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

Question

Serenity Johnson · Accepted Answer

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# 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#

James Allen · Answer

undefined To find the vertical, horizontal, and oblique asymptotes for the function $ \frac{x^2 - 9x + 20}{2x^2 - 8x} $:

1. **Vertical Asymptotes**:
   Vertical asymptotes occur where the denominator equals zero, but the numerator doesn't. So, set the denominator equal to zero and solve for $ x $. These solutions will give the vertical asymptotes.

$ 2x^2 - 8x = 0 $
   $ 2x(x - 4) = 0 $
   $ x = 0 $ or $ x = 4 $

Thus, the vertical asymptotes are $ x = 0 $ and $ x = 4 $.

2. **Horizontal Asymptotes**:
   To find horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degree of the numerator is 2 and the degree of the denominator is also 2. So, we divide the leading coefficients: $ \frac{1}{2} $.

Therefore, the horizontal asymptote is $ y = \frac{1}{2} $.

3. **Oblique Asymptote**:
   To find the oblique asymptote, perform polynomial long division. Divide the numerator by the denominator and the quotient obtained will represent the oblique asymptote.

$$ \frac{x^2 - 9x + 20}{2x^2 - 8x} = \frac{1}{2} - \frac{1}{4x} $$

Thus, the oblique asymptote is $ y = \frac{1}{2} - \frac{1}{4x} $.