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

Feb 25, 2016

Two vertical asymptotes are $2 x + 1 = 0$ and $x - 3 = 0$.

#### Explanation:

To find all the asymptotes for function y=(1-x)/(2x^2-5x−3), let us first start with vertical asymptotes, which are given by putting denominator equal to zero or (2x^2-5x−3)=0.

Let us find factors of (2x^2-5x−3) by splitting middle term in $- 6 s$ and $x$ i.e. (2x^2-6x+x−3=2x(x-3)+1(x-3)=(2x+1)(x-3).

As the factors of denominators are $\left(2 x + 1\right)$ and $\left(x - 3\right)$, two vertical asymptotes are $2 x + 1 = 0$ and $x - 3 = 0$.

As the highest degree of numerator is less than that of denominator, there is no horizontal or slanting asymptote.