How do you find vertical, horizontal and oblique asymptotes for (x^2-2x-3) /( 2x^2-x-10)x2−2x−32x2−x−10?

1 Answer
Eddie
Jul 1, 2016

vertical asymptotes at x = -2, 2.5x=−2,2.5

horizontal asymptote is y = 1/2y=12

Explanation:

(x^2-2x-3) /( 2x^2-x-10)x2−2x−32x2−x−10

looking at denominator

2x^2-x-10 = 2 {x^2 - 1/2 x - 5}2x2−x−10=2{x2−12x−5}

= 2 {(x+2)(x- 5/2)}=2{(x+2)(x−52)} so the denominator is zero at x = -2, 2.5x=−2,2.5

the numerator factorises as x^2-2x-3 = (x-3)(x+1)x2−2x−3=(x−3)(x+1) so it is zero at x = -1, 3 thus finite at x = -2, 2.5x=−2,2.5

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

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