Question

graph $y = 6\cos\left(\frac{\pi}{4}x\
ight)-2$ in the interactive widget. note that one moveable point always defines an extremum point in the graph and the other point always defines a neighbouring intersection with the midline.

Explanation:

Step1: Determine the amplitude

The general form of a cosine function is \(y = A\cos(Bx)+C\). For \(y = 6\cos(\frac{\pi}{4}x)-2\), the amplitude \(|A| = 6\).

Step2: Determine the mid - line

The mid - line is given by \(y = C\). Here, \(C=-2\).

Step3: Determine the period

The period of the function \(y = A\cos(Bx)+C\) is \(T=\frac{2\pi}{|B|}\). Since \(B = \frac{\pi}{4}\), then \(T=\frac{2\pi}{\frac{\pi}{4}}=8\).

Step4: Find the extremum points

The maximum value of \(y = 6\cos(\frac{\pi}{4}x)-2\) occurs when \(\cos(\frac{\pi}{4}x)=1\). Then \(y = 6\times1 - 2=4\). Solving \(\frac{\pi}{4}x = 2k\pi\) (\(k\in\mathbb{Z}\)), we get \(x = 8k\). When \(k = 0\), \(x = 0\) (not the extremum in the non - negative part of the graph we are considering for the widget). When \(k = 1\), \(x = 8\) (a bit far). The minimum value occurs when \(\cos(\frac{\pi}{4}x)=-1\), \(y=6\times(-1)-2=-8\).
For the positive \(x\) values near the origin, when \(x = 2\), \(y = 6\cos(\frac{\pi}{4}\times2)-2=6\cos(\frac{\pi}{2})-2=- 2\) (mid - line intersection). When \(x = 0\), \(y=6\cos(0)-2=6 - 2 = 4\) (maximum).

Answer:

To graph \(y = 6\cos(\frac{\pi}{4}x)-2\):

• The amplitude is \(6\), mid - line \(y=-2\), and period \(T = 8\).
• One extremum (maximum) point is \((0,4)\) (since when \(x = 0\), \(y=6\cos(0)-2=4\)) and a mid - line intersection point near it is \((2,-2)\) (since when \(x = 2\), \(y=6\cos(\frac{\pi}{4}\times2)-2=-2\)).