Question

1. $y = cos x + 1$, $-2pi leq x leq 2pi$

Explanation:

Step1: Determine the key features of \(y = \cos x\)

The function \(y=\cos x\) has an amplitude \(A = 1\), period \(T=2\pi\), range \([- 1,1]\), and key points \((0,1),(\frac{\pi}{2},0),(\pi,-1),(\frac{3\pi}{2},0),(2\pi,1)\) and their counterparts for negative \(x\) values (e.g., \((-\frac{\pi}{2},0),(-\pi,-1)\)).

Step2: Analyze the transformation \(y=\cos x + 1\)

The transformation \(y = f(x)+k\) (where \(f(x)=\cos x\) and \(k = 1\)) is a vertical shift. The amplitude \(A = 1\) and period \(T = 2\pi\) remain unchanged. The range becomes \([0,2]\) (since \(y=\cos x+1\), when \(\cos x=-1,y = 0\) and when \(\cos x = 1,y=2\)). The key - point transformation: if \((x,y)\) is on \(y=\cos x\), then \((x,y + 1)\) is on \(y=\cos x+1\). So, the key points are \((0,2),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2},1),(2\pi,2)\) and \((-\frac{\pi}{2},1),(-\pi,0),(-\frac{3\pi}{2},1)\).

Step3: Plot the function

• Mark the \(x\) - axis from \(-2\pi\) to \(2\pi\) and the \(y\) - axis from \(0\) to \(2\).
• Plot the key points \((-2\pi,2),(-\frac{3\pi}{2},1),(-\pi,0),(-\frac{\pi}{2},1),(0,2),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2},1),(2\pi,2)\).
• Connect the points with a smooth curve that has the shape of a cosine wave (a smooth, oscillating curve) within the interval \(-2\pi\leq x\leq2\pi\).

Answer:

Plot the function \(y = \cos x+1\) over the interval \(-2\pi\leq x\leq2\pi\) by first identifying the key points \((-2\pi,2),(-\frac{3\pi}{2},1),(-\pi,0),(-\frac{\pi}{2},1),(0,2),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2},1),(2\pi,2)\) (derived from the vertical shift of the \(y = \cos x\) key points) and then connecting them with a smooth, oscillating curve (characteristic of a cosine - type function) within the given \(x\) - interval.