Question

which function has the given properties below?
the domain is the set of all real numbers.
one x-intercept is \\((2\pi,0)\\).
the amplitude is 4.
the point \\((\frac{\pi}{2},-4)\\) is on the graph.
the y-intercept is \\((0, 0)\\).
\\(\circ\\) \\(y = -4\sin(x)\\)
\\(\circ\\) \\(y = -4\cos(x)\\)
\\(\circ\\) \\(y = 4\sin(x)\\)
\\(\circ\\) \\(y = 4\cos(x)\\)

Explanation:

Step1: Check Domain

All trigonometric functions \( y = A\sin(x) \) and \( y = A\cos(x) \) have domain \( \mathbb{R} \), so all options pass this.

Step2: Check x - intercept \((2\pi, 0)\)

• For \( y=-4\sin(x) \): \( -4\sin(2\pi)=-4\times0 = 0 \), so \((2\pi,0)\) is on it.
• For \( y = - 4\cos(x) \): \( -4\cos(2\pi)=-4\times1=-4

eq0 \), so \((2\pi,0)\) not on it.

• For \( y = 4\sin(x) \): \( 4\sin(2\pi)=4\times0 = 0 \), so \((2\pi,0)\) is on it.
• For \( y = 4\cos(x) \): \( 4\cos(2\pi)=4\times1 = 4

eq0 \), so \((2\pi,0)\) not on it. Eliminate \( y=-4\cos(x) \) and \( y = 4\cos(x) \).

Step3: Check Amplitude

Amplitude of \( y = A\sin(x) \) or \( y = A\cos(x) \) is \( |A| \). Both \( y=-4\sin(x) \) and \( y = 4\sin(x) \) have amplitude \( | - 4|=4 \) and \( |4| = 4 \), so they pass.

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

• For \( y=-4\sin(x) \): \( -4\sin(\frac{\pi}{2})=-4\times1=-4 \), so \( (\frac{\pi}{2},-4) \) is on it.
• For \( y = 4\sin(x) \): \( 4\sin(\frac{\pi}{2})=4\times1 = 4

eq - 4 \), so \( (\frac{\pi}{2},-4) \) not on it. Eliminate \( y = 4\sin(x) \).

Step5: Check y - intercept \((0,0)\)

For \( y=-4\sin(x) \): \( -4\sin(0)=-4\times0 = 0 \), so \((0,0)\) is on it.

Answer:

\( y = - 4\sin(x) \)