Question

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

Answer 1

Horizontal: `larr y=0 rarr`
Vertical: `uarr x = (-1+-sqrt 5)/2 darr`

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

As `x to +-oo, 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 `(-1+-sqrt5)/2`, the

vertical asymptotes are `x = (-1+-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 `(x-1)`.

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]}