How do you find vertical, horizontal and oblique asymptotes for #(x^2-2x-3) /( 2x^2-x-10)#?

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Answer 1 · 2016-07-01T09:33:29

vertical asymptotes at #x = -2, 2.5#

horizontal asymptote is #y = 1/2#

#(x^2-2x-3) /( 2x^2-x-10)#

looking at denominator

#2x^2-x-10 = 2 {x^2 - 1/2 x - 5}#

#= 2 {(x+2)(x- 5/2)}# so the denominator is zero at #x = -2, 2.5#

the numerator factorises as #x^2-2x-3 = (x-3)(x+1)# so it is zero at x = -1, 3 thus finite at #x = -2, 2.5#

we can therefore conclude vertical asymptotes at #x = -2, 2.5#

#lim_{x to pm oo} (x^2-2x-3) /( 2x^2-x-10)#

#=lim_{x to pm oo} (1-2/x-3/x^2) /( 2-1/x-10/x^2) = 1/2#

so the horizontal asymptote is #y = 1/2#