Question

graph the function and state the amplitude, period and midline. state the maximum and minimum y-values and the corresponding x-values on one period for x ≥ 0.

f(x) = -5 sin(x)

Explanation:

Step1: Find the amplitude

The general form of a sine function is \(y = A\sin(Bx - C)+D\). For the function \(f(x)=- 5\sin(x)\), \(A=-5\). The amplitude is \(|A|\), so \(| - 5|=5\).

Step2: Find the period

For the function \(y = A\sin(Bx - C)+D\), the period is \(T=\frac{2\pi}{|B|}\). Here \(B = 1\), so \(T=\frac{2\pi}{|1|}=2\pi\).

Step3: Find the mid - line

For the function \(y = A\sin(Bx - C)+D\), the mid - line is \(y = D\). Here \(D = 0\), so the mid - line is \(y = 0\).

Step4: Find the maximum and minimum values

We know that \(\sin(x)\) has a range of \([-1,1]\). For \(y=-5\sin(x)\), when \(\sin(x)=-1\), \(y=-5\times(-1) = 5\). When \(\sin(x)=1\), \(y=-5\times1=-5\).

• To find the \(x\) - values:
• Set \(\sin(x)=-1\). We know that \(\sin(x)=-1\) when \(x=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}\). For \(x\geq0\) in one period (\(0\leq x\leq2\pi\)), \(x = \frac{3\pi}{2}\) gives \(y = 5\).
• Set \(\sin(x)=1\). We know that \(\sin(x)=1\) when \(x=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}\). For \(x\geq0\) in one period (\(0\leq x\leq2\pi\)), \(x=\frac{\pi}{2}\) gives \(y=-5\).

Answer:

• Amplitude: \(5\)
• Period: \(2\pi\)
• Mid - line: \(y = 0\)
• Maximum \(y\) - value: \(y = 5\) at \(x=\frac{3\pi}{2}\)
• Minimum \(y\) - value: \(y=-5\) at \(x=\frac{\pi}{2}\)