Question

plot at least 6 points of the graph of the function $y = 2 \tan ( x - \frac { \pi } { 4 } ) - 1$

Explanation:

Step1: Find key points

For the tangent function \(y = A\tan(B(x - C))+D\), here \(A = 2\), \(B = 1\), \(C=\frac{\pi}{4}\), \(D=- 1\). The period of \(y=\tan x\) is \(\pi\), so the period of \(y = 2\tan(x-\frac{\pi}{4})-1\) is also \(\pi\).

We know that the vertical asymptotes of \(y=\tan x\) are \(x=(n +\frac{1}{2})\pi,n\in Z\). For \(y = 2\tan(x-\frac{\pi}{4})-1\), the vertical asymptotes are \(x-\frac{\pi}{4}=(n+\frac{1}{2})\pi\), i.e., \(x=(n+\frac{3}{4})\pi\)

When \(x=\frac{\pi}{4}\), \(y=2\tan(0)-1=-1\)

When \(x=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}\), \(y = 2\tan(\frac{\pi}{4})-1=2\times1 - 1=1\)

When \(x=\frac{\pi}{4}-\frac{\pi}{4}=0\), \(y=2\tan(-\frac{\pi}{4})-1=2\times(- 1)-1=-3\)

When \(x=\frac{\pi}{4}+\frac{\pi}{2}=\frac{3\pi}{4}\), \(y = 2\tan(\frac{\pi}{2})\) (asymptote)

When \(x=\frac{\pi}{4}-\frac{\pi}{2}=-\frac{\pi}{4}\), \(y = 2\tan(-\frac{\pi}{2})\) (asymptote)

When \(x=\frac{\pi}{4}+\frac{3\pi}{4}=\pi\), \(y=2\tan(\frac{3\pi}{4})-1=2\times(-1)-1=-3\)

Answer:

Plot the points \((0,-3)\), \((\frac{\pi}{4},-1)\), \((\frac{\pi}{2},1)\), \((\pi,-3)\) and mark the vertical asymptotes \(x =-\frac{\pi}{4}\), \(x=\frac{3\pi}{4}\) etc. on the coordinate - plane.