Question

Question #3f59b

Answer 1

See below.

Explanation:

`(x^2-6x+8)/(x^2-x-2)`

Factor numerator and denominator:

`((x-4)(x-2))/((x+1)(x-2)`

We are able to cancel `(x-2)`, This shows that this is a removable discontinuity.

So there is a hole at `x=2`

`((x-4)cancel((x-2)))/((x+1)cancel((x-2)))=(x-4)/(x+1)`

Vertical asymptotes occur where the function is undefined.

`(x-4)/(x+1)`

is undefined for `x=-1` ( division by 0 )

So, the line `color(blue)(x=-1)` is a vertical asymptote.

Domain will therefore be:

`{x in RR : x != -1}`

`(x-4)/(x+1)`

Divide by `x`:

`(x/x-4/x)/(x/x+1/x)=(1-4/x)/(1+1/x)`

as `x->-oo` , `color(white)(888)(1-4/x)/(1+1/x)->(1-0)/(1+0)=1`

as `x->oo` , `color(white)(888888)(1-4/x)/(1+1/x)->(1-0)/(1+0)=1`

The line `color(blue)(y=1)` is a horizontal asymptote.

GRAPH: