How do you find the asymptotes for (1-x)/(2x^2-5x-3)1−x2x2−5x−3?

1 Answer
Jim G.
Apr 26, 2016

vertical asymptotes x=-1/2 , x = 3x=−12,x=3
horizontal asymptote y = 0

Explanation:

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

solve : 2x^2-5x-3 = 0 → (2x+1)(x-3) = 0 2x2−5x−3=0→(2x+1)(x−3)=0

rArr x = -1/2" and " x = 3" are the asymptotes "⇒x=−12 and x=3 are the asymptotes

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

If the degree of the numerator < degree of the denominator,as is the case here then the equation is always y = 0
graph{(1-x)/(2x^2-5x-3) [-10, 10, -5, 5]}

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