Question 6.4.8: Graphing y = a cos b[(x - c)] Graph y=-cos [2(x + π/2)] over...

Use the steps in Example 7 for the procedure for graphing y=a \cos [b(x-c)] .

Step 1     Here y=-\cos \left[2\left(x+\frac{\pi}{2}\right)\right]=(\underset{\overset{\uparrow }{a = -1} }{-1} ) \cos \underset{\overset{\uparrow }{b = 2} }{2} \left[x-\underset{\overset{\uparrow }{c = -\frac{\pi}{2}} }{\left(-\frac{\pi}{2}\right)} \right]

amplitude =|a|=|-1|=1           period =\frac{2 \pi}{b}=\frac{2 \pi}{2}=\pi

phase shift =c=-\frac{\pi}{2}             Shift left because  c<0.

Step 2  The cycle begins at x=-\frac{\pi}{2}.

One cycle is graphed over the interval \left[-\frac{\pi}{2},-\frac{\pi}{2}+\pi\right]=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Step 3  \frac{1}{4}(\text { period })=\frac{1}{4}(\pi)=\frac{\pi}{4}

The x-coordinates of the five key points are

starting point =-\frac{\pi}{2},-\frac{\pi}{2}+\frac{\pi}{4}=-\frac{\pi}{4},-\frac{\pi}{2}+\frac{\pi}{2}=0,-\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{\pi}{4} ,

and  -\frac{\pi}{2}+\frac{2 \pi}{2}=\frac{\pi}{2}.

Step 4        Because  a=-1<0, we graph y=|-1| \cos \left[2\left(x+\frac{\pi}{2}\right)\right] by sketching one cycle, beginning at \left(-\frac{\pi}{2}, 1\right) and going through the points \left(-\frac{\pi}{4}, 0\right), (0,-1),\left(\frac{\pi}{4}, 0\right), \text { and }\left(\frac{\pi}{2}, 1\right). We then reflect this graph in the x-axis to obtain the graph of y=-\cos \left[2\left(x+\frac{\pi}{2}\right)\right].