How do you find vertical, horizontal and oblique asymptotes for #( -3x^2+2) /( x+2)#?

1 Answer
Nov 27, 2016

Answer:

The vertical asymptote is #x=-2#
The oblique asymptote is #y=-3x+6#
No horizontal asymptote

Explanation:

Let #f(x)=(-3x^2+2)/(x+2)#

The domain of #f(x)# is #D_f(x)=RR-{-2} #

As we cannot divide by #0#, #x!=-2#

So, the vertical asymptote is #x=-2#

The degree of the numerator is #># the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##-3x^2+2##color(white)(aaaa)##∣##x+2#

#color(white)(aaaa)##-3x^2-6x##color(white)(aaa)##∣##-3x+6#

#color(white)(aaaa)##-0+6x+2#

#color(white)(aaaaaaa)##+6x+12#

#color(white)(aaaaaaaa)##+0-10#

Therefore,

#(-3x^2+2)/(x+2)=(-3x+6)-10/(x+2)#

So, the oblique asymptote is #y=-3x+6#

To calculate the limits to #oo#, we take the terms of highest degree in the numerator and the denominator.

#lim_(x->-oo)f(x)=lim_(x->-oo)-(3x^2)/x=lim_(x->-oo)-3x=+oo#

#lim_(x->+oo)f(x)=lim_(x->+oo)-(3x^2)/x=lim_(x->+oo)-3x=-oo#

graph{(y-(-3x^2+2)/(x+2))(y+3x-6)=0 [-74.04, 74.05, -37, 37]}