Question

Question #b0b54

Answer 1

See below.

Explanation:

graph{3x/((x-2)(x+5)) [-32.46, 32.48, -16.24, 16.22]}

For the domain we need to solve the inequalities `(x-2)(x+5)>0` and `(x-2)(x+5)<0`

Since we can't have a zero in the denominator.

`(x-2)(x+5)>0=> x>2 , x<-5`

`(x-2)(x+5)<0=> x> -5 , x < 2`

This gives:

`{x in RR : -oo < x < -5}uu{x in RR : -5 < x < 2}uu{x in RR : 2 < x < oo}`

This could also be expressed:

`{x in RR : x != 2 and x != -5 }`

`y` axis intercepts occur where `x = 0`

`(3(0))/((0-2)(0+5))=0`

Vertical asymptotes occur when the denominator is 0, this is at `x=2` and `x=-5`

Equations of these are:

`x=2` and `x=-5`

As `x` increases without bound in the positive direction:

`x-> oo` , `(3x)/((x-2)(x+5)) -> 0`

And in the negative direction:

`x-> -oo` , `(3x)/((x-2)(x+5)) -> 0`

So:

`y=0` is a horizontal asymptote.