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

Dec 29, 2016

Horizontal: $\leftarrow y = 0 \rightarrow$
Vertical: $\uparrow x = \frac{- 1 \pm \sqrt{5}}{2} \downarrow$

#### Explanation:

If y = f(x)/g(x) and both f and g have the same factor h(x), then

y= (f(x)/(h(x))((g(x)/h(x).

Here. $y = \frac{\left(2\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + x - 1\right)} = \frac{2}{{x}^{2} + x - 1}$

As $x \to \pm \infty , y \to 0$.

So, y = 0 is the horizontal asymptote.

As the [zeros](https://socratic.org/precalculus/polynomial-functions-

of-higher-degree/zeros) of ${x}^{2} + x - 1 = 0$ are $\frac{- 1 \pm \sqrt{5}}{2}$, the

vertical asymptotes are $x = \frac{- 1 \pm \sqrt{5}}{2}$.

The degree of the numerator is 0 and the the degree of the

denominator is 1 that is higher. So, there is no possibility of another

asymptote,

The two graphs are for the given function and the function that is

obtained after cancelling the common factor $\left(x - 1\right)$.

The asymptote y = 0 is also marked in the graphs.

Of course, it was not possible ( for me ) to mark vertical asymptotes, using this utility.

graph{y(y-(2x-2)/((x-1)(x^2+x-1) ))=0[-10, 10, -5, 5]}

graph{y(y-2/(x^2+x-1))=0 [-10, 10, -5, 5]}