### Video Transcript

What function is represented in the figure below?

We have the graph of some function that we want to know what function that that graph is of. Letβs take a look at the features of this graph. Thereβs a vertical asymptote, the π¦-axis, and a horizontal asymptote with equation π¦ equals negative three. Can we think of a function whose graph has very similar features? Well, here is a graph of the reciprocal function π of π₯ equals one over π₯. It doesnβt match the graph in our question exactly. But thereβs some strong family resemblance. Weβre going to use graph transformations to transform this reciprocal graph into the graph in our question.

Like the graph in the question, the reciprocal graph has a vertical asymptote and a horizontal asymptote. But unlike the graph in our question, which to the left to the π¦-axis approaches the vertical asymptote by going upward, the reciprocal graph approaches this asymptote from the left by going downward. The graphs are different to the right of the π¦-axis too. The function represented on the left is increasing when π₯ is greater than zero. And so its graph comes up from below. Whereas, the reciprocal function is decreasing when π₯ is greater than zero. And so its graph comes down from above to the right of the π¦-axis.

How can we transform the reciprocal graph on the right to make it more like the graph on the left in our question? One thing we could do is to reflect this graph in the π₯-axis. Hereβs a quick sketch of what this reflected graph looks like. We can see that the reflected graph does the right thing to the left and right of the π¦-axis, going up to approach the vertical asymptote to the left of the π¦-axis and coming up from below to the right of it. What is the function associated with this reflected graph? If you reflect the graph of π of π₯ in the π₯-axis, you get to the graph of negative π of π₯. As we started off with the graph of π of π₯ equals one over π₯, the reflected graph represents π of π₯ equals negative one over π₯. Before tidying up and moving to the next step, weβll just note that instead of reflecting in the π₯-axis, we could have chosen to reflect in the π¦-axis. Looking at the symmetry of this picture, you might guess that reflecting in the π¦-axis gives you the graph of π of π₯ equals negative one over π₯ as well. And youβd be right. Reflection in the π¦-axis takes the graph of π of π₯ to π of negative π₯. As we started with π of π₯ equals one over π₯, weβd end up with π of π₯ equals one over negative π₯ which can be rewritten as π of π₯ equals negative one over π₯.

Now that weβve tidied up, we can more easily compare the graph of our improved guess, π of π₯ equals negative one over π₯, with the graph in our question. The vertical asymptote matches the graph in the question. Itβs the π¦-axis with equation that π₯ equals zero. However, the horizontal asymptote is not where it should be. It is the π₯-axis with equation that π¦ equals zero and not the line with equation that π¦ equals negative three, as in the graph in our question. We need to transform this graph by translating it down by three units to get it to match the graph in our question.

We can draw a quick sketch of what this translated graph would look like. And it seems to match up well. The question is, what function does this translated graph represent? Translating the graph of π of π₯ by π units in the positive π¦-direction takes this graph to the graph of π of π₯ plus π. Weβre translating it down by three units. So thatβs three units in the negative π¦-direction or negative three units in the positive π¦-direction. And so we subtract three from our function getting π of π₯ equals negative one over π₯ minus three. And perhaps by trying a few points on the graph in our question, we can convince ourselves that this really is the graph of the function π of π₯ equals negative one over π₯ minus three. There is no additional scaling or stretching required.

Letβs just recap how we solved this question. We started off with the reciprocal graph. That is the graph of the function π of π₯ equals one over π₯ which we thought was a good starting point because it had a vertical and horizontal asymptote like the graph in our question. Certainly, it was a better starting point than a straight line graph or a parabola. And we then applied successive graph transformations. First, a reflection and then a translation in the π¦-direction until we had a perfect match. One transformation fact that we didnβt have to use is that translation by π units in the positive π₯-direction takes the graph of π of π₯ to the graph of π of π₯ minus π. We would have had to use this fact had the vertical asymptote of the graph in our question not been the π¦-axis.