Question

the function ( g(x) ) represents ( f(x)=9 cos left(x-\frac{pi}{2}
ight)+3 ) after translating ( \frac{pi}{6} ) units left and 4 units up. which equation represents ( g(x) )? ( \bigcirc g(x)=9 cos left(x-\frac{pi}{3}
ight)-1 ) ( \bigcirc g(x)=9 cos left(x-\frac{pi}{3}
ight)+7 ) ( \bigcirc g(x)=9 cos left(x-\frac{2 pi}{3}
ight)-1 ) ( \bigcirc g(x)=9 cos left(x-\frac{2 pi}{3}
ight)+7 )

Explanation:

Step1: Horizontal translation

For a function \(y = f(x)\), translating \(a\) units to the left gives \(y=f(x + a)\).
Here, \(f(x)=9\cos(x-\frac{\pi}{2})+3\), and \(a = \frac{\pi}{6}\).
So, \(f(x+\frac{\pi}{6})=9\cos((x+\frac{\pi}{6})-\frac{\pi}{2})+3=9\cos(x-\frac{\pi}{2}+\frac{\pi}{6})+3=9\cos(x-\frac{3\pi - \pi}{6})+3=9\cos(x-\frac{\pi}{3})+3\).

Step2: Vertical translation

For a function \(y = f(x)\), translating \(b\) units up gives \(y=f(x)+b\).
Here, \(b = 4\), so \(g(x)=9\cos(x-\frac{\pi}{3})+3 + 4\).

Answer:

\(g(x)=9\cos(x-\frac{\pi}{3})+7\) (the second option)