How do you find vertical, horizontal and oblique asymptotes for #(x)/(4x^2+7x-2)#?

1 Answer
Mar 25, 2016

Answer:

vertical asymptotes , x = -2 , # x = 1/4 #
horizontal aymptote y = 0

Explanation:

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

solve: # 4x^2 + 7x - 2 = 0 → (x + 2 )(4x - 1 ) = 0 #

#rArr x = - 2 , x = 1/4" are the 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 is the case here , then the equation of the asymptote is always y = 0 .

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

Here is the graph of the function.
graph{x/(4x^2+7x-2) [-10, 10, -5, 5]}