### Video Transcript

Which of the following functions is graphed in the given figure? Is it A: 𝑦 equals negative 𝑥 plus one squared times 𝑥 plus seven, B: 𝑦 equals 𝑥 minus one squared times 𝑥 minus seven, C: 𝑦 equals 𝑥 minus seven squared times 𝑥 minus one, D: 𝑦 equals negative 𝑥 plus seven squared times 𝑥 plus one, or is it E: 𝑦 equals 𝑥 plus seven squared times 𝑥 plus one?

We can see that the graph has 𝑥-intercepts at one and seven. This means that all functions must have the form 𝑦 equals some number 𝐴 times 𝑥 minus one to the power of some number 𝑛 times 𝑥 minus seven to the power of some number 𝑚. We can eliminate several of the options which don’t have this form. We have eliminated A, D, and E. And so we’re left to choose between options B and C.

How are we going to choose between these two options? Well in option B, the value of 𝑛 is two and the value of 𝑚 is one. But in option C, it’s the other way around. The value of 𝑚 is two and the value of 𝑛 is one. On the graph, you can see that these two 𝑥-intercepts are quite different. The 𝑥-intercept at 𝑥 equals one has the graph crossing the 𝑥-axis, whereas the 𝑥-intercept at 𝑥 equals seven has the graph touching but not crossing the 𝑥-axis.

The fact that the graph touches the 𝑥-axis at 𝑥 equals seven tells us that there must be a repeated root here. And so the exponent of 𝑥 minus seven, 𝑚, cannot be equal to one. Actually, it turns up we can say more than that. We actually can say that 𝑚 is even, but we don’t need that for this problem. We already have enough information to eliminate option B, because 𝑚 there was one. And we’re left only with option C: 𝑦 equals 𝑥 minus seven squared times 𝑥 minus one.

In this question, we didn’t need to know anything about the value of 𝐴, but we can actually see from the graph that 𝐴 would be positive which makes sense given our answer where 𝐴 is one. Had 𝐴 been negative, then the graph of the function would have moved from the second quadrant to the fourth quadrant, as 𝑥 increases rather than from the third quadrant to the first quadrant as it does. If 𝐴 had been negative one, then the graph you would have seen would be the graph that we have reflected in the 𝑥-axis. And I’ve sketched in that graph using a dotted line. If 𝑦 equals negative 𝑥 minus seven squared times 𝑥 minus one had been one of the options, then we could have eliminated it by expanding that option and seeing that that implied that the 𝑦-intercept was positive whereas clearly on our graph 𝑦-intercept is negative.