Question

the black graph is y = cos x. choose the equation for the red graph. y = cos(x + π/2) y = cos(x - π/2) y = cos(x) + 1 y = cos(x) - 1

Explanation:

Step1: Recall the transformation rules for cosine functions

The general form of a cosine function is \(y = A\cos(B(x - C))+D\), where \(C\) represents the horizontal shift. If \(C>0\), the graph shifts to the right; if \(C < 0\), the graph shifts to the left.

Step2: Analyze the shift from \(y=\cos x\) (black graph with maximum at \((0,1)\)) to the red graph (maximum at \((-\frac{\pi}{2},1)\))

For the function \(y=\cos x\), when \(x = 0\), \(y = 1\). For the red - graph function \(y=\cos(x +\frac{\pi}{2})\), we use the formula \(y=\cos(u)\) where \(u=x+\frac{\pi}{2}\). When \(y = 1\), \(\cos(u)=1\), then \(u = 2k\pi,k\in\mathbb{Z}\). Let \(k = 0\), then \(x+\frac{\pi}{2}=0\), \(x=-\frac{\pi}{2}\).
For \(y=\cos(x-\frac{\pi}{2})\), when \(\cos(x - \frac{\pi}{2})=1\), \(x-\frac{\pi}{2}=2k\pi\), \(x=\frac{\pi}{2}+2k\pi\).
For \(y=\cos(x)+1\), the maximum value of \(y=\cos(x)+1\) is \(y = 2\) (since the maximum of \(\cos(x)\) is \(1\)) and it occurs when \(x = 2k\pi\).
For \(y=\cos(x)-1\), the maximum value of \(y=\cos(x)-1\) is \(y=0\) (since the maximum of \(\cos(x)\) is \(1\)).

Answer:

\(y=\cos(x+\frac{\pi}{2})\)