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

1 Answer
Mar 26, 2016

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#

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#