8.2 Graphs of the other trigonometric functions  (Page 3/9)

• Step 1. The function is already written in the form $y=A\tan \left(Bx-C\right)+D.$
• Step 2. $A=\mathrm{-2},$ so the stretching factor is $\left|A\right|=2.$
• Step 3. $B=\pi ,$ so the period is $P=\frac{\pi }{\left|B\right|}=\frac{\pi }{\pi }=1.$
• Step 4. $C=-\pi ,$ so the phase shift is $\frac{C}{B}=\frac{-\pi }{\pi }=\mathrm{-1.}$
• Step 5-7. The asymptotes are at $x=-\frac{3}{2}$ and $x=-\frac{1}{2}$ and the three recommended reference points are $\left(\mathrm{-1.25},1\right),$ $\left(\mathrm{-1,}\mathrm{-1}\right),$ and $\left(\mathrm{-0.75,}\mathrm{-3}\right).$ The graph is shown in [link] .

  

The graph has the shape of a tangent function.

• Step 1. One cycle extends from –4 to 4, so the period is $P=8.$ Since $P=\frac{\pi }{\left|B\right|},$ we have $B=\frac{\pi }{P}=\frac{\pi }{8}.$
• Step 2. The equation must have the form $f(x)=A\tan \left(\frac{\pi }{8}x\right).$
• Step 3. To find the vertical stretch $A,$ we can use the point $\left(2,2\right).$
  $2=A\tan \left(\frac{\pi }{8}\cdot 2\right)=A\tan \left(\frac{\pi }{4}\right)$

Because $\tan \left(\frac{\pi }{4}\right)=1,$ $A=2.$

This function would have a formula $f(x)=2\tan \left(\frac{\pi }{8}x\right).$