# How do you find the asymptotes for f(x)= (2x^2 +x -1 )/( x-1)?

Apr 23, 2017

$\text{vertical asymptote at } x = 1$
$\text{oblique asymptote } y = 2 x + 3$

#### 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.

$\text{solve " x-1=0rArrx=1" is the asymptote}$

Since the degree of the numerator > degree of the denominator there is an oblique asymptote but no horizontal asymptote.

$\text{using the divisor as a factor in the numerator}$

$\textcolor{red}{2 x} \left(x - 1\right) \textcolor{m a \ge n t a}{+ 2 x} + x - 1$

$= \textcolor{red}{2 x} \left(x - 1\right) \textcolor{red}{+ 3} \left(x - 1\right) \textcolor{m a \ge n t a}{+ 3} - 1$

$= \textcolor{red}{2 x} \left(x - 1\right) \textcolor{red}{+ 3} \left(x - 1\right) + 2$

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

as $x \to \pm \infty , f \left(x\right) \to 2 x + 3$

$\Rightarrow y = 2 x + 3 \text{ is the asymptote}$
graph{(2x^2+x-1)/(x-1) [-24.98, 24.98, -12.48, 12.5]}