which function has the given properties below? the domain is the set of all real numbers. one x - intercept…

Explanation:

Step1: Check domain

All given functions (y = - 4\sin(x)), (y=-4\cos(x)), (y = 4\sin(x)), (y = 4\cos(x)) have domain (x\in(-\infty,\infty)) (set of all real - numbers), so this property doesn't help to eliminate any option for now.

Step2: Check x - intercept

For (y = A\sin(x)) and (y = A\cos(x)), the x - intercepts of (y=\sin(x)) are (x = k\pi,k\in\mathbb{Z}), and the x - intercepts of (y=\cos(x)) are (x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}). When (x = 2\pi), (\sin(2\pi)=0) and (\cos(2\pi)=1). So functions (y=-4\cos(x)) and (y = 4\cos(x)) can be eliminated since they do not have an x - intercept at ((2\pi,0)).

Step3: Check amplitude

The general form of a sine and cosine function is (y = A\sin(x)) or (y=A\cos(x)), where (|A|) is the amplitude. For (y=-4\sin(x)) and (y = 4\sin(x)), (|A| = 4), which satisfies the amplitude condition.

Step4: Check the point ((\frac{\pi}{2},-4))

Substitute (x=\frac{\pi}{2}) into (y = 4\sin(x)), we get (y = 4\sin(\frac{\pi}{2})=4). Substitute (x=\frac{\pi}{2}) into (y=-4\sin(x)), we get (y=-4\sin(\frac{\pi}{2})=-4).

Step5: Check y - intercept

For (y=-4\sin(x)), when (x = 0), (y=-4\sin(0)=0). For (y = 4\sin(x)), when (x = 0), (y = 4\sin(0)=0). But from step 4, the function that passes through the point ((\frac{\pi}{2},-4)) is (y=-4\sin(x)).

Answer:

(y=-4\sin(x))