How do you find vertical, horizontal and oblique asymptotes for # f(x)= (x^3+1 )/( x+1)#?

Question

James Allen · Accepted Answer

No asymptotes

# f(x)= (x^3+1 )/( x+1) = 1 - x + x^2# so no asymptotes

Anna Allen · Answer

undefined To find the vertical, horizontal, and oblique asymptotes for the function $ f(x) = \frac{x^3 + 1}{x + 1} $, follow these steps:

1. **Vertical Asymptotes:**
   Vertical asymptotes occur where the denominator of the rational function is equal to zero and the numerator is nonzero. Set the denominator $ x + 1 $ equal to zero and solve for $ x $:
   $$ x + 1 = 0 $$
   $$ x = -1 $$
   Therefore, there is a vertical asymptote at $ x = -1 $.

2. **Horizontal Asymptotes:**
   Horizontal asymptotes can be determined by examining the degrees of the numerator and denominator polynomials. Since the degree of the numerator is equal to the degree of the denominator (both are 1), there is a horizontal asymptote. Divide the leading term of the numerator by the leading term of the denominator:
   $$ \lim_{{x \to \infty}} \frac{x^3 + 1}{x + 1} = \lim_{{x \to \infty}} \frac{x^3}{x} = \lim_{{x \to \infty}} x^2 = \infty $$
   Therefore, there is no horizontal asymptote.

3. **Oblique Asymptotes (Slant Asymptotes):**
   Oblique asymptotes occur when the degree of the numerator is one more than the degree of the denominator. Perform polynomial long division or use synthetic division to divide $ x^3 + 1 $ by $ x + 1 $ to find the oblique asymptote:
   $$
   \begin{array}{c|c}
   x^2 - x + 1 & x \
   \hline
   x^3 + 0x^2 + 0x + 1 & \
   \end{array}
   $$
   The quotient is $ x^2 - x + 1 $.
   Therefore, the oblique asymptote is the line $ y = x^2 - x + 1 $.

In summary:
- Vertical asymptote: $ x = -1 $
- Horizontal asymptote: None
- Oblique asymptote: $ y = x^2 - x + 1 $

Lincoln Anderson · Answer

undefined To find the vertical asymptote(s), set the denominator equal to zero and solve for $x$. In this case, $x + 1 = 0$ gives $x = -1$. So, there is a vertical asymptote at $x = -1$.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (1), there is no horizontal asymptote.

To determine if there are oblique asymptotes, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient will be the equation of the oblique asymptote if the degree of the numerator is exactly one greater than the degree of the denominator. In this case, the degrees are equal, so there are no oblique asymptotes.