Question

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

Answer 1

`"vertical asymptote at " x=1`
`"oblique asymptote " y=2x+3`

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 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"`

Since the degree of the numerator > degree of the denominator there is an oblique asymptote but no horizontal asymptote.

`"using the divisor as a factor in the numerator"`

`color(red)(2x)(x-1)color(magenta)(+2x)+x-1`

`=color(red)(2x)(x-1)color(red)(+3)(x-1)color(magenta)(+3)-1`

`=color(red)(2x)(x-1)color(red)(+3)(x-1)+2`

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

as `xto+-oo,f(x)to2x+3`

`rArry=2x+3" is the asymptote"`
graph{(2x^2+x-1)/(x-1) [-24.98, 24.98, -12.48, 12.5]}