Question

Specify a sequence of transformations that will yield each graph of \(h\) from the graph of the function \(f(x)=\sin x\). (a) \(h(x)=\sin \left(x+\frac{\pi}{2}\right)+1\) (b) \(h(x)=-\sin (x-1)\)

Short answer

For \(h(x)=\sin \left(x+\frac{\pi}{2}\right)+1\), the graph of \(f(x) = \sin x\) is shifted left \(\frac{\pi}{2}\) units and then upward 1 unit. For \(h(x) = -\sin(x - 1)\), the graph of \(f(x) = \sin x\) is reflected over the x-axis and then shifted right 1 unit.

Step-by-step solution

Identify transformations of h(x) from f(x) for the (a) function

First, let's analyze \(h(x)=\sin \left(x+\frac{\pi}{2}\right)+1\). Here two transformations are applied to \(f(x)\): (1) a shift to the left by \( \frac{\pi}{2} \) units, which is evidenced by \(x + \frac{\pi}{2}\) inside the parentheses, and (2) a shift upwards by 1 unit, which is noted by the +1 outside the sine function.

Identify transformations of h(x) from f(x) for the (b) function

Next, for \(h(x)=-\sin (x-1)\), there are also two transformations from \(f(x)\): (1) a reflection along the x-axis, shown by the negative sign in front of the sine function, and (2) a shift to the right by 1 unit, which is demonstrated by \(x - 1\) inside the parentheses.

Describe sequence of transformations

Now that transformations for each function have been identified, we can describe them in a sequence. For (a), firstly, move the function \( \frac{\pi}{2} \) units to the left and then, shift it upwards by 1 unit. For (b), firstly reflect the function along the x-axis and then, shift it to the right by 1 unit.