Question

consider the given functions.
$f(x)=31cos(x+\frac{pi}{2})+24$
$k(x)=31cos(x-\frac{3pi}{4})+24$
what is the transformation of function $f(x)$ to function $k(x)$?
o a shift left $\frac{pi}{4}$ units
o a shift left $\frac{5pi}{4}$ units
o a shift right $\frac{pi}{4}$ units
o a shift right $\frac{5pi}{4}$ units

Explanation:

Step1: Recall the horizontal shift formula

For a function \(y = A\cos(B(x - h))+C\), the horizontal shift is \(h\) units. If \(h>0\), shift right; if \(h < 0\), shift left.
The original function \(f(x)=31\cos(x+\frac{\pi}{2}) + 24=31\cos(x-(-\frac{\pi}{2}))+24\), and the new function \(k(x)=31\cos(x-\frac{3\pi}{4})+24\)

Step2: Calculate the shift amount

We want to find \(h\) such that \(x-(-\frac{\pi}{2})+h=x - \frac{3\pi}{4}\). Solving for \(h\):

$$h=-\frac{3\pi}{4}-\frac{\pi}{2}=-\frac{3\pi}{4}-\frac{2\pi}{4}=-\frac{5\pi}{4}$$

Another way: Let \(u=x+\frac{\pi}{2}\) for \(f(x)\) and \(v=x-\frac{3\pi}{4}\) for \(k(x)\). We want to find the change in \(x\) such that \(v = u+h\). So \(x-\frac{3\pi}{4}=(x+\frac{\pi}{2})+h\), then \(h=-\frac{3\pi}{4}-\frac{\pi}{2}=-\frac{5\pi}{4}\) (this is wrong, correct formula: if \(y = f(x)\) and \(y = f(x - h)\), for \(f(x)=31\cos(x+\frac{\pi}{2})+24\) and \(k(x)=31\cos(x-\frac{3\pi}{4})+24\), we set \(x+\frac{\pi}{2}=x - \frac{3\pi}{4}-h\) (no, correct: for \(y = A\cos(B(x - h_1))+C\) and \(y = A\cos(B(x - h_2))+C\), the shift is \(h_2 - h_1\). Here \(h_1=-\frac{\pi}{2}\), \(h_2=\frac{3\pi}{4}\), so shift \(h = h_2 - h_1=\frac{3\pi}{4}-(-\frac{\pi}{2})=\frac{3\pi + 2\pi}{4}=\frac{5\pi}{4}\) to the right.

Answer:

a shift right \(\frac{5\pi}{4}\) units