How do you find vertical, horizontal and oblique asymptotes for (2x+4)/(x^2-3x-4)2x+4x2−3x−4?
1 Answer
Apr 7, 2016
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 x2−3x−4=0→(x−4)(x+1)=0
rArr x = -1 , x = 4 " are the asymptotes" ⇒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]}