Question

What of the following is not a vertical asymptote of the function

f(x)= tan(pi x/4) / x^2-8x+15?

2
3
4
5
6

Answer 1

Rewrite the function as

`f(x) = \frac{\tan(\left(\frac{\pi x}4\right)}{x^2 - 8x + 15} = \frac{\sin\left(\frac{\pi x}4\right)}{\cos\left(\frac{\pi x}4\right) (x-3) (x-5)}`

Vertical asymptotes occur wherever the denominator is zero (but not necessarily when the numerator is also zero).

We have

`\cos\left(\frac{\pi x}4\right) (x-3) (x-5) = 0`

`\implies \cos\left(\frac{\pi x}4\right) = 0 \text{ or } x - 3 = 0 \text{ or } x - 5 = 0`

In the first case, we have

`\cos\left(\frac{\pi x}4\right) = 0 \implies \frac{\pi x}4 = \pm\frac\pi2 + 2n\pi \implies x = \pm 2 + 8n`

where
`n` is an integer. That is, any of
`x\in\{\ldots,-10,-2,\boxed{6},14,\ldots\}` and
`x\in\{\ldots,-6,\boxed{2},10,18,\ldots\}` are asymptotes.

In the latter two cases, we get
`x=\boxed{3}` or
`x=\boxed{5}`. The numerator is finite at these values
`\left(\sin\left(\frac{3\pi}4\right)=\frac1{\sqrt2}\text{ and } \sin\left(\frac{5\pi}4\right)=-\frac1{\sqrt2}\right)`.

This means
`x=4` is not an asymptote. (C)