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

1 Answer
Mar 12, 2016

Answer:

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

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation equate the denominator to zero.

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

#rArr x = -1 and x = 4 " are asymptotes " #

Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0#

If the degree of the numerator is less than the degree of the denominator, as in this case, degree of numerator is 1 and degree of denominator is 2 then the equation of asymptote is
y = 0

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