How do you find vertical, horizontal and oblique asymptotes for #(2x+4)/(x^2-3x-4)#?

1 Answer
Apr 7, 2016

Answer:

vertical asymptotes x = -1 , x = 4
horizontal asymptote y = 0

Explanation:

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve: # x^2 - 3x - 4 = 0 → (x -4)(x + 1) = 0 #

# rArr x = -1 , x = 4 " are the asymptotes" #

Horizontal asymptotes occur as # lim_(xto+-oo) f(x) to 0 #

When the degree of the numerator < degree of the denominator , as is the case here then the equation is always
y = 0.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}