Answer:
A) f(x) = tan(2x - π)
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{8.3cm}\underline{Standard form of a tangent function}\\\\$f(x)=A \tan(B(x+C))+D$\\\\where:\\\\\phantom{ww}$\bullet$ $A=$ vertical stretch\\ \\\phantom{ww}$\bullet$ $\dfrac{\pi}{|B|}=$ period\\\\\phantom{ww}$\bullet$ $C=$ horizontal shift (positive is to the left)\\\\\phantom{ww}$\bullet$ $D=$ vertical shift\\\end{minipage}}[/tex]
The parent tangent function is:
[tex]f(x)=\tan(x)[/tex]
The period of the parent tangent function is π.
A tangent function is discontinuous when cos(x) = 0, so it has vertical asymptotes whenever cos(x) = 0.
Therefore, the parent tangent function has vertical asymptotes at:
[tex]x=\dfrac{\pi}{2}+\pi n[/tex]
and so its domain is:
[tex]\left\{ x \in \mathbb{R} \;| \;x \neq \dfrac{\pi}{2}+\pi n\right\}[/tex]
If the domain of the given tangent function is:
[tex]\left\{ x \in \mathbb{R} \;| \;x \neq \dfrac{\pi}{4}+\dfrac{\pi}{2}n\right\}[/tex]
then its vertical asymptotes are when:
[tex]x =\dfrac{\pi}{4}+\dfrac{\pi}{2}n[/tex]
Therefore, its period is π/2.
[tex]\implies \dfrac{\pi}{|B|}=\dfrac{\pi}{2}[/tex]
[tex]\implies B=2[/tex]
And it has been horizontally shifted by π/2:
[tex]\implies f(x)=\tan\left(2\left(x-\dfrac{\pi}{2}\right)\right)[/tex]
[tex]\implies f(x)=\tan\left(2x-\pi\right)[/tex]
Function g(x)
[tex]g(x)=\tan(x- \pi)[/tex]
- Period = π
- Horizontal shift = π
- Vertical asymptotes = π/2 + πn
Function h(x)
[tex]h(x)=\tan \left(x-\dfrac{\pi}{2}\right)[/tex]
- Period = π
- Horizontal shift = π/2
- Vertical asymptotes = π + πn
Function j(x)
[tex]j(x)=\tan \left(\dfrac{x}{2}- \pi \right)[/tex]
- Period = 2π
- Horizontal shift = 2π
- Vertical asymptotes = π + 2πn
