How do you find vertical, horizontal and oblique asymptotes for (x^2 - 2x + 1)/(x)x2−2x+1x?
1 Answer
Apr 3, 2016
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 (x2x−2xx+1x)=x−2+1x As
xto+-oo , 1/x to 0" and y to x - 2 x→±∞,1x→0andy→x−2
rArr y = x - 2 " is the asymptote " ⇒y=x−2 is the asymptote Here is the graph of the function.
graph{(x^2-2x+1)/x [-10, 10, -5, 5]}