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

1 Answer
Dec 15, 2016

Answer:

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]}