Question

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

Explanation:

Step1: Recall the cosine function transformation rules

The general form of a cosine function is \(y = A\cos(B(x - C))+D\). Here, \(A\) is the amplitude, \(B\) affects the period (\(T=\frac{2\pi}{|B|}\)), \(C\) is the phase - shift (\(C>0\) shifts right, \(C < 0\) shifts left), and \(D\) is the vertical - shift. For \(y=\cos(x)\), when we consider \(y = \cos(x - C)\), the graph of \(y=\cos(x)\) is shifted \(C\) units to the right.

Step2: Use the point - substitution method

We know that for the black graph \(y = \cos(x)\), when \(x = 0\), \(y=\cos(0)=1\). For the red graph, when \(x = 0\), \(y=-1\).

• For \(y=\cos(x)+1\): When \(x = 0\), \(y=\cos(0)+1=1 + 1=2\).
• For \(y=\cos(x-\pi)\): Use the cosine subtraction formula \(\cos(A - B)=\cos A\cos B+\sin A\sin B\). So \(y=\cos(x-\pi)=\cos x\cos\pi+\sin x\sin\pi\). Since \(\cos\pi=-1\) and \(\sin\pi = 0\), then \(y=-\cos x\). When \(x = 0\), \(y=-\cos(0)=-1\).
• For \(y=\cos(x)-2\): When \(x = 0\), \(y=\cos(0)-2=1-2=-1\), but the shape of \(y = \cos(x)-2\) is a vertical shift of \(y=\cos(x)\) down by 2 units. The maximum value of \(y=\cos(x)\) is 1, and for \(y=\cos(x)-2\) the maximum value is \(1 - 2=-1\) and the minimum value is \(-1-2=-3\).
• For \(y=\cos(x+\frac{\pi}{2})\): Use the cosine addition formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\). So \(y=\cos(x+\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2}\). Since \(\cos\frac{\pi}{2}=0\) and \(\sin\frac{\pi}{2}=1\), then \(y=-\sin x\). When \(x = 0\), \(y=-\sin(0)=0\).

Answer:

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