How do you find vertical, horizontal and oblique asymptotes for #y=x^2/(x-1)#?

Question

Benjamin Achtenberg · Accepted Answer

vertical asymptote at x = 1
oblique asymptote at y = x + 1

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve: #x-1=0rArrx=1" is the asymptote"# Horizontal asymptotes occur when the degree of the numerator ≤ degree of the denominator. This is not the case here ( numerator-degree 2 , denominator-degree 1 ) Hence there are no horizontal asymptotes. Oblique asymptotes occur when the degree of the numerator > degree of the denominator. Hence there is an oblique asymptote. Polynomial division gives. #y=x+1+1/(x-1)# as #xto+-oo,ytox+1+0# #rArry=x+1" is the asymptote"# graph{(x^2)/(x-1) [-10, 10, -5, 5]}

James Allen · Answer

undefined To find the vertical, horizontal, and oblique asymptotes for $ y = \frac{x^2}{x - 1} $, follow these steps:

1. **Vertical Asymptotes (VA):**
   Vertical asymptotes occur where the denominator of the function becomes zero and the numerator is not zero.
   
   Set the denominator equal to zero and solve for $ x $:
   $$ x - 1 = 0 $$
   $$ x = 1 $$
   Therefore, there is a vertical asymptote at $ x = 1 $.

2. **Horizontal Asymptotes (HA):**
   Horizontal asymptotes occur as $ x $ approaches positive or negative infinity.

To find horizontal asymptotes, compare the degrees 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 degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.
   - If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, the degree of the numerator (2) is equal to the degree of the denominator (1). So, divide the leading coefficients:
   $$ \text{HA} = \frac{\text{Leading coefficient of numerator}}{\text{Leading coefficient of denominator}} $$
   $$ \text{HA} = \frac{1}{1} = 1 $$
   Therefore, there is a horizontal asymptote at $ y = 1 $.

3. **Oblique Asymptote (OA):**
   Oblique asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator.

To find the oblique asymptote, perform polynomial long division of the numerator by the denominator.
   $$ \frac{x^2}{x - 1} = x + \frac{x}{x - 1} $$

As $ x $ approaches positive or negative infinity, the term $ \frac{x}{x - 1} $ approaches zero, leaving the oblique asymptote as the linear term $ y = x $.

In summary:
- Vertical asymptote: $ x = 1 $
- Horizontal asymptote: $ y = 1 $
- Oblique asymptote: $ y = x $