Question

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

Answer 1

Two vertical asymptotes are `2x+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 `-6s` 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 `(2x+1)` and `(x-3)`, two vertical asymptotes are `2x+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.