Question

sketch at least one cycle of the graph of the function $y = 4-cosleft(\frac{x}{3}
ight)$. determine the period, phase shift, and range of the function. identify five points on the graph. use the graphing tool to graph the equation. (for any answer boxes shown with the grapher, type an exact answer. type the word pi to insert the symbol $pi$)

Explanation:

Step1: Find the period

For the function \(y = A\cos(Bx - C)+D\), the period is \(T=\frac{2\pi}{|B|}\).
In \(y = 4-\cos(\frac{x}{3})\), \(B=\frac{1}{3}\).
So, \(T=\frac{2\pi}{\frac{1}{3}} = 6\pi\).

Step2: Find the phase - shift

The phase - shift formula is \(h=\frac{C}{B}\).
In \(y = 4-\cos(\frac{x}{3})\), \(C = 0\) and \(B=\frac{1}{3}\).
So, the phase - shift \(h = 0\).

Step3: Find the range

The range of \(y=\cos(x)\) is \([- 1,1]\).
For \(y = 4-\cos(\frac{x}{3})\), when \(\cos(\frac{x}{3})=-1\), \(y = 4-(-1)=5\); when \(\cos(\frac{x}{3}) = 1\), \(y=4 - 1=3\).
So, the range is \([3,5]\).

Step4: Find five points

Let \(\frac{x}{3}=0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi\).

• When \(\frac{x}{3}=0\) (i.e., \(x = 0\)), \(y=4-\cos(0)=4 - 1=3\). The point is \((0,3)\).
• When \(\frac{x}{3}=\frac{\pi}{2}\) (i.e., \(x=\frac{3\pi}{2}\)), \(y=4-\cos(\frac{\pi}{2})=4-0 = 4\). The point is \((\frac{3\pi}{2},4)\).
• When \(\frac{x}{3}=\pi\) (i.e., \(x = 3\pi\)), \(y=4-\cos(\pi)=4+1 = 5\). The point is \((3\pi,5)\).
• When \(\frac{x}{3}=\frac{3\pi}{2}\) (i.e., \(x=\frac{9\pi}{2}\)), \(y=4-\cos(\frac{3\pi}{2})=4-0 = 4\). The point is \((\frac{9\pi}{2},4)\).
• When \(\frac{x}{3}=2\pi\) (i.e., \(x = 6\pi\)), \(y=4-\cos(2\pi)=4 - 1=3\). The point is \((6\pi,3)\).

Answer:

Period: \(6\pi\); Phase - shift: \(0\); Range: \([3,5]\); Five points: \((0,3)\), \((\frac{3\pi}{2},4)\), \((3\pi,5)\), \((\frac{9\pi}{2},4)\), \((6\pi,3)\)