# How do you identify all asymptotes or holes and intercepts for f(x)=(x^3-1)/((2(x^2-1))?

Nov 25, 2016

There is a hole at $x = 1$
A vertical asymptote is $x = - 1$
The slant asymptote is $y = \frac{x}{2}$
There is no horizontal asymptote.

#### Explanation:

Let's factorise the denominator

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

Let's factorise the numerator
${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

Therefore,

$f \left(x\right) = \frac{{x}^{3} - 1}{2 \left(x + 1\right) \left(x - 1\right)} = \frac{\cancel{x - 1} \left({x}^{2} + x + 1\right)}{2 \left(x + 1\right) \cancel{x - 1}}$

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

There is a hole at $x = 1$

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

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

A vertical asymptote is $x = - 1$

As the degree of the numerator is $>$ than the degree of the denominator, we expect a slant asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} + x + 1$$\textcolor{w h i t e}{a a a a}$∣$2 x + 2$

$\textcolor{w h i t e}{a a a a}$${x}^{2} + x$$\textcolor{w h i t e}{a a a a a a a}$∣$\frac{x}{2}$

$\textcolor{w h i t e}{a a a a a}$$0 + 0 + 1$

So, $\frac{{x}^{2} + x + 1}{2 \left(x + 1\right)} = \frac{x}{2} + \frac{1}{2 x + 2}$

The slant asymptote is $y = \frac{x}{2}$

To calculate the limit, we take the terms of highest degree.

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} {x}^{2} / \left(2 x\right) = \pm \infty$

There is no horizontal asymptote.

graph{(y-(x^2+x+1)/(2x+2))(y-x/2)=0 [-5.542, 5.553, -2.783, 2.766]}