How do you find the Vertical, Horizontal, and Oblique Asymptote given #Q(x) =( 2x^2) / (x^2 - 5x - 6)#?

Question

Serenity Jones · Accepted Answer

Vertical Asymptotes : x=6 and x=-1
Horizontal asymptotes:y=2

# Q(x)=(2x^2)/(x^2-5x-6)# The first step is factor the numerator and denominator if possible . We can factor the denominator (x^2-5x-6=(x-6)(x+1)# #q(x)=(2x^2)/((x-6)(x+1))# Vertical asymptote is a point to which x approaches make the function approaches + or - infinity . So Set denominator as 0, solve for x. #(x-6)(x+1)=0# #(x-6)=0 or (x+1)=0# #x=6 or x=-1# So vertical asymtotes are x=6 and x=-1 Horizontal asymptote is the value of the function when x approches infinity. To find horizontal asymptote we have to use the degree of numerator and denominator. Here both have the same degree y= ( leading coefficient of numerator) /(leading coefficient of denominator) is the horizontal asymptote. #y=2/1# #y=2# If the degree of numerator > degree of the denominator , then we would get the slant asymptote. Here no slant asymptote are there.

Sadie Ackerman · Answer

undefined To find the vertical asymptotes of a rational function, determine the values of $ x $ for which the denominator equals zero. In this case, set $ x^2 - 5x - 6 = 0 $ and solve for $ x $. The roots of this equation represent the values where the function has vertical asymptotes.

To find the horizontal asymptote, compare 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 at $ y = 0 $. If the degrees are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the oblique asymptote, divide the numerator by the denominator using polynomial long division or synthetic division. The quotient obtained from this division represents the equation of the oblique asymptote. If the degrees of the numerator and denominator are equal or if the degree of the numerator is less than the degree of the denominator by one, there is no oblique asymptote.

So, for the given rational function $ Q(x) = \frac{2x^2}{x^2 - 5x - 6} $:

1. Find the values of $ x $ that make the denominator zero to determine the vertical asymptotes.
2. Compare the degrees of the numerator and denominator to find the horizontal asymptote, if it exists.
3. Perform polynomial division to find the oblique asymptote, if it exists.