Originally Posted by

**Joker37** State the conditions on k such that the graph of

$\displaystyle y=\frac{1}{x^2+kx+16}$

has:

a) two vertical asympototes

b) only one vertical asymptote

c) no vertical asymptotes.

Answers:

a) $\displaystyle (-\infty ,-8)\cup(8,\infty)$

b) $\displaystyle k=\pm8$

c) $\displaystyle (-8,8)$

However if the equation above were to have two vertical asymptotes wouldn't the factorised denominator of the equation above be:

$\displaystyle \frac{1}{(x+8)(x+2)},\;\; or\;\;\frac{1}{(x-8)(x-2)},\;\; or \;\;\frac{1}{(x+1)(x+16)},\;\; or\;\; \frac{1}{(x-1)(x-16)}$

So wouldn't

$\displaystyle k=\pm10,\;\; or \;\;k=\pm17\;\;?$

**If not, why not** ?