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

Aug 10, 2017

Vertical asymptote is at $x = - 1$
Slant asymptote is $y = 4 x - 5$

#### Explanation:

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

Vertical asymptote is fomed when deniminator is zero.

$\therefore x + 1 = 0 \mathmr{and} x = - 1$ So vertical asymptote is at $x = - 1$

Here degree in numerator is $1$ greater than degree in

denominator . So we have a slant asymptote which is found by

doing long division . By long division we find Quotient as

$y = 4 x - 5$ . So slant asymptote is $y = 4 x - 5$

graph{(4x^2-x+2)/(x+1) [-101.25, 101.2, -50.6, 50.7]} [Ans]