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

Mar 26, 2016

Vertical
We see $g \left(x\right) = 0$; for $x \implies \left(0 , 4\right)$ so we find vertical asymptotes at
$x = 0 \mathmr{and} x = 4$

Horizontal
$\frac{{x}^{2} - 9 x + 20}{2 \left({x}^{2} - 4 x\right)} = \frac{1}{2} \frac{{x}^{2} - 9 x + 20}{{x}^{2} - 4 x}$
the horizontal asymptote is $y = \frac{1}{2}$

#### Explanation:

Vertical Asymptote:
A rational function h=f(x)/g(x) have a vertical asymptotes
$\forall x : g \left(x\right) = 0$.

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 $n \mathmr{and} m$
4) if $n = m + 1$, there is an oblique or slant asymptote.

Now letting $f \left(x\right) = \left({x}^{2} - 9 x + 20\right) \mathmr{and} g \left(x\right) = \left(2 {x}^{2} - 8 x\right)$

Vertical:
We see g(x) = 0; for x=> (0, 4) so we find vertical asymptotes at
$x = 0 \mathmr{and} x = 4$

Horizontal:
definition we have $n = m$ thus the ratio of the leading coefficient is the asymptote.
$\frac{{x}^{2} - 9 x + 20}{2 \left({x}^{2} - 4 x\right)} = \frac{1}{2} \frac{{x}^{2} - 9 x + 20}{{x}^{2} - 4 x}$
Thus the Horizontal is asymptote is $y = \frac{1}{2}$