How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x^2+4x+3)/(x^2 - 9)#?

Question

Lily Anderson · Accepted Answer

using lim calculus

vertical asymptote: first find domain: #x^2-9!=0# then #x!=+-3# then calculate #lim f(x) #when #x->+3# and it is #24/0 =oo# but is not the same for #x->-3# so you have vertical asymptote x=3 horizontal asymptote; calculate #lim f(x)# when #x->oo# that's 1 so you have horizontal asymptote for y=1 when you have horizontal asymptote you haven't oblique ones

Sophie Adams · Answer

vertical asymptote x = 3
horizontal asymptote y = 1

The first step is to factorise and simplify f(x). #f(x)=(cancel((x+3))(x+1))/(cancel((x+3))(x-3))=(x+1)/(x-3)# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : x - 3 = 0 → x = 3 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x #(x/x+1/x)/(x/x-3/x)=(1+1/x)/(1-3/x)# as #xto+-oo,f(x)to(1+0)/(1-0)# #rArry=1" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ). Hence there are no oblique asymptotes. graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}

Serenity Jones · Answer

undefined To find the vertical asymptotes of the function $ f(x) = \frac{x^2 + 4x + 3}{x^2 - 9} $, we need to determine where the denominator equals zero but the numerator does not. In this case, the denominator $ x^2 - 9 $ equals zero when $ x = -3 $ or $ x = 3 $. However, we need to check if these values make the numerator zero as well. Since neither $ x = -3 $ nor $ x = 3 $ make the numerator $ x^2 + 4x + 3 $ zero, these are vertical asymptotes.

To find the horizontal asymptote, we look at the degrees of the numerator and denominator. Since both have the same degree (2), we look at the leading coefficients of the terms with the highest power. For the numerator, the leading coefficient is 1, and for the denominator, it is also 1. Therefore, the horizontal asymptote is $ y = \frac{1}{1} = 1 $.

For the oblique asymptote, we perform long division or polynomial division on the function to simplify it. After dividing $ x^2 + 4x + 3 $ by $ x^2 - 9 $, we get $ 1 + \frac{13x + 3}{x^2 - 9} $. As $ x $ approaches infinity, the term $ \frac{13x + 3}{x^2 - 9} $ approaches zero, so the oblique asymptote is $ y = 1 $.