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

Question

Cora Alexander · Accepted Answer

Vertical Asymptotes are based on any factors in the Denominator while Horizontal/Slant Asymptote is based on the highest power of #x# in the Numerator and Denominator.

Vertical Asymptote: Solve for #x# for each factor in the Denominator. #x + 3 = 0# and #x - 4 = 0# So Vertical Asymptote will appear at #x = -3# and #x = 4# Horizontal Asymptote: Let's say #f(x) = (ax^n)/(bx^m)# If #n = m#, then Horizontal Asymptote is #y = a/b# (simplified). If #n < m#, then Horizontal Asymptote is #y = 0#. If #n > m#, then Horizontal Asymptote is None because it doesn't exist. Slant Asymptote only occurs if #n = m + 1#. We would have to use Long Division or Synthetic Division to find the Linear Slant Asymptote. For the problem above, multiply the bottom factors: #x/(x^2 -x - 12)# Top Power of 1 #<# Bottom Power of 2. So Horizontal Asymptote is #y = 0#

Theodore Amidon · Answer

undefined To find the vertical asymptotes of the function $ f(x) = \frac{x}{(x+3)(x-4)} $, set the denominator equal to zero and solve for $ x $. The vertical asymptotes occur at the values of $ x $ that make the denominator zero. In this case, set $ (x+3)(x-4) = 0 $ and solve for $ x $. You'll find that $ x = -3 $ and $ x = 4 $.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since both have the same degree (1), divide the leading coefficient of the numerator by the leading coefficient of the denominator. The horizontal asymptote is the resulting value, which is $ y = 1 $.

Since the degree of the numerator is less than the degree of the denominator, there is no slant asymptote for this function.