# How do you find vertical, horizontal and oblique asymptotes for f(x) =( x^3 - 4x^2 + 5x + 5)/(x - 1)?

Dec 21, 2017

$f \left(x\right)$ has one vertical asymptote $x = 1$ and no other linear asymptotes.

#### Explanation:

$f \left(x\right) = \frac{{x}^{3} - 4 {x}^{2} + 5 x + 5}{x - 1}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{{x}^{3} - {x}^{2} - 3 {x}^{2} + 3 x + 2 x - 2 + 7}{x - 1}$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{2} - 3 x + 2 + \frac{7}{x - 1}$

This function has a vertical asymptote $x = 1$.

As $x \to \pm \infty$ we have $\frac{7}{x - 1} \to 0$, so $f \left(x\right)$ is asymptotic to ${x}^{2} - 3 x + 2$, which is not a line.

So there are no other linear asymptotes.

graph{(x^3-4x^2+5x+5)/(x-1) [-10, 10, -200, 200]}