Question

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

Explanation:

Step1: Recall the transformation rules

For a function \(y = A\cos(B(x - C))+D\), \(C\) is the horizontal shift (right if \(C>0\), left if \(C < 0\)), \(D\) is the vertical shift (up if \(D>0\), down if \(D < 0\)).

Step2: Check the vertical shift

The original function \(y=\cos x\) has a maximum value of \(1\) (at \(x = 0\)). The red - graph has a maximum value of \(2\) at \(x=\pi\). The vertical shift \(D\) is such that if \(y=\cos(x)+D\), when \(y = 2\) (maximum of red - graph) and maximum of \(y = \cos x\) is \(1\), then \(D=1\) (since \(y=\cos(x)+1\) has a maximum of \(1 + 1=2\)).

Step3: Check the horizontal shift

We know that the maximum of \(y=\cos x\) is at \(x = 0\), and the maximum of the red - graph is at \(x=\pi\). Using the formula \(y=\cos(x - C)+1\). When \(x=\pi\), \(y = 2\). Substitute into \(y=\cos(x - C)+1\), we get \(2=\cos(\pi - C)+1\), then \(\cos(\pi - C)=1\). We know that \(\cos\theta=1\) when \(\theta = 2k\pi,k\in\mathbb{Z}\). Let \(k = 0\), then \(\pi - C=0\), so \(C=\pi\).

Answer:

\(y=\cos(x-\pi)+1\)