QUESTION IMAGE
Question
the black graph is y = cos x. choose the equation for the red graph. y = cos(x) + 2 y = cos(x) - 1 y = cos(x - π) y = cos(x + π/2)
Step1: Analyze vertical shift
The black graph \(y = \cos x\) has a \(y -\)intercept at \((0,1)\). The red graph has a \(y -\)intercept at \((0,0)\).
For a function \(y=\cos(x)+k\), when \(x = 0\), \(y=\cos(0)+k=1 + k\). If \(y = 0\) at \(x = 0\), then \(1 + k=0\), so \(k=-1\).
Step2: Check other options
- For \(y=\cos(x)+2\), when \(x = 0\), \(y=\cos(0)+2=1 + 2=3\).
- For \(y=\cos(x-\pi)\), using the formula \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), \(y=\cos(x-\pi)=\cos x\cos\pi+\sin x\sin\pi=-\cos x\). When \(x = 0\), \(y=-\cos(0)=-1\).
- For \(y=\cos(x+\frac{\pi}{2})\), using the formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\), \(y=\cos(x+\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2}=-\sin x\). When \(x = 0\), \(y = 0\), but the shape of \(y =-\sin x\) is different from the red - graph ( \(y =-\sin x\) is a sine - type curve, while the red - graph is a cosine - type curve shifted vertically).
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\(y=\cos(x)-1\)