How do you find vertical, horizontal and oblique asymptotes for #(3x^2+x-4) / (2x^2-5x)#?

1 Answer
Apr 4, 2016

Answer:

vertical asymptotes x = 0 , #x = 5/2 #
horizontal asymptote # y = 3/2 #

Explanation:

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

solve : # 2x^2 - 5x = 0 → x(2x-5) = 0 #

#rArr x = 0 , x = 5/2 " are the asymptotes " #

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

divide all terms on numerator/denominator by #x^2 #

#((3x^2)/x^2 + x/x^2 - 4/x^2)/((2x^2)/x^2 -(5x)/x^2)= (3+1/x-4/x^2)/(2-5/x) #

As # x to+-oo , 1/x , 4/x^2" and " 5/x to 0 #

#rArr y = 3/2 " is the asymptote " #

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{(3x^2+x-4)/(2x^2-5x) [-10, 10, -5, 5]}