Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2-9x+20)/(2x^2-8x)`?

Answer 1

Vertical
We see `g(x) = 0`; for `x=> (0, 4)` so we find vertical asymptotes at
`x=0 and x=4`

Horizontal
`(x^2-9x+20)/(2(x^2-4x))= 1/2(x^2-9x+20)/(x^2-4x)`
the horizontal asymptote is `y=1/2`

Explanation:

Vertical Asymptote:
A rational function h=f(x)/g(x) have a vertical asymptotes
`AA x: g(x) = 0`.

Horizontal Asymptote: The location of the horizontal asymptote is determined by looking at the degrees of the numerator `n` and denominator `m`.
1) If `n < m`, the x-axis, y=0 is the horizontal asymptote.
2) If `n=m`, then `y=a_n / b_m` the ratio of the leading coefficients is asymptot.
3) If `n>m`, there is no horizontal asymptote.

Oblique Asymptote: Still looking at the degrees `n and m`
4) if `n=m+1`, there is an oblique or slant asymptote.

Now letting `f(x)=(x^2-9x+20) and g(x)=(2x^2-8x)`

Vertical:
We see `g(x) = 0; for x=> (0, 4)` so we find vertical asymptotes at
`x=0 and x=4`

Horizontal:
definition we have `n=m` thus the ratio of the leading coefficient is the asymptote.
`(x^2-9x+20)/(2(x^2-4x))= 1/2(x^2-9x+20)/(x^2-4x)`
Thus the Horizontal is asymptote is `y=1/2`