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

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Answer 1 · 2016-12-15T08:33:58

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

By actual division,

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

Reorganizing,

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

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

$(y-2x+7)(x+2)=18$ , 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]}

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