Question

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

Answer 1

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

Explanation:

By actual division,

`y = 2x-7+18/(x+2)`

Reorganizing,

`(y-2x+7)(x+2)=18`.

This represents a hyperbola.(https://socratic.org/precalculus/functions-defined-and-notation/asymptotes)

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

`(y-2x+7)(x+2)=18`, and so, the asymptotes are.

`y-2x+7 = 0 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, `(-2, -11)`.

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