Transformed tangent functions
Tangent has an unbounded range and vertical asymptotes, so it is analysed branch by branch. Factor the inner expression to read the period, phase shift and asymptote positions
\[f(x)=a\tan\left(n(x-h)\right)+k,\quad T=\frac{\pi}{n}\]
- Use the half-turn period Tangent has base period π, so a positive inner frequency n gives T=π/n. \(\displaystyle n=2\Rightarrow T=\frac{\pi}{2}\)
- Locate asymptotes Set the tangent input to an odd multiple of π/2 and solve for x. \(\displaystyle x=h+\frac{(2m+1)\pi}{2n}\)
- Locate branch centres Set the tangent input to an integer multiple of π; the centre has y-value k. \(\displaystyle x=h+\frac{m\pi}{n}\)
- Use the full real range Tangent has no amplitude because its range is all real numbers \(\displaystyle \operatorname{range}(f)=\mathbb{R},\quad\text{tangent has no amplitude}\)
Worked example
For the function shown, state its period, one vertical asymptote, one branch centre and range
\[f(x)=-3\tan\left(2\left(x-\frac{\pi}{6}\right)\right)+4\]
- \(\displaystyle T=\frac{\pi}{2}\)The tangent period is π divided by the positive inner frequency 2.
- \(\displaystyle x=\frac{\pi}{6}+\frac{\pi}{4}=\frac{5\pi}{12}\)For m=0, the asymptote immediately to the right of the centre x=h is h+π/(2n)
- \(\displaystyle x=h=\frac{\pi}{6},\quad f(h)=k=4\Rightarrow\left(\frac{\pi}{6},4\right)\)The branch centre occurs when the tangent input is zero
- \(\displaystyle \operatorname{range}(f)=\mathbb{R},\quad\text{there is no amplitude}\)Vertical scale changes the branch shape, but tangent still takes every real value
Common mistakes
Using the sine/cosine period
✕\(\displaystyle T=\frac{2\pi}{n}=\pi\text{ when }n=2\)✓\(\displaystyle T=\frac{\pi}{n}=\frac{\pi}{2}\text{ when }n=2\)
Tangent has base period π, so its transformed period is π/n rather than 2π/n.Assigning an amplitude to tangent
✕\(\displaystyle f(x)=2\tan(2x)+1\Rightarrow A=2\)✓\(\displaystyle \operatorname{range}(f)=\mathbb{R},\quad\text{no amplitude}\)
Tangent is unbounded, so it has no amplitude even when its branches are vertically scaledLeaving asymptotes unshifted
✕\(\displaystyle f(x)=\tan\left(2\left(x-\frac{\pi}{8}\right)\right)\Rightarrow x=\frac{(2m+1)\pi}{4}\)✓\(\displaystyle x=\frac{\pi}{8}+\frac{(2m+1)\pi}{4}\)
The phase shift h moves every centre and asymptote horizontallyDifficulty levels
introIdentify a basic tangent branch, centre or intercept from its rule
standardFind one active transformation and the resulting period or asymptote
challengeAnalyse a tangent rule with several transformations branch by branch
examMatch or justify a tangent rule using period, asymptotes, centres and range
Questions about this topic
Why is the tangent period π instead of 2π?+
Tangent has the same value after an angle increases by π, so its base period is π
Why does tangent have no amplitude?+
Tangent is unbounded near its vertical asymptotes and its range is all real numbers. Amplitude is defined for bounded oscillations
How are tangent asymptotes found?+
Set the input to an odd multiple of π/2, where tangent is undefined, then solve the resulting equation for x.
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 10: Transformed circular functions