as a diagonal line, we describe it as one horizontal and one vertical. component, then we have the vector p and we have the measures of the vector p, which is what basically gives us 10 i and four j.

Now what happens to the vector q? So with the vector q we can see that it's four i, so we're going to take another 4. across and it's negative 9j, so it's going to go 9 down, so the plus and minus just tell you it's going down. left or right, up or down, so now we can label the vector q so that it goes straight, not left and then down. so it will look like this, we have vector p and at the end of vector p we have vector q. Now we can do more calculations with this, for example we could have vector p plus vector q and do so. that we take the vector p 10i plus 4j and we take the vector q 4i minus 9j and I'm going to add those vectors so that 10i plus 4i is 14i and 4j we subtract 9j is negative 5 j now we can also show this vector in thediagram so p plus q we're going to start at the beginning of p I'm going to finish at the end of q so again it's not like a complicated wave of instructions we do 10 different instructions you go to the right and then up and then to the right and then down Below, what you're actually doing is going directly from the start of p to the end of q, so p plus q is what I'm labeling in pink, so again there's a distinction between what the vector actually does. which is a direct root and the way it's written, which could give you vertical and horizontal measurements to give you an idea of ​​where the vector is actually pointing.

Now you can also do multiples of these, let me do the vectors above, so if p plus q is equal to 14i plus five j, then we can say that two lots of p plus two lots of q, you know it's going to be something like 2 lots of 4i minus 5j, in fact, you can also write the original expression using parentheses so you can write them this way, you can also expand the square brackets so that the square bracket is a good way to show what you are going to do: 2 times 14 is 28 2 times 5 is 10, so 2 many people q would be 28 i minus 10 j if we were to take that out, then what would happen is we would get something like this and therefore it would extend further and be twice as long.

Now the vectors i and j are called position vectors and we can use them to draw shapes and also demonstrate things about shapes, so we are looking at a parallelogram, let's try to draw the parallelogram first, so these position vectors, they all come from the origin, where they come from, you know the point from which we are measuring everything, so let's call the origin o and try to draw all the different vectors so that a a goes minus 2i, that is, 2 to the left and plus 4j, so than 4 up, this is where point a will be, so the position vectors just have If we get an idea of ​​the direction that vector a is going, we'll see vector b, so it's 8 i , so 8 to the right and 3 j, so 3 up again, the vector only gives us the direct root, so two components only gives us the length in both directions so let's look at c so c is negative i and then it's plus 10 j so it's going to be something like here and again it gives us the direct root the last one for d so it's minus 11 i and plus 11 j, so it's going to be somewhere around here and it's going to come up to here so we can see what it looks like. a little messy right now, so we've gone from the origin in four different directions and we're saying this is a parallelogram, so let's see if it forms something that roughly looks like a parallelogram, so let's go from a to b, from b to c, to d and from a to d, and we can see that you know it could be possible. that this is going to be a parallelogram now that I've sketched it out, I don't have a grid here that you can't see, so it's also vaguely precise, but you can get an idea of ​​where all the different vectors we're going to look at are and how it describes where from the form comes.

The next step is how we prove that it is a parallelogram. So for the proof, what we need to do is show that the opposite sides are equal, so a b must be the same vector as c a d or it will be in the same order, so from d to c, so here I go from left to right, so be careful in which direction the letters are, if they are the same length and the same direction then they must be parallel the same to both sides so we need to have a today would have to be the same as b to c so we need to see what are these vectors so let's look at a to b first then to go from a to b we can't go the direct route there is no vector label there but there is a vector from a to o which was negative a and there is a vector from o to b which was b positive, so if we had the labels now they become negative and then they become positive at b.

We also had vector c and vector d. Also, we'll need that later, so what's negative is going to be good, we already have a is negative 2i plus 4j, so the negative version of that we're going to reverse the signs so that let t i minus four j and then when I add b to what is a i plus three j, so if we add all of this plus the like terms two i plus eight i is negative 10i 4j plus 3j is negative 1j, then that's what it is a b, so now let's take a look to see if it has the same direction and the same length as d a c, so to start at d and end at c we can't take the direct route, but we can go through the origin because that's what they give us the position vectors so we can go back. down d to get to the origin and then go up c to get to negative c d flip the signs towards d it's going to be 11i minus 11j and then c what positive c then it's the same as it's written negative i plus 10 j so collect terms similar again 11i minus i is 10 i and minus 11 j plus 10 j will be negative 1 j and what you can see here is that the vectors are equal, they are both 10 i minus j, so if a b and d c are equal to the same thing, that means that, therefore they are parallel, so we have shown a set of parallel sides and we can also use a parallel notation on this, now these are definitely parallel and also have the same length, so what?

What we need to do now is follow the same process to see if aad and b a c also have the same length. Now I'm not going to do that because I think we've gone over this enough now, so you should know what to do next. then you can call it yourself and you should be able to get the same answer for a d and b c but it is actually a parallelogram saying "prove that it is a parallelogram" which will be a parallelogram if you say something like "it is a parallelogram then" Not being could be an option and then you would prove it by showing that these answers are different, just be careful with the wording of the question because you made a mistake here and they didn't seem to be parallel and it says prove. is parallel you should know that you have made a mistake you need to go over it again you are solving something else to do with vectors they are collinear lines all this means is a straight line so again we can try to draw If you have a grid, you can do it precisely, starting from the origin a is one to the right and two up b is seven to the right and two down and c is ten to the right and four down, then we have the vectors for these o a a forwards a direct path o to b and o to c now looking at b and c themselves if called linear then they will be a straight line so let's see if they really are a straight line so joining a to c we should make b stop in the middle and we can see that since I drew this accurately you can see that in reality they are all in a straight line, however, this is not always the case.

We're going to have some graph paper to do this accurately, so we need another way to show that they're all in a straight line and then it's going to be by looking at the values ​​of the vectors, so what we're going to do is we want to see the vector a b and we want to see the vector bc and there should be something in the two different sections of the line that tells you that they are straight, so let's look at b first to start at a and end at b and there is no vector between a and b but we could go from a to or that would be minus a and let's go from o to b which is plus b the negative vector of a is negative i minus two j and the positive vector of b is seven i minus two j add the i's you get six i and add the j's you get minus four j now let's look at b c and c is there something we can convert so b to c we can't go directly but can we? backtrack down b to the origin and then go forward to c to get to c negative b is negative seven i plus 2j and positive c is 10i minus 4j, so collecting like terms minus 7i plus 10i will be 3 i and then look at the js 2j minus 4j is going to be negative 2 j, so look at the two answers, they seem to be different, so how can we say they are in a straight line when the vectors are different?

Well, what we can do is try to factor. Now if we factor b, we can take two out of both and that will give us three i minus two j and that's actually the same answer that we had for bc, so we can see that the actual inj bit has the same length and the same address . It's just that a b is also multiplied by two, which means that a b is twice as long as bc, so we are showing that a straight line does not need to have the same direction and the same length, a straight line is only going to be in the same address so everyone should look at the address but the addresses don't match, take a look and see if you can factor to make the addresses match, the only real way is you can't have any. j ion bits outside the support the support bit has to be exactly the same at the end we can simply write a b and b c they have three i minus two j as a factor therefore they are a straight line and are called linear position the vectors give a vector in terms of a type of length x and y.

You can always have vectors written like this diagram where we have the length of the vector and we have an angle of the vector. Now, one thing you might want to do is I want to change this to a position vector, so we might want to know how far away is what our vector i is going to be. I want to know how far our vector j will go, so how do we do it? Well, you can notice that you could draw this as a triangle, so we have a hypotenuse 11.7, so that's the total length of the vector that we want to get j and i, so we can write i here and we could write j on the other one. side of the triangle will be the same length, we also know the angle which is 42 degrees, now since it will be a right angle because in a coordinate grid you know that i and j are at right angles to each other as the x and y axes are then we can use trigonometry for this we can label the sides again we have the hypotenuse we have the adjacent and we have an opposite so we can use sokatoa we're going to want a triangle formula for this and we can't find them both at once we can only find one at a time so let's try to find the i first.

I noticed the label on both eyes, let's make sure the one on the wide section is going to be a j, so with i means we have the adjacent and we have the hypotenuse as a clue, so it's going to be a question about the car, so writing can form the triangle and we will look for which one is adjacent, so if we don't have the adjacent we only have the cosine and the hypotenuse so we need to multiply them together so that it is the cosine of the angle 42 degrees multiplied by the hypotenuse 11.7 and if we do that calculation we will get the kind of distance to the left that i goes so cos 42 multiplied by 11.7 gives us eight point six nine to two decimal places now let's take a look at j so let's paste again the same thing we need a formula triangle for j look at the j which is the opposite we have the hypotenuse of the clue, it will be a solar question, we are looking for the opposite, so it will be sine times the hypotenuse, so sine 42 times 11.7 now, if you just look at two calculus notes, They are quite similar, so you can use your previous calculations on the calculator screen by pressing the back button and swapping because the sign you should get is 7.83 to two decimal places, so now we can write this as a position vector, for so i is 8.69 and j is 7.83 now. the last thing you need to do here will be positive or negative because it all depends on where the angle is measured from now, if the angle starts to be measured from the positive side x then it will already give you its negative or positive so no need to think in it if it is from negative .

Of the four possible directions you can measure from the x axis, if you go from the positive side counterclockwise you will get a positive or negative result if it goes in the other directions then you need to look if it is positive or negative yourself if the angle is measured from the y axis then what you need to do is take it away from 90 so that you have the kind of equivalent angle from the x axis these have to be measured from the x axis now think about the quadrants of one of these we are is the top left, which means that x will be negative, so i will be negative and j will be positive, so our final answer for the position vectors is negative 8.69 i plus 7.83 j what we have here is a column vector and you can rewrite it as a position vector, it would be 8.23 ​​i plus 11.33 j, so with a column vector the top one is x, the bottom y is y x is i and y is j now what would happen if you wanted this to be some kind of full length and angle like the last bit we saw?

How do we know the angle of this direction? How can we find it? Again it will be about triangles. We can imagine that we have vector i or pair i of vector 8.23, we have part j 11.33, so the total length of the vector will be from the start of i to the end of j and we call This is the magnitude that we label r. We're also going to want the angle of this which we'll call theta. So,how do we find these two things? Yes it's really easy. We have two sides of a triangle. What is the hypotenuse of a? triangle is the Pythagorean theorem so we're going to square the two sides, we're going to add the answers and then we're going to take the square root of that answer to get the hypotenuse so you can type all that into the calculator in one line.

Actually, I'm going to do it in two, so I'm going to write the bits squared and add them together and then I'm going to take the square root of the answer. You will write it on one line, first you write the square root and then. you put brackets every day on everything that has a square root when you do that you should get the answer which is 14. now it's 40.00 so to one or two decimal places it will be 14. so that's the magnitude of the vector, it's the length of the vector, so we have a length of x, a length of y, and then the total lengths combined.

So what about the angle? How do we find the angle? Now again we're going to use trigonometry, so we have the hypotenuses and the adjacent ones are 8.23 ​​and the opposite is 11.33, so we're going to need a formula triangle. Let's use sockatoa. Now we have o and a as clues, so it will be a Torah question and we are looking for the angle to see. we have or over a, that means the angle term is equal to the opposite 11.33 over the adjacent eight point two three, so solve the equation you need to do ten to the power of negative one on both sides to get theta on its own. it's going to be ten to the power of negative one eleven point three three over eight point two three that can be written on the calculator remember to open the brackets that's really important divide the numbers by each other close the bracket and you should get 54 degrees and that's actually 54 to 1 decimal as well, so now we know from the column vector 8.23 ​​over 11.33, we know the total length of the vector, the magnitude of the vector which is 14 and we know that the direction it is going is 54 degrees from the X axis.

That's why in some videos I have inexplicable scratches