How do you find the vertical, horizontal or slant asymptotes for f(x)=(x^3)/((x-1)^2)?

Jan 3, 2017

The vertical asymptote is $x = 1$
No horozontal asym`ptote.
The slant asymptote is $y = x + 2$

Explanation:

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

As you cannot divide by $0$, $x \ne 1$

The vertical asymptote is $x = 1$

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

The denominator is

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

We do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$∣${x}^{2} - 2 x + 1$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2} + x$$\textcolor{w h i t e}{a a a a a a}$∣$x + 2$

$\textcolor{w h i t e}{a a a a a}$$0 + 2 {x}^{2} - x$

$\textcolor{w h i t e}{a a a a a a a}$$+ 2 {x}^{2} - 4 x + 2$

$\textcolor{w h i t e}{a a a a a a a a a}$$+ 0 + 3 x - 2$

So,

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

Therefore,

${\lim}_{x \to - \infty} f \left(x\right) - \left(x + 2\right) = {\lim}_{x \to - \infty} \frac{3 x}{x} ^ 2 = {\lim}_{x \to - \infty} \frac{3}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) - \left(x + 2\right) = {\lim}_{x \to + \infty} \frac{3 x}{x} ^ 2 = {\lim}_{x \to + \infty} \frac{3}{x} = {0}^{+}$

So,

The slant asymptote is $y = x + 2$

graph{(y-(x^3/(x-1)^2))(y-x-2)(y-100(x-1))=0 [-27.35, 30.37, -11.67, 17.2]}