How do you find the vertical, horizontal or slant asymptotes for #f(x) = (x^3-8)/(x^2-5x+6)#?

1 Answer
Oct 13, 2016

Answer:

Vertical asymptotes: #x=3# and #x=2#

Horizontal asymptotes: None

Slant asymptotes: #y=x+5#

Explanation:

The function #f(x) = (x^3-8)/(x^2-5x+6)# has vertical asymptotes at #x=3# and #x=2#.

Vertical asymptotes:

In order to work out whether a rational function, #(P(x))/(Q(x))#, has any vertical asymptotes, we simply set the denominator equal to #0#. If we can solve the equation, then we have vertical asymptotes, if not, then we don't.

In this case:

#(x^2-5x+6) = 0#

Using any method to solve this equation tells us that #x=3# and #x=2#. Therefore, we know that our two vertical asymptotes exist at #x=3# and #x=2#

Horizontal asymptotes

Horizontal asymptotes occur when the polynomial of the denominator of a rational function has a higher degree than the polynomial of the numerator. If so, then the #x#-axis will be the horizontal asymptote. (The horizontal asymptote may change via translation)

In this function, this doesn't occur, so there are no horizontal asymptotes.

Slant asymptotes

Slant asymptotes occur when the polynomial of the denominator of a rational function has a lower degree than the polynomial of the numerator. In order to find our slant asymptote, we must divide the numerator by the denominator.

If we divide the numerator by the denominator, we get the slant asymptote as #y=x+5#. (Remember that the slant asymptote is the polynomial part of the answer, not the remainder.)

And here's your graph plotted on Desmos. (Although it seems like the graph crosses the first horizontal asymptote, the graph is actually undefined for that part.)