How do you find the horizontal asymptote for #(3x-2) / (x+1) #?

Question

Serenity Johnson · Accepted Answer

Horizontal asymptotes are always trickier than vertical asymptotes.

To find the horizontal asymptotes we must look at the highest powers in the numerator and the denominator. The highest powers are both #x^1 = x#. When the highest powers in the numerator and the denominator are equal, the asymptote will occur at the ratio between the coefficient of the highest power in the numerator (3) and that of the denominator (1). The horizontal asymptotes is therefore #3/1 = 3# Verification: Plug in 3 for y. #y = (3x - 2)/(x + 1)# #3 = (3x - 2)/(x + 1)# #3(x + 1) = 3x - 2# #3x + 3 = 3x - 2# #0x = -5# #x = -5/0# #x = O/# So, x has no solution because division by 0 is undefined. This confirms that our horizontal asymptote is at #y = 3#. This also defines an asymptote: an asymptote is a vertical or horizontal line on the graph of a function, where the function is undefined. The function will get extremely close, but will never touch this line. Practice exercises: Find the horizontal asymptote of #y = (3x^2 + 4x + 5)/(2x^2 - x - 1)# Challenge problem By intuition, what do you think would be the horizontal asymptote in #y = (3x + 7)/(3x^2 + 11x + 6)#?

James Allen · Answer

undefined To find the horizontal asymptote of the function $ \frac{3x - 2}{x + 1} $, you divide the leading terms of the numerator and denominator. 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, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degree of the numerator (1) is less than the degree of the denominator (1). Therefore, the horizontal asymptote is at $ y = 0 $.