Question

What are the asymptote(s) and hole(s) of `f(x) = (2x-4)/(x^2-3x+2)`?

Answer 1

`"hole at " x=2`
`"vertical asymptote at " x=1`
`"horizontal asymptote at " y=0`

Explanation:

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

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

The removal of the factor (x - 2) indicates a hole at x = 2

The graph of `2/(x-1)` is the same as `(2x-4)/(x^2-3x+2)`

`"but without the hole"`

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve "x-1=0rArrx=1" is the asymptote"`

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-(3x)/x^2+2/x^2)=(2/x^2)/(1-3/x+2/x^2)`

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

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