Question

the black graph is ( y = sin x ).
choose the equation for the red graph.
( y - 2 = sin x )
( y + 2 = sin x )
( y = sin(x + pi/2) )
( y = sin(x - pi/2) )

Explanation:

Step1: Recall the rules of function translation

For a function \(y = f(x)\), a horizontal translation is of the form \(y=f(x - h)\) (shift right by \(h\) units) or \(y = f(x+h)\) (shift left by \(h\) units), and a vertical translation is of the form \(y=f(x)+k\) (shift up by \(k\) units) or \(y=f(x)-k\) (shift down by \(k\) units).

Step2: Analyze the given graphs

The black graph is \(y = \sin x\) which has a zero - crossing at \((0,0)\). The red graph has a zero - crossing at \((\frac{\pi}{2},0)\).
Let \(y=\sin(x - h)\). When \(y = 0\), \(0=\sin(x - h)\), then \(x - h=n\pi\), \(n\in\mathbb{Z}\). For the basic function \(y = \sin x\), when \(n = 0\), \(x = 0\). For the red graph, when \(n = 0\), \(x=\frac{\pi}{2}\). Substituting into \(x - h=0\) (since \(\sin(0)=0\)), we get \(\frac{\pi}{2}-h = 0\), so \(h=\frac{\pi}{2}\).
If we consider the general form of a horizontal shift \(y=\sin(x - h)\), substituting \(h=\frac{\pi}{2}\) gives \(y=\sin(x-\frac{\pi}{2})\).
We can also check using the formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\). So, \(\sin(x-\frac{\pi}{2})=\sin x\cos\frac{\pi}{2}-\cos x\sin\frac{\pi}{2}=-\cos x\). But we can also use the transformation rule directly.
For vertical - shift equations \(y\pm2=\sin x\) (i.e., \(y=\sin x\mp2\)), the mid - line of \(y = \sin x\) is \(y = 0\). If we shift \(y=\sin x\) up by 2 units (\(y=\sin x + 2\)) or down by 2 units (\(y=\sin x-2\)), the zero - crossing of \(y=\sin x\) (where \(y = 0\)) would not be at \(x=\frac{\pi}{2}\) for these vertical - shift functions.

Answer:

\(y=\sin(x - \frac{\pi}{2})\)