### Video Transcript

Sketch the graph of the function π of π₯ is equal to one divided by π₯ plus two and then find the π¦-intercept.

In this question, weβre asked to sketch the graph of the function π of π₯ is equal to one divided by π₯ plus two. And since this function is the quotient of two polynomials, we can see this is a rational function. We then need to determine its π¦-intercept. Letβs start by determining how weβre going to sketch the graph of this function.

Letβs start by looking at our function. We can see that this function is very similar to the reciprocal function. In fact, the only difference is weβre adding two to our values of π₯. Weβre adding two to all of the input values. And we can recall if we add two to all of the input values of a function, then its curve is translated two units to the left. So, the graph of π¦ equals π of π₯ is a translation two units to the left of the graph of π¦ is equal to one divided by π₯. So, letβs start by recalling the shape of the graph of the reciprocal function, π¦ is equal to one over π₯.

We can recall that its shape look something like the following, where the π₯-axis is a horizontal asymptote of this curve and the π¦-axis is a vertical asymptote of the curve. So, its horizontal asymptote is π¦ is equal to zero, and its vertical asymptote is π₯ is equal to zero. We want to translate this graph two units to the left. And one way of doing this is to think what happens to the asymptotes. If we translate the horizontal asymptote two units to the left, it will still be the line π¦ is equal to zero. However, if we translate the vertical asymptote two units to the left, it will be the line π₯ is equal to negative two.

So, we can use this to sketch our curve. The π₯-axis will be a horizontal asymptote of our curve, and the line π₯ is equal to negative two will be a vertical asymptote. Finally, since weβre only translating the curve, the shape of the curve will be exactly the same as the reciprocal graph, π¦ is equal to one over π₯. This then gives us the following sketch of π¦ is equal to π of π₯, which is equal to one divided by π₯ plus two.

And itβs worth noting we canβt directly find the π¦-intercept by translating the graph two units to the left. This is because itβs hard to find the coordinates of the point which when translated two units to the left ends up on the π¦-axis. Instead, itβs just easier to directly find the π¦-intercept by using the given equation.

We can just recall that the π¦-intercept occurs when our value of π₯ is equal to zero. So, we substitute π₯ is equal to zero into our function π of π₯ to find its π¦-intercept. π evaluated at zero is equal to one divided by zero plus two, which gives us a value of one-half. So, we can add this onto our diagram. And itβs worth noting this agrees with our sketch since this is above the π₯-axis; the π¦-intercept is positive.

Therefore, by using our knowledge of transformations and by using our knowledge of how to sketch the graph π¦ is equal to one over π₯, we were able to sketch a graph of the function π of π₯ is equal the one divided by π₯ plus two. And we were also able to find that its π¦-intercept was the point zero, one-half.