### Video Transcript

The graph shows π¦ equals π divided by π₯ minus three minus two. We can see that the intersection of its asymptotes is at three, negative two and the point 0.5, negative 1.5 and 1.5, negative one are below and above the graph, respectively. Determine the interval in which π lies.

We have two asymptotes: one at π₯ equals three and the other at π¦ equals negative two. The asymptote π₯ equals three is vertical and thatβs because if you take a look at the denominator, it can not equal zero. So the value that would make it equal zero would be three, so we can not have any points on π₯ equals three.

Now, the other is from the transformation of this minus two at the end of our equation; it shifts the entire graph down two. So instead of having the normal horizontal asymptote at π¦ equals zero, it will shift it down to π¦ equals negative two. So in order to figure out what π is actually equal to, we have some points that are above and below our graph. So if we know that a point on the graph is in between that, we can use inequalities to figure out what π would be equal to.

Specifically, weβre looking at this little pink section. So thinking about what π₯-values there could be, the π₯-value below the graph is 0.5 and the π₯-value above the graph is 1.5. So a value in between that, that would be easy to use, it would be one. So taking our equation, letβs plug in one for π₯. So now we have one minus three on the bottom, which is equal to negative two.

Now, as for the π¦-values, itβs between negative one and a half and negative one. So if we are determining an interval, we can create an interval using these values. So we want our equation, our graph- weβre looking at the points between negative one and a half and negative one, so greater than negative one and a half, but less than negative one.

So to solve, letβs first add two to both sides. Now, we need to multiply the sides by negative two. However, when you multiply by a negative, weβre gonna have to flip our sign. So π would be less than negative one, but greater than negative two. And a more common way to write that would be negative two is less than π is less than negative one.