How do you find the vertical, horizontal or slant asymptotes for #(x-4)/(x^2-3x-4)#?

Question

Chloe Alexander · Accepted Answer

The vertical asymptote is #x=-1#
The horizontal asymptote is #y=0#
There is no slant asymptote

We can factorise the denominator #x^2-3x-4=(x+1)(x-4)# #:.(x-4)/(x^2-3x-4)=cancel(x-4)/((x+1)cancel(x-4))=1/(x+1)# So the vertical asymptote is #x=-1# as we cannot divide by #0# Limit #1/(x+1)=0^-# #x->-oo# Limit #1/(x+1)=0^+# #x->+oo# So the horizontal asymptote is #y=0# And the intercept with the y axis is #(0,1)# As the degree of the numerator is #<# the degree of the denominator, there is no slant asymptote graph{1/(x+1) [-12.66, 12.65, -6.33, 6.33]}

James Allen · Answer

undefined To find the vertical asymptotes of a rational function, we look for values of $ x $ that make the denominator zero, but not the numerator. These values are the vertical asymptotes.

To find horizontal or slant asymptotes, we examine the degrees of the numerator and denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the $ x $-axis (y = 0). If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator by exactly one, there is a slant asymptote, which can be found using polynomial long division.

For the given function $ \frac{x - 4}{x^2 - 3x - 4} $, the vertical asymptote can be found by setting the denominator equal to zero and solving for $ x $.

$$ x^2 - 3x - 4 = 0 $$

This quadratic equation can be factored as $ (x - 4)(x + 1) = 0 $, so the vertical asymptotes occur at $ x = 4 $ and $ x = -1 $.

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the $ x $-axis (y = 0). There is no slant asymptote because the degree of the numerator is not greater than the degree of the denominator by exactly one.