Question

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

Explanation:

Step1: Recall the cosine function transformation rules

The general form of a cosine function is \(y = A\cos(B(x - C))+D\), where \(A\) is the amplitude, \(B\) affects the period (\(T=\frac{2\pi}{|B|}\)), \(C\) is the phase - shift (horizontal shift, shift to the right if \(C>0\) and to the left if \(C < 0\)), and \(D\) is the vertical shift (shift up if \(D>0\) and down if \(D < 0\)).

Step2: Analyze the vertical shift

For \(y=\cos(x)\), when \(x = 0\), \(y=\cos(0)=1\). For the red graph, when \(x = 0\), \(y = 0\).
If we consider the function \(y=\cos(x)+k\), substituting \(x = 0\):

• For \(y=\cos(x)-1\), when \(x = 0\), \(y=\cos(0)-1=1 - 1=0\).
• For \(y=\cos(x)+2\), when \(x = 0\), \(y=\cos(0)+2=1 + 2=3\).

Step3: Analyze the phase - shift (optional for this problem)

For \(y=\cos(x-\pi)\), when \(x = 0\), \(y=\cos(-\pi)=-1\).
For \(y=\cos(x+\frac{\pi}{2})\), when \(x = 0\), \(y=\cos(\frac{\pi}{2})=0\), but we also need to check the general shape. The function \(y=\cos(x)\) has a maximum at \(x = 0\). The function \(y=\cos(x+\frac{\pi}{2})=-\sin(x)\) (using the identity \(\cos(A + B)=\cos A\cos B-\sin A\sin B\), \(\cos(x+\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2}=-\sin x\)) which is a sine - like function (not the same shape as a vertically - shifted cosine for our simple vertical - shift - based analysis here).

Answer:

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