### Video Transcript

Determine the function represented by the graph shown. Is it A) 𝑦 equals two to the 𝑥, B) 𝑦 equals negative two to the negative 𝑥, C) 𝑦
equals negative two to the 𝑥, or D) 𝑦 equals two to the negative 𝑥?

We’re going to examine the graph that we’ve been given and use features of that graph
to eliminate the three incorrect options, leaving us with the correct option, which
will be our answer. One thing we notice from the graph is that the 𝑦-intercept is negative one. And this means that when we input zero into the function that this graph represents,
we should get an output of negative one.

So let’s go through our options one by one. Starting with option A, 𝑦 equals two to the 𝑥: when 𝑥 is zero, 𝑦 is two to the
zero, which is one. Remember if you raise any number to the power of zero, you get one. So the graph of this function has a 𝑦-intercept at one and not negative one. And so we can eliminate this option.

We have eliminated option A. Let’s move on to option B. When 𝑥 is zero, 𝑦 is equal to negative two to the power of negative zero. Negative zero is just zero. And so we have negative two to the power of zero. And two to the power of zero is one. So 𝑦 is negative one.

We have to be slightly careful here about the order of operations. We need to perform the exponentiation before the negation. And had we got our order of operations wrong, we would have ended up with a value of
one rather than negative one. But getting the order of operations right, we see that the 𝑦-intercept of the graph
of this function, which of course is the value you get when you substitute 𝑥 equals
zero in, is negative one.

This is of course the 𝑦-intercept in the graph that we were given. And so B is certainly a possibility for the function represented. This doesn’t mean that the answer is option B for sure. It just means that we can’t eliminate option B as a possibility.

So we move on to option C. And again, we try 𝑥 equals zero to find the 𝑦-intercept of the graph of this
function. And much like in option B, being careful of the order of operations, we see that the
𝑦-intercept of this function is negative one.

C remains a possibility. And we move on to option D. Substituting 𝑥 equals zero, we find that the 𝑦-intercept of the graph of this
function is one. The graph that we’re given has a 𝑦-intercept of negative one. And hence, the function in option D isn’t the function represented and we can
eliminate it.

By considering the 𝑦-intercept of the graph that we have, we’ve managed to eliminate
options A and D. And so the only two possibilities remaining are options B and C. And we’re going to have to decide between them somehow. We could do this by substituting another value of 𝑥 in.

Looking at our graph, we can see that as 𝑥 increases, the graph of the function gets
closer and closer to the 𝑥-axis. Let’s pick a value of 𝑥 — say six — and think about what’s the corresponding
𝑦-values are for our two options B and C.

For option B, we substitute in six. And we get that 𝑦 is then negative two to the power of negative six, which we can
rewrite as negative one over two to the six. And as two to the six is 64, this is negative one over 64. Looking at our graph, this is a plausible value of 𝑦 when 𝑥 is six. Negative one over 64 is less than zero and so this point will lie below the
𝑥-axis. But it’s also very close to zero. So the point will lie very close to the 𝑥-axis, which is what we see.

Doing the same for option C, we get 𝑦 equals negative two to the six. And as two to the six is 64, we get that 𝑦 is equal to negative 64. This doesn’t seem to match up with what we see on our graph. A 𝑦-value of negative 64 would be so far away from the 𝑥-axis that it wouldn’t even
appear on our graph.

Whereas the value of negative one over 64 was plausible for option B, for option C
the value of negative 64 is not plausible. And so we eliminate option C. This means that our answer must be the only option that we haven’t eliminated yet,
option B.

It’s worth noting that there wasn’t anything special about six. It’s just that as 𝑥 gets greater and greater, negative two to the power of negative
𝑥, which was our function in option B, gets closer and closer to zero. And so its graph gets closer and closer to the 𝑥-axis, whereas negative two to the
power of 𝑥 which was our function in option C as 𝑥 increases gets further and
further away from zero and so its graph gets further and further away from the
𝑥-axis.