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

1 Answer
Dec 29, 2016

Answer:

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]}