# Question Video: Sketching a Rational Function to Find the Asymptotes Mathematics • 10th Grade

Sketch the graph of the function π(π₯) = (1/(π₯ + 2)) β 1, and then find the horizontal asymptote of π(π₯). Find the vertical asymptote of π(π₯).

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### Video Transcript

Sketch the graph of the function π of π₯ is equal to one divided by π₯ plus two minus one and then find the horizontal asymptote of π of π₯. Find the vertical asymptote of π of π₯.

In this question, weβre given a function π of π₯ which is equal to one divided by π₯ plus two minus one. And we want to determine both the horizontal asymptote of the graph of this function and the vertical asymptote of the graph of this function. And weβre asked to do this by sketching the graph of the function. So to do this, letβs start by looking at the graph weβre asked to sketch, one divided by π₯ plus two minus one. And one way of doing this is to note π of π₯ is very similar to the standard reciprocal function π of π₯ is equal to one divided by π₯. In fact, we can transform the reciprocal function onto the function π of π₯ by using the following transformations.

We note that π of π₯ plus two minus one is equal to π of π₯. We find these values directly from the given function π of π₯. We can then recall if we add two to the input values of the function, then weβre going to translate the curve horizontally two units to the left. Similarly, if we subtract one from the outputs of the function, then weβre going to translate the curve vertically one unit down. Therefore, we can sketch the curve π¦ is equal to π of π₯ by sketching the curve π¦ is equal to one over π₯ and then applying these two transformations.

To do this, letβs start by recalling what the graph of the reciprocal function π¦ is equal to one over π₯ looks like. Its shape looks like the following. As the values of π₯ approach β, the outputs of the function approach zero from the positive direction. And as the values of π₯ approach negative β, the outputs of the function approach zero from the negative direction. The π₯-axis is a horizontal asymptote as shown. Similarly, as the values of π₯ approach zero from the positive direction, the outputs of the function approach β. And as the values of π₯ approach zero from the negative direction, the outputs of the function approach negative β.

The π¦-axis is also a vertical asymptote of the function. We want to translate the graph of this function two units left and one unit down. And the easiest way to do this is to first translate its asymptotes. Of course, vertically translating a vertical line wonβt change its position. Similarly, horizontally translating a horizontal line wonβt change its position. So we only need to translate the vertical asymptote two units left and the horizontal asymptote one unit down. Translating the line π₯ equals zero two units left gives us the line π₯ is equal to negative two and translating the line π¦ is equal to zero one unit down gives us the line π¦ is equal to negative one. These will be the asymptotes of our function π of π₯.

The general shape of the function will remain unchanged because weβre only translating it horizontally and vertically. It will still have the same shape as the reciprocal curve. However, before we sketch this curve, it can be useful to determine things like the π¦- and π₯-intercepts of the curve to determine its orientation on the plane. We can find the π¦-intercept by substituting π₯ is equal to zero into π of π₯. We see that π evaluated at zero is one-half minus one, which we can evaluate as negative one-half. So our curve π¦ is equal to π of π₯ must intercept the π¦-axis at negative one-half. In fact, this is enough to now sketch our curve since we know its general shape and π¦-intercept. This gives us the following sketch.

And itβs worth noting we could find the coordinates of the π₯-intercept by setting π of π₯ equal to zero and solving. We would get that the value of π₯ is negative one. This sketch now helps us justify why the horizontal asymptote and vertical asymptote of the function are the lines π¦ is equal to negative one and π₯ is equal to negative two, respectively. Since weβre just translating the reciprocal curve, weβre just translating its asymptotes. So translating the reciprocal curve one unit down will translate its horizontal asymptote one unit down. We will have the horizontal asymptote π¦ is equal to negative one. Similarly, translating the reciprocal curve two units left will translate its vertical asymptote two units left onto the line π₯ is equal to negative two.

Therefore, we were able to sketch the graph of the function π of π₯ is equal to one divided by π₯ plus two minus one and find its horizontal and vertical asymptotes. We showed its horizontal asymptote was the line π¦ is equal to negative one and its vertical asymptote was the line π₯ is equal to negative two.