# How do you find the asymptotes for y=(2x^2-3x+4)/(x+2)?

Dec 15, 2016

Slant asymptote+ $y - 2 x + 7 = 0$
Vertical asymptote: $x + 2 = 0$
Illustrative graph is inserted.
:

#### Explanation:

By actual division,

$y = 2 x - 7 + \frac{18}{x + 2}$

Reorganizing,

$\left(y - 2 x + 7\right) \left(x + 2\right) = 18$.

Note that the equation of a hyperbola with asymptotes as the pair

of lines

(y-m x-c)(y-m' x-c') = 0 is

(y-m x-c)(y-m' x-c') = k and here it is

$\left(y - 2 x + 7\right) \left(x + 2\right) = 18$, and so, the asymptotes are.

$y - 2 x + 7 = 0 \mathmr{and} x + 2 = 0$

Also, the only conic with asympiotes is hyperbola.

The center of the hyperbola is the point of intersection of the

asymptotes, $\left(- 2 , - 11\right)$.

graph{((y-2x+7)(x+2)-18)(y-2x+7)(x+1.99)(x+2.01)=0[-20 20 -40 20]}