How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(x^2 + 1)/ (x-2)^2#?

Question

Caleb Alexander · Accepted Answer

#VA: x=2#
#HA: y=1# #f(x)= (x^2 +1) / (x-2)^2# #f(x)= (x^2 +1) /((x-2)(x-2))# *(https://tutor.hix.ai) #VA: x=2# #HA: y=1/1=1#

Serenity Johnson · Answer

undefined To find the vertical asymptotes of the function $ f(x) = \frac{x^2 + 1}{(x - 2)^2} $, we need to look for the values of $ x $ where the denominator becomes zero, but the numerator doesn't. In this case, the denominator becomes zero when $ x = 2 $. Therefore, there is a vertical asymptote at $ x = 2 $.

To find the horizontal asymptote, we examine the degrees of the numerator and the denominator. Since the degree of the numerator (2) is equal to the degree of the denominator (2), we divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. In this case, it's $ y = 1 $.

To find the slant asymptote, we divide the numerator by the denominator using long division or polynomial division.

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

Performing polynomial division:

$$ x^2 + 1 \, \text{divided by} \, x^2 - 4x + 4 $$

We find that the quotient is $ 1 $ and the remainder is $ 4x - 3 $. Since the degree of the remainder (1) is less than the degree of the denominator (2), the slant asymptote is a linear function. Therefore, the slant asymptote is $ y = x + 1 $.