Video Transcript
Consider the graph below. Which function does the plot in the
graph represent? (A) 𝑦 equals cos 𝑥. (B) 𝑦 equals two cos 𝑥. (C) 𝑦 equals negative sin 𝑥. (D) 𝑦 equals sin 𝑥. Or (E) 𝑦 equals negative cos
𝑥.
In this example, we have been given
a graph and need to decide which of the options represents it. Since all of the options include
sine or cosine or constant multiples of them, we should begin by reviewing the
properties that these functions have. Let us first recall the
𝑦-intercepts of sine and cosine, that is, their values when 𝑥 equals zero. We have cos zero is equal to one
and sin zero equals zero. Comparing this to the given graph,
we see that the 𝑦-intercept is at zero. This means that 𝑦 equals cos 𝑥,
option (A), cannot be correct and is not the function in the graph.
In fact, we can see that this also
applies to the other options that are multiples of cos 𝑥. This is because for option (B) two
cos zero is equal to two and for option (E) negative cos zero equals negative
one. That is to say, neither of the
𝑦-intercepts are equal to zero. And since we already know that the
𝑦-intercept is at zero, neither of these options can be correct.
We now have two remaining options:
𝑦 equals sin 𝑥 and 𝑦 equals negative sin 𝑥. Both options have their
𝑦-intercepts at zero, so we have to consider other properties of sin 𝑥. One thing in particular is the
behavior of sin 𝑥 as 𝑥 increases from zero. This can be seen by considering a
table of some of the values. From this, we can see that sin 𝑥
increases from zero to one as 𝑥 increases from zero to 𝜋 over two. Considering the graph shown, we can
see that the opposite occurs. As 𝑥 increases from zero to 𝜋
over two, the value of the graph goes to negative one.
On the other hand, if we consider
some of the first values of negative sin 𝑥, we can see the following. As we can see, the entries are the
same as the first table except they’ve been multiplied by negative one. As it happens, this behavior indeed
corresponds with what we can see in the graph. In particular, the value of the
graph at 𝜋 over two is indeed negative one. Therefore, option (C), 𝑦 is equal
to negative sin 𝑥, is the solution.
