How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (2x+4) /( x^2-3x-4)#?

1 Answer
Sep 25, 2016

Answer:

The vertical asymptotes are #x=4# and #x=-1#. The horizontal asymptote is #y=0#. There are no oblique asymptotes.

Explanation:

#f(x)=(2x+4)/(x^2-3x-4)#

Factor the denominator.

#f(x)=frac{2x+4}{(x-4)(x+1)}#

To find the vertical asymptotes, set the denominator equal to zero and solve.
#(x-4)(x+1)=0#
#x-4=0color(white)(aaa)x+1=0#
#x=4color(white)(aaa)x=-1color(white)(aaa)#These are the VA's.

To find the horizontal asymptotes, compare the degree of the numerator to the degree of the denominator.

  • If the deg of the num > deg of den, there is an oblique asymptote.

  • If the deg of the num = deg of den, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.

  • If the deg of the num < deg of den, the HA is #y=0#.

In this example, the degree of the numerator is #color(red)1# and the degree of the denominator is #color(blue)2#
#f(x)=(2x^color(red)1+4)/(x^color(blue)2-3x-4)#

The degree of the numerator < degree of the denominator, so the HA is #y=0#.

There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.