Question

How do you identify all asymptotes or holes and intercepts for `f(x)=(x^2+1)/((x-1)(2x-4))`?

Answer 1

`"vertical asymptotes at "x=1" and "x=2`
`"horizontal asymptote at "y=1/2`

Explanation:

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-1)(2x-4)=0`

`rArrx=1" and "x=2" are the asymptotes"`

`"horizontal asymptotes occur as"`

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

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

`f(x)=(x^2/x^2+1/x^2)/((2x^2)/x^2-(6x)/x^2+4/x^2)=(1+1/x^2)/(2-6/x+4/x^2)`

as `xto+-oo,f(x)to(1+0)/(2-0+0)`

`rArry=1/2" is the asymptote"`

Holes occur when there are common factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{(x^2+1)/(2x^2-6x+4) [-20, 20, -10, 10]}