# How do you find the asymptotes for f(x) = (3x^2 + 15x +18) /( 4x^2-4)?

Jan 3, 2017

The vertical asymptotes are $x = 1$ and $x = - 1$
No slant asymptote
The horizontal asymptote is $y = \frac{3}{4}$

#### Explanation:

We need

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

We factorise the denominator

$4 {x}^{2} - 4 = 4 \left({x}^{2} - 1\right) = 4 \left(x - 1\right) \left(x + 1\right)$

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

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

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

As the degree of the numerator is $=$ to the degree of the denominator, there is no slant asymptote

To find the horizontal asymptotes, we look at the limits of $f \left(x\right)$ as $x \to \pm \infty$

${\lim}_{x \to \pm \infty} f \left(x\right)$

$= {\lim}_{x \to \pm \infty} \frac{3 {x}^{2}}{4 {x}^{2}}$

$= \frac{3}{4}$

The horizontal asymptote is $y = \frac{3}{4}$

graph{(y-(3x^2+15x+18)/(4x^2-4))(y-3/4)=0 [-14.24, 14.23, -7.12, 7.11]}