# How do you find the vertical, horizontal or slant asymptotes for (x + 1)/(x^2 + 5x − 6)?

Dec 29, 2016

The vertical asymptotes are $x = 1$ and $x = 6$
No slant asymptote
The horizontal asymptote is $y = 0$

#### Explanation:

We start by factorising the denominator

${x}^{2} + 5 x - 6 = \left(x - 1\right) \left(x + 6\right)$

Let $f \left(x\right) = \frac{x + 1}{\left(x - 1\right) \left(x + 6\right)}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{1 , - 6\right\}$

As we cannot divide by $0$, $x \ne 1$ and $x \ne - 6$

So,

The vertical asymptotes are $x = 1$ and $x = 6$

As the degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

For the horizontal asymptote, we calculate the $\lim$ of $f \left(x\right)$ as $x \to \pm \infty$

lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(- )

lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->oo)1/x=0^(+ )

Therefore,

The horizontal asymptote is $y = 0$

graph{(y-(x+1)/(x^2+5x-6))(y)=0 [-10, 10, -5, 5]}