Question

Question #ef8d7

Answer 1

`"vertical asymptote at "x=2`
`"slant asymptote "y=x+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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve "x-2=0rArrx=2" is the asymptote"`

Horizontal asymptotes occur when the degree of the numerator `<=` the degree of the denominator. This is not the case here hence there are no horizontal asymptotes.

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is a slant asymptote.

`"dividing out gives"`

`x(x-2)+2(x-2)+5`

`rArrf(x)=(x^2+1)/(x-2)=x+2+5/(x-2)`

`"as " xto+-oo,f(x)tox+2`

`rArry=x+2" is the asymptote"`
graph{(x^2+1)/(x-2) [-40, 40, -20, 20]}