Values of c for which the function has a horizontal/vertical asymptote

Question

Given $$f(x) = \frac1{x^2 + 2x + c}$$ with $c \in \mathbb{R}$, find values of $c$ for which $f$ has a horizontal asymptote.

Also for what values of $c$ does the function have no vertical asymptote, one vertical asymptote and two vertical asymptotes?

Thanks.

Answer 1

HINT

For horizontal asymptotes consider

• $\lim_{x\to \pm \infty} f(x)$

For vertical asymptotes consider the roots for

• $x^2 + 2x + c=0$

Answer 2

There's a horizontal asymptote at $y=0$ $\forall c$.

This is because your function cannot equal $0$ as everything being affected by $x$ is on the denominator.

With vertical asymptotes, I'll summarise how you can easily work out how many each will have:

Let $x^2+2x+c=(x+p)(x+q)$ The discriminant ($b^2-4ac$) for these equations is $4-4c$

If $4-4c=0$, we get one vert asymp, this is just when $c=1$

If $4-4c>0$, we get two vertical asymptotes, when $c<1$

If $4-4c<0$, we get no vertical asymptote, when $c>1$.

The reason this method works, is because setting the discriminant to $>0,=0,<0$ will tell you an equation will have $2,1,0$ distinct real solutions respectively.