the function ( g(x) ) represents ( f(x)=9 cos left(x-\frac{pi}{2}\right)+3 ) after translating (…

the function ( g(x) ) represents ( f(x)=9 cos left(x-\frac{pi}{2}\right)+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}\right)-1 ) ( \bigcirc g(x)=9 cos left(x-\frac{pi}{3}\right)+7 ) ( \bigcirc g(x)=9 cos left(x-\frac{2 pi}{3}\right)-1 ) ( \bigcirc g(x)=9 cos left(x-\frac{2 pi}{3}\right)+7 )

the function ( g(x) ) represents ( f(x)=9 cos left(x-\frac{pi}{2}\right)+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}\right)-1 ) ( \bigcirc g(x)=9 cos left(x-\frac{pi}{3}\right)+7 ) ( \bigcirc g(x)=9 cos left(x-\frac{2 pi}{3}\right)-1 ) ( \bigcirc g(x)=9 cos left(x-\frac{2 pi}{3}\right)+7 )

Answer

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)