Question

sketch the graph of the function.
$f(x)=4+\cos(x)$

Explanation:

Step1: Analyze the parent function

The parent function is \(y = \cos(x)\). The general form of a cosine function is \(y = A\cos(Bx - C)+D\). For \(y=\cos(x)\), \(A = 1\), \(B = 1\), \(C = 0\), \(D=0\). The amplitude \(|A|=1\), period \(T=\frac{2\pi}{|B|}=2\pi\), phase - shift \(x=\frac{C}{B}=0\), and vertical - shift \(D = 0\). The key points of \(y = \cos(x)\) are: when \(x = 0\), \(y=\cos(0)=1\); when \(x=\frac{\pi}{2}\), \(y=\cos(\frac{\pi}{2}) = 0\); when \(x=\pi\), \(y=\cos(\pi)=-1\); when \(x=\frac{3\pi}{2}\), \(y=\cos(\frac{3\pi}{2})=0\); when \(x = 2\pi\), \(y=\cos(2\pi)=1\).

Step2: Analyze the transformation

For the function \(y = 4+\cos(x)\), using the form \(y = A\cos(Bx - C)+D\), here \(A = 1\), \(B = 1\), \(C = 0\), \(D = 4\). The amplitude remains \(|A| = 1\), the period \(T=\frac{2\pi}{|B|}=2\pi\) (no change in period and amplitude). The vertical - shift is \(D = 4\).

Step3: Find the key points of the transformed function

For \(y=4+\cos(x)\):

• When \(x = 0\), \(y=4+\cos(0)=4 + 1=5\)
• When \(x=\frac{\pi}{2}\), \(y=4+\cos(\frac{\pi}{2})=4+0 = 4\)
• When \(x=\pi\), \(y=4+\cos(\pi)=4-1 = 3\)
• When \(x=\frac{3\pi}{2}\), \(y=4+\cos(\frac{3\pi}{2})=4 + 0=4\)
• When \(x = 2\pi\), \(y=4+\cos(2\pi)=4 + 1=5\)

Plot these key points \((0,5)\), \((\frac{\pi}{2},4)\), \((\pi,3)\), \((\frac{3\pi}{2},4)\), \((2\pi,5)\) and connect them with a smooth cosine - like curve. The graph of \(y = 4+\cos(x)\) is the graph of \(y=\cos(x)\) shifted up by 4 units.

Answer:

Plot the points \((0,5)\), \((\frac{\pi}{2},4)\), \((\pi,3)\), \((\frac{3\pi}{2},4)\), \((2\pi,5)\) and draw a smooth cosine - shaped curve through them.