How do you find the vertical, horizontal or slant asymptotes for #f(x) = (3x^2 + 4)/(x+1)#?

Question

Chloe Alexander · Accepted Answer

Vertical asymptote at #x= -1#, Slant asymptote: #y=3x-3# ,
Horizontal asymptote: Absent

#f(x) =(3x^2+4)/(x+1)# Vertical asymptote : Denominator is #0 :. x+1 = 0 or x = -1# Vertical asymptote at #x= -1# Degree i.e maximum power of #x# in numerator is #1# more than denominator, so there is no horizontal asymptote. Slant asymptote: Degree i.e maximum power of #x# in numerator is #1# more than denominator , so there is slant asymptote. By long division of (f(x) we get dividend as #y=3x-3# and remainder #7 :. y = 3x-3# is the slant asymptote. graph{ [-80, 80, -40, 40]} [Ans] = (3x^2+4)/(x+1)

Sadie Ackerman · Answer

undefined To find the vertical, horizontal, or slant asymptotes for $ f(x) = \frac{3x^2 + 4}{x+1} $:

1. **Vertical Asymptotes:** Set the denominator equal to zero and solve for $ x $. Vertical asymptotes occur where the function is undefined.

$ x + 1 = 0 $  
   $ x = -1 $

Therefore, the vertical asymptote is $ x = -1 $.

2. **Horizontal Asymptotes:** Compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degree of the numerator (2) is less than the degree of the denominator (1). So, there is a horizontal asymptote at $ y = 0 $.

3. **Slant Asymptotes (Oblique Asymptotes):** If the degree of the numerator is exactly one more than the degree of the denominator, there may be a slant asymptote. To find it, perform polynomial long division or synthetic division.

$ \frac{3x^2 + 4}{x+1} = 3x + 2 - \frac{2}{x+1} $

The slant asymptote is the equation of the linear term, which is $ y = 3x + 2 $.