### Video Transcript

Which of the following is the graph
of π¦ equals sin of π₯ minus 90?

We will begin by recalling some of
the key features of the main sine function graph. The function π of π₯ equals sin of
π₯ has a period length of two π radians or 360 degrees. All five graphs provided show the
scaling along the π₯-axis in 90-degree increments. The main sine function has an
amplitude of one, which is the distance from the midline to the peak or the trough
of the curve. All five graphs provided show sine
functions with an amplitude of one.

When graphing the main sine
function, we usually start with the point zero, zero then plot four more important
points between zero and 360. These coordinate points include 90,
one; 180, zero; 270, negative one; and 360, zero. This shape repeats itself every
period of 360 degrees to the left and to the right. The graph of π of π₯ minus 90 is
found through a transformation of the main sine function. We recall that a horizontal
translation or phase shift of negative π is represented by π of π₯ plus π.

In our example, the value of π is
negative 90. This means our horizontal shift of
negative π is negative negative 90, which equals positive 90 and translates all
points to the right 90 degrees. A horizontal translation does not
change the amplitude or the period of the sine function. So, we are looking for the five
points of the main sine function simply shifted 90 to the right, with no period or
amplitude change.

Letβs take a closer look at graph
(A). We can trace one period of the sine
function, starting with the point zero, one. We notice this graph has been
shifted vertically exactly one unit up. A vertical translation is
represented by adding a value to the function instead of adding a value directly to
π₯. This appears to be the graph of sin
of π₯ plus one. So, we eliminate option (A).

Moving on to graph (B), we notice
that the period length has been changed to 180, which is half the period we are
looking for. This sort of transformation is
referred to as a period change or horizontal stretch by a scale factor of one over
π, where the function is written as π of π times π₯. To take half the period, we use π
equals two. So, this is the function of sin of
two π₯. Thus, we eliminate option (B) and
move on to graph (C).

After highlighting one period of
graph (C), we notice it is the main sine function of π of π₯ equals sin of π₯. Since graph (C) shows no
transformations at all, we eliminate this option. Graph (D) has a period change. Instead of 360 degrees, it seems
this graph has a period of about 120 degrees. This graph also appears to have a
horizontal shift to the left 30. We are looking for a sine graph
with a period of 360 and a horizontal shift to the right 90. So, we eliminate option (D).

This only leaves us with graph (E)
to consider. We begin by tracing one period of
sine. This graph has been shifted to the
right 90 and still has a period length of 360 and an amplitude of one. Therefore, we have demonstrated
that option (E) is the graph of the function π¦ equals sin of π₯ minus 90.