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How To Graph Equations - Linear, Quadratic, Cubic, Radical, & Rational Functions

May 05, 2020
Today we are going to focus on

graph

ing

linear

equations

, including those with inequalities,

quadratic

equations

, we are going to talk about transformations, graphing

radical

functions

,

cubic

functions

, absolute value equations,

rational

expressions, exponential and even logarithmic equations, so let's start with the basic, let's say. if we want to graph y is equal to 2x plus 3. now there are many ways to graph it so the first way we're going to use is um let's make a table let's keep things simple so let's say if you have in your table x and let's to plot some points, so let's plot the negative point 1 0 and 1.
how to graph equations   linear quadratic cubic radical rational functions
We don't really need that many points, so in negative one if we replace negative one with x, then two times negative one plus three that's negative two plus three that is one if you connect zero you should get three if you connect one two plus three is five and then you just have to plot those points so here is a negative one on the x axis and here y is one which is our first point when x is zero and y is three , we have this point here. I'm going to plot another point because my graph doesn't go to five, so I'm going to plot minus two if we replace the negative. 2 for x that's 2 times negative 2 that's negative 4 plus 3 we get negative 1. so that negative 2 is somewhere around here and then you could connect those points with a line.
how to graph equations   linear quadratic cubic radical rational functions

More Interesting Facts About,

how to graph equations linear quadratic cubic radical rational functions...

This is just a rough sketch. My graph is not so perfect. but that's the general idea of ​​how to plot or how to graph a

linear

equation using the table. You just have to choose some points and connect them, but now let's see if we can graph another question like this but without making a table. Let's say if we have y equals negative two x plus one, okay, this equation is in slope-intercept form mx plus b b is one, that's your y-intercept, so that's where your graph will start at one m is the slope, the slope is negative two, which is the same as negative two over one, this is your climb and this is the run, the climb over the run, so every time you move one unit to the right, you have to go down two units, so that one unit to the right goes down to the next.
how to graph equations   linear quadratic cubic radical rational functions
The point is here and if we go one year to the right to here is our next point and one unit to the right down will be somewhere around here and then we just have to connect those points with a line, so that's the slope . intercept method, you simply plot the y-intercept and then use the slope to find each successive point, so now let's try another example like this, but using fractions, so let's say that if you have this equation, y is equal to negative three quarters of x plus five, so one two three four five one two three four let's go up to eight five six seven eight so the y-intercept is five so that will be the first point now the slope is minus three over four so minus three is the rise and four is the race then we're going to travel four units to the right and we're going to go down three units, so our next point is here and then we're going to do the same thing, we're going to travel four units to the right to eight. and down three, so we'll go down by a negative one and then we'll just connect those points with a line, so go ahead and try this example, feel free to pause the video and give this problem a try and when you're ready, just unpause . okay, so let's start, let's start by plotting the y-intercept which is at minus two and the slope is two over three, so our climb is two and the run streak we'll go up two units and then three units to the right. and then the same two over three, so we have these three points to plot and then once you feel like you have enough points, you really only need at least two to make a good line, but once you have enough points, you can plot it like this.
how to graph equations   linear quadratic cubic radical rational functions
This is how you can graph linear equations in slope-intercept form, but now let's talk about how to graph them when it's in standard form. The standard form is a x plus b y equals c, so let's say if we have this equation, 2x plus 3y equals 6. Now what you want to do is find the x and y intercepts to find the x intercept by replacing 0 with y so that this becomes zero, so two x equals six, if you divide both sides by two you get x equals three so you get point three point zero so that's the you get three and it's equal to six and if you divide both sides by three and it's equal to two then your y intercept is 0 point 2. and that's what you want to do if you have an equation in standard form and then once you have those two intersections, just plot those two points and connect them with a line, so one two three four one two three four, okay, so the first point is zero two, the next one is at three zero and then just connect those two points with a line, that's the easiest way to graph an equation in standard form, so let's try another example, let's say we have 3x minus. 2y is equal to 12.
Go ahead and try this and then save it when you're ready, so let's find the x-intercept. If we replace 0 for y, 2y disappears, so we will just get that 3x is equal to 12. and if we divide both sides by 4 12 divided I mean by 3 12 divided by 3 is 4. then we get dot 4 comma 0. and for the y intercept we're going to substitute 0 for point zero minus six, so we have zero minus six and four zero and we're going to connect those two points with a line. I don't have a ruler, so my graph won't look perfectly straight, but you get that point, so now you know how to plot.
In standard form, how would you graph an equation that looks like this and is equal to 3? How would you graph that? All you have to do if y equals a number is just a horizontal line at 3. This is how you graph it and if you get a question it looks like this, let's say x is equal to 2, it's just a vertical line at 2 .so it looks like this, this is how you can graph a linear equation if y or x is equal to a constant, but now let's talk about graphing inequalities. If you have this question, say that y is greater than two x minus four and y is less than or equal to negative three over 2 x plus 3.
By the way, if it is in standard form, be sure to convert it to slope-intercept form, which is what we see here note that the slope-intercept form is like this y is equal to mx plus b you want y to be on one side of the inequality this can make it much easier to graph it okay so go ahead and plot these equations in the meantime, so let's plot the first one so that the y-intercept is negative four and the slope is two, so every time we move one unit to the right, we'll go up two, one unit to the right, until the slope is two over one and let's note that y is greater than two x minus four but not greater than or equal to if it is just greater than we need to use a dashed line now for this it will be a solid line because it is less than or equal to negative three over two x plus three, but for this one make sure to use a dashed line, so now let's graph the other equation so that the y intercept is at three and the slope is minus three over two, which means that as you move two units forward the right, the stroke is two, the rise is minus three, so we need to go down three, so over two will go down three and then over two will go down three again and then over two will go down three and it will be a continuous line, okay, now we need to know which region to shade. there are four regions on the graph this is the first this is the second this is the third and this is the fourth ignore the x and y axes look at the four regions created by the green and yellow line when you shade it it will be one of those four regions, like this So first let's focus on this equation and it's greater than two x minus four, so the green line is two x minus four, the larger line that I'm going to shade white is anywhere above that line, so all that region that is greater than the line than the green line now stops the yellow line and is less than or equal to, so it is below the yellow line.
I'm going to shade that in purple so this is the region below the yellow line now I need to shade the region where it's true for both and it's true in this region this is where we have both the purple and the white line so here it is where is the answer in a test. This is the only region you should shade and erase the others. the marks you made and this is how you choose the correct region to shade, so let's try another example, let's say if x is greater than negative three and is less than or equal to five and y is greater than x minus three, then x is in negative three which will be a vertical line and it will be a dashed line because it is greater but not equal and y is less than 5, so it is a horizontal line at 5 but it is a continuous line and y is greater than x minus three, so the slope is intercept, I mean the intercept, the y intercept is at negative three and there's a one in front of x, so the slope is one, so every time you move one to the right, you go up one and it's going to be a dashed line. now we need to know where to shade it so let's look at this first equation x is greater than negative three is greater than negative three to the right anywhere to the right of this function now y is less than 5 is less than 5 below this line anywhere place below this line and y is also greater than x minus three is greater than x minus three above this line, so we need to choose the region where all the arrows are located, if you notice that is true inside this triangle, so this is the region that we shade because all three equations are true in that region and this is how to do it right, so now let's move on to the

quadratic

equations, so let's say if you have this graph, y is equal to x squared plus 2x plus 3. note that it is in standard form ax squared plus bx plus c.
Now for this example, we're going to make a data table and we're going to plot points, so if you want to make a data table, you first need to find a vertex to make your life a lot. easier and the x corner of the vertex is negative b over 2a, that's where the axis of symmetry is, so let's say if you have a graph like this, this equation will get this x coordinate and this is useful because let's say if you plot this point and you choose this point, they will have the same value and and these two points will have the same value and, so if you want five points, you really need to find three and two of them will be the same as the other two. so a is one, this is a and b are two, so let's use the equation x is equal to negative b, which is negative two over two times a or two times one, so the vertex is at x is equal to negative 1. now Let's choose five points, but our center point will be negative vertex one.
We're going to choose two points to the right of that point and two points to the left. Two points to the left of negative one are negative two and negative three. and two points to the right are zero and one, so now let's connect these numbers, let's start with negative one, if we plug negative one in for x, we get negative one squared plus two times negative one plus three, so that's one minus. two plus three, which is two, if we plug in, let's say 0 for x, it will simply be 0 squared plus 2 times 0 plus 3, which is just 3.
So 0 and minus 2 will have the same value of y because they are equally distant from the vertex , let's plug one in, it's easy to plug in one and then negative three, so one squared plus two times one plus three will be 6. So negative 3 will also have the same value of y 6. Okay, so let's make our graph, the most of the graph is at the top, so 1 2 3 4 5 6. and most of our x values ​​are on the left side, so one two three four one two three four so the first point is that negative zero one two that is here, that is the vertex.
Then we have point zero three and negative two three and then we have point six which is here somewhere and negative three point six, so now we can graph it and it will look like this. I can't believe I missed this point so it's a rough sketch the axis of symmetry is this line here is simply x is equal to negative one is the x coordinate of the vertex so that's the axis of symmetry aos if you need to find the minimum value the minimum value is the y coordinate of the vertex is right there, that's where the minimum is located.
Let's try another example like this for practice, so let's say if we have y is equal to 2x squared minus four x plus one, then let's find the vertex first is negative b, which is negative negative four over two a a is two so this is a this is b so this is equal to one so let's make our table and we need two points to the left of one which is zero and negative one and two points to the right two and three so let's start by substituting one for x so two for one squared minus four times one plus one so this is two minus four plus one and that's equal to negative one now it's easier if we substitute zero instead of two, so two times zero squared minus four times zero plus one is one , the next point we're going to connect is negative one, by the way, it's also going to be one, so this is negative one squared is one times negative two. four times negative one is four, so we'll get seven for this point, which means thatthis will also be seven, so now let's make our graph so that it goes up to at least seven and most of our x values ​​will be on the right side, so our vertex is at one minus one, which is here, our next point is that zero one and we have another one at two point one and then three point seven, which is up to here and minus one point seven. which is right where I put the seven so I can then graph it and here is the axis of symmetry is x is equal to the x coordinate of the vertex, this is how you can plot quadratic equations using a table, but now let's say if we don't want to make a table, Let's say if you want to graph it in a simpler way, let's say you have y is equal to x squared minus two x plus three find the vertex as usual, it's negative b over 2a, so it's negative negative two over two times one, so x is one, now you want to find the y coordinate at that point, so if you plug one for for x you get one squared minus two times one plus three, that's four minus two, which is two, so we got the dot one point two so our graph is initially here so this is what you can do let me show you a pattern let's say if you plug in 1 for x you will get one for y correctly if you plug in 2 for x you will get 4 for y If you plug in 3 for x you will get 9 for y so once you have your vertex and this is 1x squared, by the way, because it was 2x squared.
You have to duplicate everything to find the next point after the vertex move one to the right and then up and do the same for the left side. one to the left up one because one squared is equal to one now two squared is four, so starting from your vertex as you move two to the right go up four one two three four same thing here two to the right four and that is a quick way we can plot this equation without needing a data table, so now we're going to focus on graphing these equations but in vertex form, so let's talk about how to convert a quadratic equation from standard form to vertex form .
Now there's something called completing the square so here's what you need to do you have x squared plus 6x we're going to focus on that number six take half of that number and then square half of six is ​​three and we're going to do it to square we still have negative three now notice we added nine on the right side so we change the equation to balance that we need to add nine to the left side or subtract nine from the right side since I don't want to change the left side I'm just going to Put -9 on the right side, so if you add 9 and subtract 9 you haven't changed the value of the equation.
Imagine that you have a bank account, let's say you have a thousand dollars in your bank account if you add 100 and subtract from it. 100 is still the same, so the two equations are equal to each other, but this can be factored out if you factor it. This is the shortcut. It's going to be what you see here x and whatever this sign is plus whatever number it is before squaring it. 3 squared will always work that way and then you can combine these two numbers, so now you have it in vertex form, which is x minus h squared plus k, the vertex is h point k, so in this form you can see easily which is The vertex is like this in our example, notice that h and negative h have the opposite sign, so you need to change it from 3 to negative 3.
This is h and k, the sign is the same, they are both positive, so whatever this number is, that is k. don't change it to be minus 3 minus 12. I'm going to go two by two 2 4 6 8 10 12. so that's minus 12 2 4 6 8 and so on so that we have the point minus 3 minus 12 which is somewhere place in this region and since 1 squared is equal to 1, as we move one to the right, it goes up one, so it will be somewhere here at negative two, negative slash, eleven, one to the right, towards up one and then as you go two to the right, you should go. up four so from negative 12 is going to be at negative eight and then two to the right up four now if we go three to the right three squared is nine we need to go up nine so that will take us to negative three and three to here on the right , which will be at negative six with the value of y, it will also be negative three if we go four to the right, four squared is sixteen, so we will be at the vertex that is at negative three, so four to the right will be x. be one minus three plus four is one and minus twelve plus sixteen is four so we're going to be on a point four which is somewhere around here and if we go four to the left minus three minus four that's minus seven and the value of y still it should be the same, it should be around uh four as well and now we can graph it so it's a rough sketch.
All my sketches are rough, so that's what you can do if you have it in vertex shape. Let's try another example, so let's say that if you have y is equal to negative x minus 3 squared plus 4. Now this negative sign tells you that the graph will open downward, so our vertex will change negative three to plus three, but don't change the four, which is our vertex three, point four, so Plot that point, so here are three points, four, and we're going to need more points on the right side. Now we know that the graph is going down, so as we move one to the right, it will go down one and one to the left, one two. to the right will go down four from the vertex, so it will be at zero on the x-axis. two to the left will also go down four now if we go down three to the right from the vertex, which will be here we have to go down nine, so we should be at minus five, which is here and if we go three to the right, we will be on the y axis and we will also go down five, this is what this graph looks like. something like this and this is how you graph it now, what about this one?
Let's say if there is a 2 in front of the parenthesis, then how do we modify our process so that the vertex we know is negative 1? You change the plus one to minus one, negative comma. two this is going to stay um k it's going to be the same my lines are never straight so notice that it's positive 2, that means the graph is going to open up, so let's plot the vertex so it's at minus 1 point negative 2, which which ended up here somewhere now we won't move 1 to the right and 1 up because our main function is y is equal to 2x squared there is a 2 in front so if you enter 1 for x you will get 2 for y if you connect on 2 for x you get a for y, so as we move one to the right it won't go up one, it will go up 2 because it will be multiplied by two, so one to the right up two, that should end it. here somewhere one to the left until starting from the vertex, as we go two to the right, we need to go up eight instead of 4. 2 squared is 4, but we have to multiply it by 2.
So if we go up 8, we should be at 6 from minus 2. and 2 to the left we need to go up 8 will also be at the same point and this is how you graph it and the axis of symmetry if you have to find it is x is equal to this negative value so here it is the axis of symmetry, it is symmetrical about that line and this graph has a minimum as long as you have a positive number here the graph always has a minimum and the minimum value is simply the y coordinate of the vertex is negative two when it opens downwards , let's say if you have negative all. for quadratic equations now let's move on to absolute value functions, the absolute value of x looks like this, it's shaped like a v and it has a slope of one, so let's say if we want to graph the absolute value of find the vertex, so we need to move three to the right because we have negative three inside and four above, kind of like in the last problem, the vertex is three point four, you change what's inside but you don't change what there is the outside now the slope is one for this problem, so when you move one to the right you need to go up one to the left of the other.
It is symmetric like the quadratic function, but notice that y equals x instead of y equals x. squared so when x is one and is one when x is two and is two then the slope will be one and then keep moving one to the right and up one and then just plot the graph that's how you can graph functions of value absolute, but now let's say that if we have y is equal to negative two x plus one minus three, now the fact that it is negative on the outside means that it will open downward instead of upward and the fact that we see two the slope is now two instead of one, so as we move one to the right, it will go down and the slope will be constant and will always be 2.
So first let's plot the vertex which is minus 1 minus 3. So that's it. here somewhere minus one minus three and we said it's going to go down, so as we move one to the right, it's going to go down two, let me use a different color and one to the left and down two and once again, one down. the right. down two one left down two then it will look like this, this is how you can graph absolute value functions, so let's move on to

cubic

functions, let's say if you have y is equal to x cubed, the main function for this graph looks like something like this which is just a sketch, so let's say if you want to graph this equation, the center point or where the origin should be is at uh one point two, you change minus one but not both, so the change moves one. to the right and it moves up two um what you see inside is the horizontal displacement that is one to the right this is the vertical displacement that corresponds to it, so the origin is now at a point two now we can draw a rough sketch if we want, we can do something like that if we want, but let's actually get points and make this graph more precise so that our main function is y equals x cubed one to the third power is one, two to the third power is eight, so to get the first two points as you move one to the right, go up one, go one to the left, go down one now, if you move two to the right, you need to go up eight and we don't really have room for that, but it will probably be somewhere above. here and if we move two to the left, we need to go down eight.
Now we can plot this because it's a place that will be at minus six, so down 8 will be up to here, so now we can get a more precise sketch. It's going to look like this, so let's try one more example with cubic functions, so let's say if you have y is equal instead of putting in minus 2, I'm going to put in minus half because I don't want to go up 16. Let's say if we have this and it's minus one, We know that the center will be at minus three minus one, so it will move three units to the left and one down, so it will be at minus 3 minus 1, which is in this area and let's say if we move one unit to the right, by the way, notice that it's negative positive x cubed looks like this negative x cubed looks like this, so we have a decreasing function instead of an increasing function, so as we move one to the right, this is It's supposed to be q by the way, one to the third is one, but we have half, so we need to move down half, so one on the right is going to go down by half because this number one on the left is going to go up. half now two raised to the third is eight but multiplied by half is four so as we move two to the right it will go down four so from negative one to negative five and as we move two to the left this point should be here let me fix that one on the left should have been half up, two on the right, that's four down, so it's here at minus five and two on the left, we need to go up five, that's four up, so one two three four should be around here somewhere, so now let's see if we can make our graph, so we have to be the function should decrease instead of increase just because of the negative sign, so that's a rough sketch of that graph, so now let's review the

radical

functions.
Let's say if we have this function, the square root of x, the main function looks like this and if we have, let's say, negative square root x, the negative in front causes it to flip over the . axis so it looks like this and let's say if you have a negative on the inside, it reflects across the y axis, so it looks like this and if you have a negative on both the inside and outside, it reflects across the origin, so it looks like this is how I like to see this, let's say here x is positive and y is positive so x and y are positive in quadrant one so it goes towards quadrant one here x is positive but y is negative so that's the quadrant four x is positive as you go to the right but y is negative as you go down here to see where the graph is I'm going to go with a radical x, but now let's apply it to a problem, so let's start with a simple one, let's say if we have y is equal to the square root of move three to the left. anduntil then let's start again here now the main function is the square root of x, the square root of 1 is 1 and the square root of 4 is 2 and the square root of 9 is 3.
So as we move one a to the right goes up one as you move four to the right goes up two so it's going to be somewhere around here and as you move nine to the right from the original point it goes up three so now we're at minus three if we move nine on the right we should be at positive six and going up three we need to be at y is equal to one so that's our graph start from here now one thing I should have mentioned is the domain and the range what is the domain of this function and what is the range the domain is the x values ​​the range is the y values ​​if you want to find a domain parse it from left to right so it starts at the lowest x values ​​at minus three there is nothing before that and you get to the highest x value infinity, then its domain is from negative three to infinity and includes negative three, the range is the y values, the lowest y value is that negative two, which is that number and this arrow keeps going up, but it goes up slowly and slowly, but it goes up to infinity, so the range is from minus two to infinity.
Now just to review the linear equations, like let's say y is equal to two x plus three, I'm just going to draw a rough sketch so it starts at three with a slope of two let's say it looks like this the domain of a linear function is simply negative infinity to infinity the range is the same now the next one we considered was a quadratic function let's say if you have 3 plus x squared the graph will start at 3 and open up, it has shifted up 3 units, so the domain for any quadratic function x can be anything, it will always be negative from infinity to infinity, however the range will change.
Note that the lowest value of y is 3 but it goes to infinity, so the range is from 3 to infinity. Now, if you have a descending parabola, let's say y is equal to five minus x squared, your graph will start at five and open up. down has shifted up five units, so its domain for any quadratic function where it has x squared as its highest degree will always be from negative infinity to infinity, although the range is from, let's see that its lowest y value It's negative infinity because these arrows will go all the way, their highest y values ​​are five, so from low to high negative infinity up to five it stops at five, so for your domain you're looking at it from left to right from the range that is looking at it from below. up or bottom up now the next thing we considered was an absolute value function, so let's say if you have 2 x minus 3 plus 1. then the graph shifts three units to the right up, it starts here and the slope is two so I'm just going to draw a rough sketch of its domain for any absolute value function it's from negative infinity to infinity unless you have a piecewise function its range notice that its lowest y value is one but it goes up to infinity, so it's one to infinity and let's say if it opens down, let's see you have three minus the absolute value of x plus two, so it shifts two units to the left and it shifts three units up, so which starts here, but there's a negative in front of the absolute value, so it goes like this, it goes down with a slope of one, so its domain is all for this function as well, but the range, the highest value is three , the lowest value is negative infinity, so it starts from low to high negative infinity and stops at three. now for cubic functions, let's say if we have . will increase instead of decrease, so it will look like this for this type of function. its domain is also everything, there are no restrictions on x, its range is also everything, and it goes down to negative infinity and to infinity, so for any domain and range of cubic function they are just real numbers, so now let's go back to radical functions, let's look at our next example, let's say that if we have the square root of 5 minus 2 minus x, the first thing we want to plot is the first point. so if you set inside equal to zero, two minus x equals zero, you'll get x equals two, so it shifts two units to the right and five units up, so it starts somewhere here , the question is in which direction.
Are you going to go towards quadrant one towards quadrant two towards quadrant three or towards quadrant four because knowing the direction is important because we need it if we don't want to make a table so notice that we have a negative in front of the radical and a negative in front of x, see that then x is negative and y is negative that we can use our general transformation: the square root of one is one, the square root of four is two, and the square root of nine is three, so as we go one unit to the left, we go down one and since this is x this is and then as we travel four units to the left it will go down two so it will be over here and as we travel 9 units to the left starting from this point it will go down 3, so this will end here now, if you prefer, You can always make a table, so I'm going to do an example by making the table so you can see which points to choose and there's the graph.
Let's say if you have y is equal to the square root of x plus one minus three and you want to make a table, the first point you want to insert is negative one and it will be negative three, that's why that's the offset. moves one to the left and three down now from that point let me put this number here you want to go one to the right so let's enter zero if you enter zero for x the square root of one is one and one minus three is minus two now your next point you don't want to connect 1 2 3 and 4. you want to connect numbers that are perfect squares because you can take the square root.
The reason I plug in 0 is because I can take the square root of 1. the next number I'm going to choose is 3 because 3 plus 1 is 4 and the square root of 4 is 2. 2 minus 3 is uh negative 1. now The next point I'm going to insert is 8 because 8 plus 1 is 9, the square root of 9 is 3, and 3 minus 3 is 0. If I want another point, I'm going to substitute 15 because 15 plus 1 is a perfect square of 16. The square root of 16 is 4 4 minus 3 is 1. so you want to insert numbers that are perfect squares where the radical would simplify to a whole number and that will make your life easier and basically the way I was plotting points for These other functions are based on this pattern of numbers. so let me show you so we know that you move one to the left down three, so this point is this one minus one minus three and for the radical x radical one is one, so as we move one to the right, will raise a notice that we have point zero minus two and the square root of four is two, so we start from this point, if we move four to the right, which will put us at 3, it will increase by 2.
So notice that we have this point 3 minus 1 and then the square root of 9 is 3. So as we go 3 to the right, one, two, three, four, five, six, seven, eight, nine, we'll go up three, so it'll be here, which is eight point zero so notice this technique gives you the points without actually making a table and if you can master this technique you can graph equations at a very fast rate so our graph looks like this now keep in mind let's say that if we put a 2 in front everything will be doubled, so instead of going one to the right as you go you are going to go up two instead of going up one as you go four to the right you are going to go up instead of going up two you need raise four which is somewhere around here and as you go go nine to the right instead of raising three you need to raise six which will be somewhere around here so your graph will look like this if it were and equals two square root of here, let's say if there is a two, it would be two square root of one, which is two if we connect it. a four will be four if we replace nine will be six square root of nine is 3 times 2 is 6. so these numbers, the amount we increase the y values ​​will simply double, so now let's move on to a different type of radical a radical with a odd root instead of an even root the square root of x what we considered was this there are two invisible ones that are not seen so if it is even it looks like this but when it is odd it looks like this, however, the negative cube root of It will look like this.
By the way, keep in mind that this is the same as the cube root of negative x, so even if you reflect it across the x-axis or across the y-axis, it will be the same because this graph is symmetric about the origin, so there are only two ways to graph the cube root of x this way if it is positive or this way if there is at least one negative, if there are two negatives they can cancel and it will be the same as this if you have a negative inside and outside, but now that we know what the main function looks like, let's try some problems, so let's say if you want to graph the cube root of x plus one plus two.
First you want to know where the graph starts so it moves one left and two up, so end here. Now let's look at our pattern of numbers that we're going to use to plot this equation, so the cube root of one is one, moving one to the right will go up one and the cube root of eight is two, so moving eight to the right will go up two, so one to the right will go up one and one to the left. one because we know the general shape is like this, it's increasing now as we go to the right, so starting from minus one we'll stop at seven, it'll increase two, so it'll be around here somewhere and as we go to the left will take us to minus 9 then it will go down 2 so it will be around here somewhere and now we can plot it so it looks like this this is how you can graph the cube root functions now let's say if you have 1 over x the main function for 1 over x looks like this and if you have 1 negative over x it looks like this it reflects on the y or plot the vertical asymptote, the vertical asymptote is when the denominator is equal to zero, so if you set the bottom equal to zero you get x equals two, that's the way, the vertical asymptote now there's also a horizontal asymptote that is and is equal to zero, which is here whenever the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is at and is equal to zero now If you want to get points for this graph, you actually only need one point, but let's make a table.
I would substitute 3 for , so it should be somewhere around here, so our graph starts from the asymptote and then connects to this point and then follows the horizontal asymptote, so it looks like this. You can add more points if you want and this graph looks like this, this is how you graph one over x minus two, so let's see if we have y equals one over x plus three plus four the vertical asymptote is at x equals minus three based on this term here if you set it equal to zero now the horizontal asymptote for this fraction is zero because it is bottom heavy the degree of the denominator is is higher than the numerator but it is zero plus this number four, so is actually and is equal to four, it's shifted up four units, so let's make our graph so that the vertical asymptote is at negative three and a horizontal asymptote is at four and let's connect some dots, so I'm going to connect a point to the right of negative three, so it's going to be negative two and one point to the left of negative three, that's negative four if I plug in negative 2, negative 2 plus 3 is 1 1 on top. 1 is 1. if I plug in negative 4 negative 4 plus 3 is negative 1 and then I got those points so negative 4 negative 1 is over here and negative two um you know I forgot to add four to each number so that the y value should be three and the value of y here should be five.
I can't forget this 4. But I applied it to the right point for some reason at negative 2, it should be at 5, so somewhere around here now we can make our graph so that this. will look like this and this graph will look like this, as you can see, it has shifted left 3 up 4. Now let's say if we have a negative value, let's say y is equal to negative one over x minus one. plus two, then the vertical asymptote, if you set x minus one equal to zero, x equals one, the horizontal asymptote will be y equal based on this number, so here is the horizontal asymptote and the vertical asymptotethat's it Now notice that we have a negative sign, so instead of the graph looking like this, it will be like this, so let's connect dots, let's make a table, so I'll connect 2 for x and 0 for x 2 minus 1 is 1 but multiplied by minus 1 divided by 1 is negative 1 plus 2, so that's 1.
If we replace 0, it's going to be negative 1 over negative 1 plus 2 and we'll get 3 if we replace 0 with x, so it has the dot 2 comma 1 which is here and 0 comma 3 which is here, so this graph will look like this and this one will look like this, so now you know how to graph principal functions one over x functions. Let's say if you have one over x squared it is very similar to one over x but both curves are above the x axis if you have a negative one over x squared it will be below the to the y axis, the other one over x was symmetrical around the origin, so let's try an example, let's say if we have negative one over x is equal to two, the horizontal asymptote is simply y is equal to three, so for these types of

rational

functions always plot the asymptotes first, so let's make a table.
I'm going to replace one for x and three, one unit to the left and one unit to the Okay, if I connect three, three minus two. that's one one squared is one minus one over one is negative one plus three that's two if I connect one one minus two is negative one but if you square it it becomes positive one minus one over one is negative one plus three is two notice that they have the same y value, which means they are symmetrical around the y axis, but in this case It's going to be symmetric about the vertical asymptote because it's shifted units to the right, so if we plot one two, that's here.
I mean that's three common, two, not one, two, one, two, they're here, but we can see the symmetry around the vertical asymptote, so it looks like this, so let's go over some different types of rational functions, let's say if you have x squared plus five. x plus 6 divided by x squared plus x minus 2. Now, this function may seem complicated, but it really isn't. We've pretty much covered this material. What you need to do is factor everything to factor x squared plus five, x plus six. of two numbers that multiply up to six but add to the middle term five two times three is six two plus three is five so it will be x plus two times x plus three to factor um this one looks for two numbers that multiply up to negative two but add to one to be positive two and one negative two plus negative one is one, so it's x plus two x minus one.
Notice that x plus two cancel, so if you set x plus two equals zero, x equals negative two is an integer, the vertical asymptote is based on this because x minus one cannot cancel, so x is equal to one is the vertical asymptote. Now you may be wondering what the horizontal asymptote is. Now notice that the degree of the denominator is 2 and the degree of the numerator is also 2 when it was down when a degree of the denominator was greater than that of the numerator like one over x or one over x squared the horizontal asymptote was and is equal to zero plus if there was a constant here if it was plus three, then add three, but we don't have that three there, so as long as the degree is the same, divide the coefficients one divided by one is one, so the horizontal asymptote is and is equal to one for this problem so now let's graph it so first let's graph the vertical asymptote at one next let's plot the horizontal asymptote at one so now at this point we know the graph is going to look like 1 over x it will be very similar so we just need a few points one point I would connect is the full negative 2 because that's important if we replace the negative 2 we just need to plug it into the surviving equation so it's negative 2 plus negative 3 negative 2 minus 1. 2 negative plus 3 is uh 1 and negative 2 minus three is negative three so we get the negative point one third.
I would also plug in a number to the right of the vertical asymptote like we did before, like two, um, let's see if I can fit it in here, two plus three is five and two minus one is one, so just get five and plug a number into the left, so let's say zero zero plus three is three and zero minus one is negative one, so we get negative three, so negative 2 point negative 1 3 somewhere around here and 2 point 5 is up here somewhere and 0 minus 3 is down here so we can see what this graph will look like as it starts from an asymptote and connects to these points and goes on to the other one, however we said this is a hole so it should be a circle open i never drew a closed circle but ignore the yellow dot there should be a hole there now for the other one we know what the general shape will be it will look like this you can add more dots if you want but that's a rough sketch of the graph so Let's try another function, let's say if we have y is equal to x squared plus let's make it 8x squared divided by four minus two x squared plus one, so let's take a two from the bottom. be 2 minus x squared and we're going to set that part equal to 0. and we're going to solve for x, so x is equal to plus or minus radical 2. so we have two vertical asymptotes in this problem now to find the horizontal asymptote note that the degree of the numerator which is x squared is the same as that of the denominator so what we are going to do is divide the coefficients eight divided by negative two so the horizontal asymptote is at negative four but we have a constant in front, so shifts up one, so negative four plus one, so it's at negative three, so now we have the vertical asymptote and the horizontal asymptote, the square root of 2 is about 1.4, so the asymptote vertical is somewhere between 1 and 2. and we have another one at minus 1.4 and a horizontal asymptote, it is at minus three, now we know what the graph will look like on the right side and on the left side, it can be like this or it can be like this , it can be like this or it can be like this in the middle a lot of things could happen, you can get something like this, you can get a curve that looks like this or sometimes it can even cross the horizontal asymptote, so for the middle we have a point connection to figure out what's going to happen, so let's make our table for this one.
I would definitely use a board. Let's connect a point to the left of radical two, so let's connect two positives for x, so it will be six. Well, two squared is 4. 8 times 4 is 32 and let's see at the bottom we have 4 minus 2 times 2 squared, so it's like 4 minus 8, which is negative 4. So we should get negative eight, so two commas minus eight, so that's it. all we need is one point in this region, so we know the graph looks like this in this area, so let's connect another point, let's say negative two, notice that it's squared, so regardless of whether you connect two or less two, the values ​​of and can will be equal, so because they are both square, it will be symmetrical or symmetrical, so at negative 2 it will be like negative 8 again, so it will be equal, so the fact that it is symmetric means we'll probably have something like this or something like this so first let's plug in zero if we plug in zero for eight over four minus two, so it's eight over two, which is four, so it's over here somewhere so we can see what's happening.
This graph will look like this in the center, so between two vertical asymptotes you can I'm not sure what's going to happen, but if you see that both terms are even and have an even exponent, you know that there will be some kind of symmetry, it may be like this or it could be like this if it were odd. it could be like this in the center between the two vertical asymptotes or it could be like this, you just have to connect dots to be sure, so that's it for that particular function. Now there is one more thing we must mention and it is rational. functions that involve a slanted or oblique asymptote, so notice that the degree of the numerator which is 2 is greater than the degree of the denominator which is 1. when the numerator is 1 degree greater than the denominator which is we are going to have a slight asymptote now what we need to do first is factor everything before we start um if we multiply these two numbers 2 by 5 it's 10 and two numbers that multiply up to positive 10 but they add up to negative three um it doesn't happen it doesn't work because two times five will add up to seven Negative two and negative five will add up to negative seven, so we can't factor the numerator, if we could, we should know that the vertical asymptote will be x equals 2.
Now, whenever the numerator is greater than the denominator, there are no horizontal asymptotes, but there are We have a slope, so we need to do long division to find a slope asymptote, but we don't have to complete the entire long division process, there is a certain point. where we can stop where that would be enough and we can have the answer 2x squared divided by x is 2x after dividing multiply 2x by which is like plus 4x you get x and you lower the 5. x divided by x is 1. once you get this constant here, you can stop so that the slight asymptote is two x plus one, so now we can graph it. in two we have the vertical asymptote and to plot the sloped asymptote, notice that we have a linear equation in slope-intercept form that we covered at the beginning of this video, so we're going to plot the y-intercept, which is 1. and the slope. is two because there's a two in front of separate the graph into four regions so we can have a graph that looks like this, it may look like this, it may be here or it could be here, it will be in two of those four regions and the only way to find out which region it will be is to connect points, we actually only need to connect two points, that is, one point to the right of the vertical asymptote and one point to the left, so let's say if we connect three for x two times three squared of three. squared is nine times two is eighteen minus three times three is negative nine plus five so that's nine plus five that's fourteen over three minus two is one so we get fourteen so in three is clearly above is well above the slight asymptote is all the way up 14, our graph doesn't go that high, so we're going to say it's in this region somewhere where we don't have enough room to plot it, usually if it's in that region, it's likely to be in this region. but let's find out for sure, so let's plug in a number less than two, plug one in, so two times one squared is two minus three times one, which is three plus five over one minus two, so it's negative one plus five. , that is, four over minus one. so it's negative four, so in one it's in negative four, so it's in this region, let's go ahead and plug in 0 2 because 0 is an easy number to enter, it's going to be 5 over negative 2, which is like negative 2.5 just to make This graph is a little more precise, so our graph will look like this.
The more points you add the more accurate your sketch will be but it is up to you how many points you want to add but usually on a test you can get a multiple choice question and all you need is one point to the left and one point to the right of the vertical asymptote and you can eliminate the incorrect answers to get the correct answer or you can just use the graphing calculator, so now let's To focus on graphing exponential equations, let's say that if we have y is equal to two x plus three minus one, now we What you want to do is set the exponent equal to two numbers zero and one, so that if you solve for x you get negative three. and negative two, these are the points you need to enter into your x and y table, which I would recommend for this type of problem, so if you enter negative three for x, negative three plus three is zero, two to the power of zero is one, whatever. elevated thing. to the zero power is one and then one minus someone is zero if you plug in negative two for x minus two plus three is one and two to the power of one is two two minus one is one now this number that you see here is the horizontal asymptote for a function exponential, so let's graph it to have it at negative one and there are no vertical asymptotes for exponential functions.
The next thing you do after plotting the horizontal asymptote is plot your two points minus three zero and minus two one, so Graph that starts from the horizontal asymptote and follows those two points and that's how you can graph it. You can add more points if you want, but I think that sketch is good enough, so let's try another one, so let's say we have and it's the same. a two minus three to the power of x plus one, so set the exponent x plus one equals zero and one and solve for x so that x equals negative one and one and zero for x if we plug in minus negative one one plus one is zero and three to the power of zero is one two minus one is one if we plug in zero zero plus one is one three to the power of zero I mean three to the one is three so we have two minus threewhich is negative one this number is the horizontal asymptote is the number is the consonant that does not have an x ​​so the horizontal asymptote this time is that two and if we plot our points minus one point one and zero minus one, so start from the asymptote horizontal and connect the two points so that you can graph that particular exponential equation, so the last thing we're going to look at is graphing logarithmic equations, so let's say if you have log base 2 x plus 1 minus 3. then exponential functions are like Opposite of logarithmic functions exponential functions have horizontal asymptotes logarithmic functions have vertical asymptotes now what you want to do is set the inside part of the log function equal to three things instead of two things set it equal to zero which will give you the vertical asymptote, always set it equal to one and set it equal to whatever the base of this number is, so that the first one we get x is equal to negative one. that's the vertical asymptote, the next point we get zero and here we get one, so let's make our table with these two points zero and one, if you plug in zero for x, you get zero plus a log of one, regardless of what the base is, is always zero so you have zero minus three is negative three if we replace one one plus one is two then we have log in base two of two canceling two only gives you one and one minus three is negative two, that's why I chose those numbers because you get a good integer to deal with makes it easier to graph, so the first thing you do is plot the vertical asymptote, which is negative one, and then connect your dots so that we have our first point at zero minus three, which is here and then one minus two, so start. from the vertical asymptote and then follow the two points and that's how it graphs.
Let's try our final example, so let's say y is equal to 2 minus log base three x minus one, so set the inside equal to zero one and the base in this case. which is three, so the vertical asymptote is x is equal to one and our other points are two and four, so let's make our table so that if we plug two in for , so two minus zero. is two if we replace four four minus one is three log base three of three is one and then two minus one is one then our vertical asymptote is that one and we have the point two point two that is here and the next one is four point one, for so the graph has to start this way if it's going to connect to those two points and that's how it's plotted, so that's it for this video.
Now you know how to graph almost any equation, there are other equations too, but for the most part, we cover the most common equations that you will see throughout your trigonometry and algebra calculus and precalculus course, so if you see any new equation that you need to graph, you can apply the principles you learned in this video. equations too, so that's it and thank you for watching this video and have a great day.

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