Answer
amplitude=$3$
period =$\displaystyle \frac{\pi}{3}$
graph (black):
.
Work Step by Step
See p. 424
For
$y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$,
$(k>0)$
Amplitude: $|a|$, Period: $\displaystyle \frac{2\pi}{k}$,
Horizontal shift: $b$
------------------
$y=-3\cos 6x$
$a=-3 , k=6, b=0$
So,
amplitude=$|-3|=3$
period =$\displaystyle \frac{2\pi}{6}=\frac{\pi}{3}$
To graph,
begin with $f(x)=\cos x,$ (red, dashed)
horizontally compress by factor $6$ and
reflect across the x-axis (black, solid line).
$g(0)=-3$
$g(\displaystyle \frac{\pi}{12})=-3\cos(\frac{\pi}{2})=0$
$g(\displaystyle \frac{\pi}{6})=-3\cos(\pi)=3$
$g(\displaystyle \frac{\pi}{4})=-3\cos(\frac{3\pi}{2})=0$
$g(\displaystyle \frac{\pi}{3})=-3\cos(2\pi)=-3.$
