How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= 1/(x^2-2x+1)#?

Question

Sadie Alexander · Accepted Answer

vertical asymptote x = 1
horizontal asymptote y = 0

The denominator of f(x) cannot be zero as this would be 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^2-2x+1=0rArr(x-1)^2=0rArrx=1# #rArrx=1" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by the highest power of x, that is #x^2# #(1/x^2)/(x^2/x^2-(2x)/x^2+1/x^2)=(1/x^2)/(1-2/x+1/x^2)# as #xto+-oo,f(x)to0/(1-0+0)# #rArry=0" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2 ) Hence there are no oblique asymptotes. graph{(1)/(x^2-2x+1) [-10, 10, -5, 5]}

Sadie Ackerman · Answer

undefined To find the vertical, horizontal, and oblique asymptotes of the function $ f(x) = \frac{1}{x^2 - 2x + 1} $, follow these steps:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $. Any values of $ x $ that make the denominator zero will be vertical asymptotes.
2. Horizontal asymptote: Examine the behavior of the function as $ x $ approaches positive and negative infinity. 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, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.
3. Oblique asymptote: If the degree of the numerator is one greater than the degree of the denominator, perform polynomial long division to find the equation of the oblique asymptote.

Let's apply these steps to the given function:

1. Vertical asymptotes: 
   Set $ x^2 - 2x + 1 = 0 $.
   Solve $ x^2 - 2x + 1 = (x - 1)^2 = 0 $.
   $ x = 1 $ (double root).
   So, there is a vertical asymptote at $ x = 1 $.

2. Horizontal asymptote:
   As $ x $ approaches positive or negative infinity, the function behaves like $ \frac{1}{x^2} $.
   Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $ y = 0 $.

3. Oblique asymptote:
   Perform polynomial long division:
   $ \frac{1}{x^2 - 2x + 1} = \frac{1}{(x - 1)^2} $.
   There's no linear term in the numerator to divide by $ (x - 1)^2 $.
   Therefore, there is no oblique asymptote.

In summary:
- Vertical asymptote: $ x = 1 $
- Horizontal asymptote: $ y = 0 $
- There is no oblique asymptote.