Question

graph $y = 4\cos\left(\frac{1}{5}x\
ight)+2$ 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: Find the amplitude

The general form of a cosine function is \(y = A\cos(Bx - C)+D\). For \(y = 4\cos(\frac{1}{5}x)+2\), the amplitude \(A = 4\).

Step2: Find the period

The period formula for \(y=\cos(Bx)\) is \(T=\frac{2\pi}{|B|}\). Here \(B = \frac{1}{5}\), so \(T=\frac{2\pi}{\frac{1}{5}}=10\pi\).

Step3: Find the mid - line

The mid - line is given by \(y = D\). For \(y = 4\cos(\frac{1}{5}x)+2\), \(D = 2\).

Step4: Find key points

• When \(x = 0\), \(y=4\cos(0)+2=4\times1 + 2=6\) (maximum point since \(\cos(0) = 1\))
• When \(x=\frac{5\pi}{2}\), \(y = 4\cos(\frac{1}{5}\times\frac{5\pi}{2})+2=4\cos(\frac{\pi}{2})+2=2\) (mid - line intersection)
• When \(x = 5\pi\), \(y=4\cos(\pi)+2=4\times(- 1)+2=-2\) (minimum point since \(\cos(\pi)=-1\))
• When \(x=\frac{15\pi}{2}\), \(y = 4\cos(\frac{1}{5}\times\frac{15\pi}{2})+2=4\cos(\frac{3\pi}{2})+2=2\) (mid - line intersection)
• When \(x = 10\pi\), \(y=4\cos(2\pi)+2=4\times1+2 = 6\) (maximum point since \(\cos(2\pi)=1\))

Plot these key points \((0,6)\), \((\frac{5\pi}{2},2)\), \((5\pi,-2)\), \((\frac{15\pi}{2},2)\), \((10\pi,6)\) and connect them with a smooth cosine - like curve.

Answer:

Plot the points \((0,6)\), \((\frac{5\pi}{2},2)\), \((5\pi,-2)\), \((\frac{15\pi}{2},2)\), \((10\pi,6)\) and draw a smooth curve through them for the graph of \(y = 4\cos(\frac{1}{5}x)+2\)