How do you find the asymptotes for #(3x^2+x-4) / (2x^2-5x)#?

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Answer 1 · 2016-02-01T11:24:43

Horizontal asymptote:
#y=3/2#

Vertical asymptote:
#x=0#
#x=5/2#

Here is a graph of the expression in question.
graph{(3x^2 + x - 4)/(2x^2 - 5x) [-10, 10, -5, 5]}
You can try to simplify the expression first.

#(3x^2 + x - 4)/(2x^2 - 5x) = 3/2 + frac{17x - 8}{4x^2 - 10x}#

When #x# is very large (or small), observe that the second term tends to zero. Therefore,

#lim_{x->-oo} frac{3x^2 + x - 4}{2x^2 - 5x} = 3/2# and

#lim_{x->oo} frac{3x^2 + x - 4}{2x^2 - 5x} = 3/2#.

Hence, there is a horizontal asymptote with equation #y=3/2#.

Also note that the expression is not defined for #x=5/2# and #x=0#, as they result in division by zero.

Hence, there are 2 vertical asymptotes, with equation #x=5/2# and #x=0#.