Question

How do you identify all asymptotes or holes for `f(x)=(x^2-x)/(2x^2+4x-6)`?

Answer 1

hole at `(1,1/8)`
vertical asymptote at x = 3
horizontal asymptote at `y=1/2`

Explanation:

Holes occur when a duplicate factor is cancelled in the numerator/denominator of f(x).
The first step is therefore to factorise and simplify f(x).

`f(x)=(xcancel((x-1)))/(2(x+3)cancel((x-1)))=x/(2(x+3))`

set `x-1=0rArrx=1`

substitute x = 1 into 'simplified' version of f(x)

`rArrf(1)=1/(2(1+3))=1/8`

`rArr"there is a hole at"(1,1/8)`

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: `2(x+3)=0rarrx=-3" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`f(x)=(x/x)/((2x)/x+6/x)=1/(2+6/x)`

as `xto+-oo,f(x)to1/(2+0)`

`rArry=1/2" is the asymptote"`
graph{(x^2-x)/(2x^2+4x-6) [-10, 10, -5, 5]}