Question

How do you find the asymptotes for `y= (2x^3-5x+3) / (x^2-5x+4)`?

Answer 1

Vertical asymptote at `x=4` and slanting asymptote at `y=2x`

Explanation:

In `y=(2x^3-5x+3)/(x^2-5x+4)`, the factors of denominator are `(x-4)(x-1)` and numerator is also divisible by `x-1`, as such

`y=(2x^3-5x+3)/(x^2-5x+4)=((x-1)(2x^2+2x-3))/((x-4)(x-1))=(2x^2+2x-3)/(x-4)`

Hence, as denominator will be zero for `x=4`, we will have vertical asymptote at `x=4`.

Further as degree of numerator in `(2x^2+2x-3)/(x-4)` is greater than that of denominator and ratio of highest degrees by one in the two is `2x^2/x=2x`, we will have a slanting asymptote at `y=2x`

graph{(2x^3-5x+3)/(x^2-5x+4) [-15, 15, -5, 5]}