Hyperbolas at the Origin - Video Tutorials & Practice Problems

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1

concept

Introduction to Hyperbolas

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Hey, everyone and welcome back. So up to this point, we have talked about circles ellipses and Parabolas and we're now going to move on to the hyperbola shape in this series on conic sections. Now, in my opinion, the hyperbola is both the most unique and oftentimes the most difficult of all four shapes that we've covered. And the reason for this is because there's a lot to remember about hyperbola but don't sweat it because over the course of this video and the next few videos on conic sections, I think you're going to find that hyper actually correlate very similarly to a lot of concepts that we've already learned. So without further ado let's get right into this. Now, the equation for hyperbola is nearly identical to the equation for an ellipse. The only difference is is there is a minus sign rather than a plus sign. So if we take a look at the horizontal hyperbola, for example, notice in this equation for the hyperbola, it's very similar to what we've learned previously about the ellipse. The only real difference is that we have a plus sign for the ellipse and a minus sign for the hyperbola. Now, even though the equations for the ellipse and hyperbola are similar, the shapes are actually quite different because visually a hyperbola appears as two parabolas that are facing away from each other. Now, something else that I'll mention this ellipse that we see drawn on this graph. You're not gonna actually need to ever draw this ellipse when dealing with hyperbola. This is just here. So you can see the similarities between the Ellipse equation and the hyperbola equation. And what the A and B values really mean. Now looking at this horizontal hyperbola notice that we can see what the A value is for the ellipse that we drew in here. The A value is 123 units long. It was talked about in previous videos how A represents the semi major axis of an ellipse. Whereas B which is 12 units long represents the semi minor axis. Now when it comes to the hyperbola, this a value tells you the distance to get from the center of the hyperbole to either of the two curves. So as you can see this is the shortest distance to get to this curve and the shortest distance to get to that curve. Now, this B value is a little bit less intuitive when it comes to the hyperbola because we know that this B value describes the semi minor axis for the ellipse. But for the hyperbola, it actually serves a bit of a different purpose. It can oftentimes help in describing the height of the hyperbola. But it's also going to be very critical to use this B value when it comes to graphing the hyperbola, which we'll talk about in later videos. This is the main idea behind the horizontal hyperbola. And this is what the equation would look like if we use this A and B value because remember you have to square both the terms in the denominator. But now let's take a look at the vertical hyperbola. Notice for the vertical hyperbola, we have an A value of 123 except the A value is now up and down to get the either of the curves for the hyperbola rather than left and right and notice how the B value which is 12 units long is now left and right as opposed to being up and down. Now for the ellipse, we've talked about how when you have a vertical ellipse that the semi major axis squared is going to be underneath the Y value this time and the semi minor axis would be under the X value. We'll notice for the hyperbola we have a similar situation where we have the A squared underneath the Y squared and the B squared underneath the X squared. So basically all we did was take the X and the Y and switch them in this version of the equation. So the equation for the vertical hyperbola is going to be Y squared over nine minus X squared over four is equal to one. And again, this is because you have to square the A and the B value when you put them in this equation. Now, there's one more big difference I want to mention between the ellipse and the hyperbola. And that is the major axis A because when dealing with the ellipse, we're used to this a value always being the largest value that we see. Well, when it comes to the hyperbola, the A is actually the first value that you see rather than the largest. So for these equations, in this example, we saw that A was the biggest number, but this A is not always going to be biggest for the hyperbola. This A is simply always going to come first in the equation. Now, you may have felt like this was a lot of information and it was, but let's actually try an example to see if we can really solidify this altogether. So in this example, we have some equations for hyperbola and we're asked to match each equation with the corresponding graph. Now we're gonna start by looking at this first equation equation A. Now, one of the first things I noticed is that we have the Y squared in front and remember this front term that you see is always going to be the A squared term. So you can see that A squared is equal to 16 this number here, which means that A is the square root of 16, which is four. So what I see here is that we have a Y squared and that our A value is four since we have the Y squared in front, that means we're going to have a vertical hyperbola and our A value is going to be four. Now, if I look at these three graphs, the only vertical hyperbola I see is this one on the far right side. So if I take a look at the graph number three here, I can see that our A value is four. So if I start from the center, I can go 1234 units up and 1234 units down. This does check out which means that graph number three is to match with equation A. But now let's take a look at equation B for equation B I can see that our A squared value which always comes first is going to be four, which means A is going to be the square root of four, which is two. So I can see that we have an X squared up front which means we're going to have a horizontal hyperbola. And I can see that our A value is two. Now if we go ahead and start from the center of this hyperbola on graph number two, I can go 12 units to the right and 12 units to the left. Well, this is a problem because notice that our A values go out side of the curve. So this is not the correct hyperbola that we're looking for. But now let's try graph number one. If we draw out our A value, we can start in the center and go 12 units to the right and 12 units to the left. And this lands us directly where our curve is. So this graph does check out now just by process of elimination, it's clear to see that equation C is going to match with graph two. But let's actually see if we can understand why. So I can see that the first term that we have here is one and this is underneath X squared X word means we're going to have a hyperbola that opens left and right. And this one means that our A squared value is equal to one, meaning A is going to be the square root of one, which is just one. So if I start in the center of this hyperbole that we have here, I can go one unit to the right and one unit to the left, this is going to be our A values and that does match with the curve that we have here. So we can see that equation C matches with graph number two. So this is the main idea behind the graphing and equation of hyperbole. Hope you found this video helpful and thanks for watching

2

Problem

Problem

Given the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$, find the length of the $a$-axis and the $b$ -axis.

A

$a=25$, $b=9$

B

$a=9,b=25$

C

$a=5,b=3$

D

$a=3,b=5$

3

Problem

Problem

Given the hyperbola $x^2-\frac{y^2}{4}=1$, find the length of the $a$ -axis and $b$ -axis.

A

$a=1$, $b=4$

B

$a=4$ , $b=1$

C

$a=1$, $b=2$

D

$a=2$, $b=1$

4

Problem

Problem

Given the hyperbola $\frac{y^2}{100}-\frac{x^2}{139}=1$, find the length of the $a$ -axis and the $b$ -axis.

A

$a=100$, $b=139$

B

$a=139$, $b=100$

C

$a=\sqrt{139}$, $b=10$

D

$a=10$, $b=\sqrt{139}$

5

concept

Foci and Vertices of Hyperbolas

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5m

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Hey, everyone and welcome back. So up to this point, we've been talking about the various shapes that you get from conic sections. And in the last video, we looked at hyperbola. Now in this video, what we're going to do is see how we can find the vertices and foci of a hyperbola. Now, I will warn you that this process is pretty tedious because there are some new equations that you'll need to know in order to solve these types of problems. And you are going to be asked to solve these at some point in this course, but don't sweat it because in this video, we're going to be going over some examples and scenarios that I think will this process seem a lot more clear? And I'll also mention that I think you'll find there are a lot of similarities in the problem solving with hyperbola that we did with ellipses in previous videos. So without further ado let's get right into this now, just like an ellipse, every hyperbola has two vertices and two foi and they are on the major axis. Now, the vertices are the points of the hyperbola that are closest to the center and this distance from the center to either vertices is a distance of a. Now to understand this a bit better, let's actually take a look at this example, we have down here where we're asked to find the vertices and fite of the hyperbola in the graph. Now, in order to find the vertices, we said they are a distance a from the center and we can see the center is at the origin here 00. Now recall that A squared is going to be the first thing that we see in our denominator assuming we have the standard form of this fraction minus that fraction is equal to one. And we can, we can see that we do have the standard form here. So that means a squared is going to be equal to this first thing in the denominator which is four, that means to calculate A, we just need to take the square root of four, which is two. So if A is equal to two, then we can start here at the center of our hyperbola. And we can go 12 units to the left and 12 units to the right, starting from the center. This puts our vertices at negative 20 and positive 20. And that is how you can calculate the vertices of a hyperbola. Now you also will need to know how to find the foi and the thing that makes the foi or the focus points unique is that these are going to be a situation where for any point on the hyperbola, the difference of the distances from each focus point to any point on the hyperbola is going to be constant. Now to make sense of this, let's say that we have this hyperbola down here. And let's say that the F I end up right about there and right about here, if these are the two F side, what you can do is find a point, any point somewhere on the curve of the hyperbola. And if you take this distance and you subtract off that distance, the number that you get will be constant for any point on the hyperbola. So let's say this distance is five and that distance is 25 minus two is three. And that means any point that you look at anywhere on the hyperbola, the distance the two distances from the fully to there. If you find their difference, it will always come out to three. So that's what makes the fori unique. Now, in order to calculate the faux I, you will need the distance C and C can be calculated from this equation. Now notice that this equation looks a lot like the equation we had for the ellipse where C squared was a scored minus B squared. The only difference is for the Foi by Hyperbola C scored is equal to A squared plus B squared. Now you might notice at this point, there is a lot of similarities for the equations between the hyperbola and the ellipse. The main differences are the fact that we have a plus sign here rather than a minus sign for the hyperbola. And then we have a minus sign rather than plus sign for the standard equation of a hyperbola. So just keep in mind that there are going to be differences in the science and how you use the equations generally. But a lot of the problem solving ends up looking pretty similar. Now, let's actually see if using this equation, we can calculate the F I on this hyperbola. So C squared is equal to A squared plus B squared. And recall that B squared is going to be equal to the second thing we see in the denominator in the standard form. So that means the B squared is equal to nine, meaning that B is going to be the square root of nine, which is three. So in order to calculate C, what we can do is we can recognize that A squared is going to be two squared, which is four and B squared is going to be three squared which is nine. So C squared is equal to four plus nine, which is 13. And I think the square root on both sides of this equation, we'll get that C is equal to the square root of 13 and the square root of 13 is approximately equal to 3.6. So what that means is that if we start at the center of our hyperbola, here I can go approximately 3.6 units to the right, which will get me to this point at square root 13 0. And I can go approximately 3.6 units to the left getting me at this point which is negative square root 13 0. And this is how you can find the vertices as well as the foi for a hyperbola. Now, one more thing that I want to mention is how the vertices and F I will look different depending on the hyperbola's orientation because we just looked in example of a horizontal hyperbola and notice how the vertices ended up on the X axis at a zero negative A zero and notice how the foresite also ended up on the X axis at C zero, negative C zero. So what we can conclude is that when we have a horizontal hyperbola, the vertices and fite end up on the X axis. Now, if we instead had a vertical hyperbola, the vertices and fite would actually end up on the axis. So you would end up seeing the vertices end up around here and there at zero A zero negative A. And you find the foresite would end up just outside of these two curves just like we had with the horizontal hyperbole. Except these would be at zero C and zero negative C. So when it's a vertical hyperbola, the vertices and foresite on the y axis as for a horizontal hyperbola, they're on the X axis. So that is how you can find the vertices and foresite of a hyperbola. Hope you found this video helpful. Thanks for watching.

6

Problem

Problem

Determine the vertices and foci of the hyperbola $\frac{y^2}{4}-x^2=1$.

Find the equation for a hyperbola with a center at $\left(0,0\right)$, focus at $\left(0,-6\right)$ and vertex at $\left(0,4\right)$.

A

$\frac{y^2}{16}-\frac{x^2}{20}=1$

B

$\frac{y^2}{20}-\times\frac{^2}{16}=1$

C

$\frac{y^2}{4}-\frac{x^2}{\sqrt{20}}=1$

D

$\frac{y^2}{\sqrt{20}}-\frac{x2}{4}=1$

8

concept

Asymptotes of Hyperbolas

Video duration:

5m

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Hey, everyone and welcome back. So in the last video, we talked about how you can find the foci and vertices for hyperbola and continuing on this shape. We're going to now discuss how you can find the asymptotes of hyperbola. Now, this might sound a bit complicated because up to this point, we haven't really dealt with asymptotes for any of these shapes. And I will warn you that the asymptotes of a hyperbole can be a bit complicated when you first learn it. But what we're going to learn in this video is that even though the process is a little bit tedious, there actually is a pattern that forms when you deal with the asymptotes of a hyperbola. And we're going to find out that that pattern that forms allows us to find a general equation that makes solving the asymptotes of a hyperbola super quick and straightforward. So without further ado let's get right into this. Now, the asymptotes are required when graphing hyperbole. Now, in order to find the asymptotes, you can use the A and B values and recall how we've discussed how A is the major axis and B is the minor axis for a hyperbola. But what these A and B values do is they form a box where you can take lines and draw them through the corners of this box. And these lines that you draw are going to be the asymptotes of your hy. Now, to understand this a bit better, let's take a look at this example we have down here, we're asked to find and draw the asymptotes of the given hyperbola and to start things off, I'm first going to see if I can find my A and B values. Now we're called that A squared is equal to the first thing we see in the denominator assuming that we have standard form, which we do. So I can see that A squared is going to be nine and define A, I just need to take the square root on both sides giving me that A is the square root of nine, which is three. Now to find B, what we're called the B squared is equal to the second thing we see in the denominator in this case, it's four. So to find B I just take the square root on both sides here giving me that B is the square root of four, which is two. So we can see that our A values which since we have a vertical hyperbola, they're going to be up in. So we have our A values here and there, these also correspond with the vertices points and then I can see that our B points are going to be to the left two units and to the right two units. Now, from here, we can now draw this box that we talked about where I take a box that connects these four points together. Now, once you have this box drawn, you can draw the asymptotes through the corners of the box and to do this, well, one line is going to go through the corners and you do have to do this pretty delicately because it's gonna go through the corner of the box. And it's going to also cross through the center assuming that your hyperbole is at the origin like we have here and it's going to go out the other side through this corner. And this is the first Asymptote. Now the second Asymptote is going to go through the other direction. So this is how you can graph the asymptotes and define the equation for each of the asymptotes. Well, we could do that by recognizing that we have two lines that we formed. And the equation for a line as we've discussed in previous videos is Y is equal to MX plus B where M is the slope of the line and B is the Y intercept. Now, I can see both of these lines cross through the center, meaning that our Y intercept is going to be zero since it's right at the origin of the graph and our slope well, our slope is going to be rise over run. Now, I can see for this, this line that slopes up, our rise is going to be three units up. So we're going to have three as the rise and that's divided by the run, which I can see we have to run two units to the right 12. So that means our run is going to be two. So our slope is 3/2 and our Y intercept is zero. So plugging everything into this equation, what I'll have is that Y is equal to three halves times X. And that's going to be the slope and the equation for this line. Now, for the other line, we can find the equation for this Asymptote pretty quickly because notice that this is the same line we have here. It just slopes down now. So all we're going to do is change the slope to negative, meaning that we're going to have Y equals negative three halves X. So this is how you can find the equations for the asymptotes of your parabola. Now, this might have seemed like it was a long and tedious process because we first had to draw this graph, find the A and B points and then the lines and then the line equation. But you're going to find that there's actually a pattern that forms if you look at the equations that you got because notice that our rise was three. Well, that's just associated with what we have for A and notice that our run was two. Well, that's just B so it turns out that when dealing with vertical hyperbola, the equation for the asymptotes are always going to be assuming you're at the origin, they're always just going to be plus or minus A over B and we can see here that that's exactly what we got. We got a positive 3/2 X and A negative 3/2 X. So that is how you can quickly find the asymptotes for a vertical hyperbola. Now, the question becomes, what if we're dealing with a horizontal hyperbola instead? Well, it turns out for a horizontal hyperbola, all you need to do is flip the A and the B. So rather than having plus or minus A over B times X, we have plus or minus B over a times X, all we need to do is just flip this. So in this situation above, we had A is equal to three and B is equal to two. And it turns out for this hyperbole, it's the same situation. So we have the A is equal to three and B is equal to two. But since we are dealing with a horizontal hyperbole, at this time, the equations are going to be 2/3 X because we now flip the, the B into A and then it's going to be negative 2/3 X. So notice how we were able to find the Asym equation much more quick when we knew what this formula was now to make sure that we can actually graph this as well. Let's see what we end up with. So we see that A is equal to three and this time it's a horizontal hyperbola. So that means we need to go three units to the left and then three units to the right. And we can also see that our B values too. So we need to go up two units and down two units just like before we'll go ahead and connect these four points with a box. And then this will allow us to draw the asymptotes of our shape. One Asymptote is going to go through this corner and then the next Asymptote is going to go through the other corners of the box. And these would be the two asymptotes and notice that the slope is gonna be two thirds. So we can see that we rise two units and go to the right three units. So the rise is two and run is three. And then we can see here that the rise is going to be negative 23 for the other line. And then we have to go down two units which is negative two and then run three. So everything checks out with these asymptotes as well. So that is how you can find the asymptotes of a hyperbola. Hope you found this video helpful. Thanks for watching and let me know if you have any questions.

9

Problem

Problem

Find the equations for the asymptotes of the hyperbola $\frac{x^2}{64}-\frac{y^2}{100}=1$.

A

$y=\pm\frac45x$

B

$y=\pm\frac54x$

C

$y=\pm\frac{16}{25}x$

D

$y=\pm\frac{25}{16}x$

10

Problem

Problem

Find the equations for the asymptotes of the hyperbola $\frac{y^2}{16}-\frac{x^2}{9}=1$.

A

$y=\pm\frac{9}{16}x$

B

$y=\pm\frac{16}{9}x$

C

$y=\pm\frac34x$

D

$y=\pm\frac43x$

11

concept

Graph Hyperbolas at the Origin

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4m

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Hey, everyone, welcome back. So in the previous video, we talked about finding the asymptotes of a hyperbola. And in this video, we're going to see if we can take all this knowledge that we've learned about hyperbola and combine them into graphing hyperbola from just the equation. Now, this process has a lot of steps to it. But if you've watched the previous videos up to this point, hopefully all these steps will make a lot of sense and don't worry because I'm going to be explaining them all in this video. So without further ado, let's get right into this. Now, here, we have an example where we're asked to graph the hyperbole and identify the foi given this equation. Now, our first step should be to determine whether or not the hyperbola is horizontal or vertical. And I can do that by just looking at the equation that we have, we're called that the first term that you see in the denominator is the A squared term. And this A squared term is underneath X squared. Since we see an X that shows up first, that means this hyperbola will be oriented here on the X axis So because of this, we could say that we are dealing with a horizontal hyperbola. So that's the first step. Now, our next step should be to identify the vertices. And because we have a horizontal hyper, we only are going to care about the X coordinate for the vertices here. So we're going to use this version of the coordinates now to find out what A is well, a squared is equal to nine. So if A squared is equal to nine, that means A is going to be the square root of nine and the square root of nine is equal to three. So our vertices are going to be at 30 and at negative 30. And on our graph down here, well, three is between two and four. So we're going to have a point right there and then a point at negative three, which is right about there. Now step three here tells me that we need to find our B points and to find our B points, we need to look at this other value that we have in the denominator. This corresponds with B and we see that B squared is equal to 64 which means B is going to be the square root of 64 the square root of 64 turns out to be eight because eight times eight is 64. So that means that our B values are going to now follow this pattern. And by the way it's this pattern because since we had our vertices on the horizontal axis, our B values are going to be on the vertical axis. So we're going to have 08 because our B values eight and zero negative eight. So if I go ahead and draw the B values, one's going to be up here at eight and one's going to be down here at negative eight. Now, our fourth step is going to be to find the asymptotes and this is going to be a two step process. Step A is going to be to draw a box through the vertices and B points. So basically, with all the points that we found so far, I can connect these all with a rectangular box. So this is what the box would look like. And step B for drawing the asymptotes is to draw two lines through the corners of the box. So if I draw lines through the corner, one line is going to look like this and another line is going to look like that. So these are the two lines that we need to draw, which are also the asymptotes for our hyperbola. And our last step which is step five is going to be to draw branches at the vertices approaching the asymptotes, which are these lines. So if I started the vertices here, I can draw a curve that looks like this and that looks like that and notice how it approaches these lines that we see and another curve is going to be drawn like this and like that as well. So this is going to be what the hyperbola looks like. So now that we have our two branches or curves, we have successfully graphed this hyperbola. And as a last step, we're asked to also find the faux side of this hyperbola. Well, to do this, we're called that we need to use this equation which is C squared is equal to A squared plus B squared. Our C squared is going to equal A squared and we set up here that A squared is equal to nine. So it's going to be nine plus B squared, which we said was equal to 64. It's gonna be nine plus 64 9 plus 64 comes out to 73 which I'm going to write over here. So we're going to have the C squared is equal to 73. And then we're going to take the square root on both sides of this equation giving us that C is equal to the square root of 73. And the square root of 73 is approximately equal to 8.54. This is just the approximate value for the square root of 73. But to put in this value here, we can say that our FCI are going to be on the X axis because it's a horizontal hyperbole. So we're going to have the square root of 73 and zero. And then we're going to have negative square root of 73 and zero as our pesi. And since we said that the C value is approximately equal to 8.54 and it's on the X axis, that means one of our foresite is going to be right about there and another fori is going to be right about here. So these are the fori for the hyperbola and that is how you can graph hyperbola if you're only given the equation. So I hope you found this video helpful. Thanks for watching and let's move on.

12

example

Graphing Hyperbolas at the Origin Example 1

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3m

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Welcome back everyone. Let's give this example a try. So in this example, we are given this equation and we're asked to graph the following hyperbola. Now our first step is going to be to determine the orientation of the hyperbola. And I can see that our X squared comes first. And since the X comes first, that means we're going to be oriented on the X axis. So we are dealing with a horizontal hyperbola. Now, the next thing we need to do is find the vertices and since we know that it's gonna be along the X axis, we only care about this X coordinate. So we're going to use this these coordinates right here. Now, first off, I need to find a and recall that A squared is associated with the first thing we see in the denominator or 16, I can take the square root on both sides to get a by itself which a will be the square root of 16, which is four. So that means that our vertices are going to be at positive 40 and at negative 40. And if we go over to our graph, the vertices will be right about here and right about there. So that's how we can draw the vertices. Now, the next thing we need to do is find the B points and for the B points, now these are going to be opposite the direction of the vertices since since the vertices were left and right, the B points are going to be up and down. And so we're going to use this coordinate and we need to calculate B. So we can see that B squared is going to be associated with the second number in the denominator, which is 20. And this is kind of interesting because if we take the square root on both sides, the square root of 20 doesn't come out to a nice number like this one did. So we're gonna end up with B equal to the square root of 20. Now, you could reduce this down to two times the square root of five. But for these problems, I'm just going to keep them in this whole form because I think it's going to be a little easier to deal with. So B is gonna be square root of 20 the square root of 20 on a calculator comes out to around 4.47. So this is about 4.47. And so what that means is our B points are going to be up to right about there, about halfway between four and five. And then the negative one is going to be down here right around between negative five and negative four. So the B points are going to be at square root 20 0 and at negative square root 20 0. Now, the next step is going to be to find the asymptotes and to find the asymptotes, we first need to draw a box. Now I'll draw a box through the vertices and through the B points, which is going to look something like this. So this is what the box is going to look like. And then what we need to do from here is draw lines through the corners of the box. So one of the lines is going to go through this corner and another line is going to go through that corner. So we now have the asymptotes drawn for this hyperbola. And that was step four. And our fifth step is going to be to draw the branches of the vertices that approach the asymptotes. So one of these branches is going to start the vertices and go like this and it's going to approach the Asymptote that we drew and it's going to extend down here as well. And then the other branch that we need to draw is going to start this vertic and it's gonna go up like this and start to touch the approach, touching the Asymptote and it's going to go down as well there. So this right here is going to be the graph of our hyperbola and that was step five. Now, we also need to find the foi and to find the foi, well, since we're dealing with a horizontal hyperbola, the fite are going to be to the left and to the right. And what we need to do is use the equation C squared is equal to, to A squared plus B squared and A squared plus B squared. Well, we determined that A squared is what's under the denominator first, which is 16 and B squared is what's under this denominator, which is 20. So while that C squared is 16 plus 20 which is 36 and then what I need to do is take the square root on both sides, which will give AC and I go ahead and write C over on this side. C is gonna be the square root of 36 which is six. So that means that our F side are going to be at six zero and negative 60. And we're going to have to go a little bit off our graph to do this. But if we imagine that this line extends back here to negative 61 of the vertices would be right about there. And if this graph were to extend to the right at positive six, which you say is right about here, then that's where our other vertices is going to be. So these are going to be the two vertices as well as the graph. And that is going to be the answer for this example. So I hope you found this video helpful. Let me know if you have any questions.