Find the Vertical and horizontal asymptotes of a function?

Question

(x^2-2x+1)/(x^3-x^2+x-1), Please help me find the vertical and horizontal asymptotes.

Answer 1

Hi,

Vertical asymptotes occur when there are factors in the denominator that can be set equal to zero and solved that are not also in the numerator. When the same factor is in both the numerator and denominator, there is not a vertical asymptote, but instead there is a hole in the graph at the x value that is that factor's solution.

x² - 2x + 1

------------------- =

x³ - x² + x - 1

(x - 1)²

-------------------

(x² + 1)(x - 1)

The factor (x - 1) is in both the numerator and denominator, so when x - 1 = 0 and solves to x = 1, there is not a vertical asymptote there, but there is a hole in the graph there.

When the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is always at y = 0.

This rational expression has no vertical asymptote, a hole in the graph when x = 1, and a horizontal asymptote at y = 0. <==ANSWER

You can verify this by looking at the graph. The graph does not shoot up or down toward infinity near x = 1 like a vertical asymptote would do. Instead, if you look at the simplified rational expression,

x - 1

---------, when x = 1 this expression equals 0, so the graph is a

x² + 1

continuous line through the missing point (1,0).

I hope that helps!! :-)

Answer 2

Vertical is when the one on the bottom = 0

x^3 - x^2 + x -1 = 0

(x-1) (x^2 - x +1) = 0

x = 1

Horizontal is when you let x = +/- infinity

inf^2/inf^3 = 1/inf = 0

-inf^2/-inf^3 = 1/-inf = 0

y=0

The vertical asymptote is x=1, the horizontal is y = 0.