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

Sep 29, 2016

vertical asymptote at x = 1
oblique asymptote is y = 2x + 3

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $x - 1 = 0 \Rightarrow x = 1 \text{ is the asymptote}$

Horizontal asymptotes occur when the degree of the numerator ≤ degree of the denominator. This is not the case here (numerator-degree 2 , denominator- degree 1 ) Hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator which is the case here.

Use $\textcolor{b l u e}{\text{polynomial division}}$ to obtain.

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

as xto+-oo,(f(x)to2x+3+0

$\Rightarrow y = 2 x + 3 \text{ is the oblique asymptote}$
graph{(2x^2+x-1)/(x-1) [-40, 40, -20, 20]}