Question

graph the function and state the amplitude, period and midline. graph the function and state the amplitude, period and midline. then give the maximum and minimum y - values and the corresponding x - values from the first full period starting at x = 0.

$f(x)=\frac{1}{2}\cos(x)$

Explanation:

Step1: Find the amplitude

For the function \(y = A\cos(x)\), the amplitude is \(|A|\). For \(y=\frac{1}{2}\cos(x)\), \(A = \frac{1}{2}\), so the amplitude \(a=\frac{1}{2}\).

Step2: Find the period

The general form of a cosine function is \(y = A\cos(Bx - C)+D\). For \(y=\frac{1}{2}\cos(x)\), \(B = 1\). The period of \(y=\cos(Bx)\) is \(T=\frac{2\pi}{|B|}\). Since \(B = 1\), the period \(T = 2\pi\).

Step3: Find the mid - line

The general form is \(y=A\cos(Bx - C)+D\). For \(y=\frac{1}{2}\cos(x)\), \(D = 0\). The mid - line is \(y = 0\).

Step4: Find the maximum and minimum values

The maximum value of \(y=\cos(x)\) is \(1\) when \(x = 2k\pi,k\in\mathbb{Z}\). For \(y=\frac{1}{2}\cos(x)\), when \(x = 0\) (in the first period \(x\in[0,2\pi]\)), \(y=\frac{1}{2}\cos(0)=\frac{1}{2}\).
The minimum value of \(y = \cos(x)\) is \(- 1\) when \(x=(2k + 1)\pi,k\in\mathbb{Z}\). For \(y=\frac{1}{2}\cos(x)\), when \(x=\pi\) (in the first period \(x\in[0,2\pi]\)), \(y=\frac{1}{2}\cos(\pi)=-\frac{1}{2}\).

Answer:

• Amplitude: \(\frac{1}{2}\)
• Period: \(2\pi\)
• Mid - line: \(y = 0\)
• Maximum \(y\) - value: \(\frac{1}{2}\) at \(x = 0\)
• Minimum \(y\) - value: \(-\frac{1}{2}\) at \(x=\pi\)