Question

How do you find the asymptotes for `f(x) = (2x-2) /( (x-1)(x^2 + x -1))`?

Answer 1

`"vertical asymptotes at "x~~-1.62" and " x~~0.62`

`"horizontal asymptote at " y=0`

Explanation:

`"the first step is to factorise and simplify f(x)"`

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

`"solve " x^2+x-1=0" using the quadratic formula"`

`x=(-1+-sqrt(1+4))/2`

`rArrx=-1/2+-1/2sqrt5`

`rArrx~~-1.62" and " x~~0.62" are the asymptotes"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" ( a constant)"`

divide terms on numerator/denominator by the highest power of x, that is `x^2`

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

as `xto+-oo,f(x)to0/(1+0-0)`

`rArry=0" is the asymptote"`
graph{2/(x^2+x-1) [-10, 10, -5, 5]}