hw: practice for test 2 : chapter 6 due before 2pm\nscore: 17/19 answered: 17/19\nquestion 4\nsketch a graph…

Explanation:

Step1: Identify amplitude

The general form of a cosine - function is $y = A\cos(Bx - C)+D$. For the function $f(x)=3\cos(\frac{1}{3}x)$, the amplitude $|A| = 3$. This means the maximum value of the function is $y = 3$ and the minimum value is $y=-3$.

Step2: Calculate the period

The period of a cosine function $y = A\cos(Bx - C)+D$ is given by $T=\frac{2\pi}{|B|}$. Here, $B=\frac{1}{3}$, so $T=\frac{2\pi}{\frac{1}{3}}=6\pi$.

Step3: Find key - points

When $x = 0$, $f(0)=3\cos(0)=3$. When $x=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=3\cos(\frac{1}{3}\times\frac{3\pi}{2})=3\cos(\frac{\pi}{2}) = 0$. When $x = 3\pi$, $f(3\pi)=3\cos(\frac{1}{3}\times3\pi)=3\cos(\pi)=-3$. When $x=\frac{9\pi}{2}$, $f(\frac{9\pi}{2})=3\cos(\frac{1}{3}\times\frac{9\pi}{2})=3\cos(\frac{3\pi}{2}) = 0$. When $x = 6\pi$, $f(6\pi)=3\cos(\frac{1}{3}\times6\pi)=3\cos(2\pi)=3$.

Step4: Sketch the graph

Plot the key - points $(0,3),(\frac{3\pi}{2},0),(3\pi,-3),(\frac{9\pi}{2},0),(6\pi,3)$ and connect them with a smooth cosine - shaped curve. The graph oscillates between $y = 3$ and $y=-3$ with a period of $6\pi$.

Answer:

Sketch a cosine - shaped curve with amplitude 3 and period $6\pi$ passing through the key - points $(0,3),(\frac{3\pi}{2},0),(3\pi,-3),(\frac{9\pi}{2},0),(6\pi,3)$.