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

1 Answer
Apr 3, 2016

Answer:

vertical asymptote x = 0
oblique asymptote y = x - 2

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

→ x = 0 is the asymptote

Horizontal asymptotes occur when the degree of the numerator ≤ degree of the denominator. This is not the case here hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here.

divide numerator by x

# (x^2/x - (2x)/x + 1/x) = x - 2 + 1/x #

As # xto+-oo , 1/x to 0" and y to x - 2 #

#rArr y = x - 2 " is the asymptote " #

Here is the graph of the function.
graph{(x^2-2x+1)/x [-10, 10, -5, 5]}