Question

For the following function: g(x) = (3x+9) / (x^2-x-12) (a) Find the domain. Answer .................. (b) horizontal asymptotes? (c) vertical asymptotes ? (d) is there discontinuity ?

Answer 1

What you start with is to ensure that you denominator isn't equal to zero. You do not want to have a division by zero:
So you set:
`x^2-x-12!=0` solving the equation it gives you that must be:
`x!=4` and `x!=-3`
But this gives you an interesting twist!
Your function can be written as:
`g(x)=[3(x+3)]/[(x+3)(x-4)]=3/(x-4)`!!!

So your function has a point of discontinuity which is where the denominator is equal to zero, i.e., `x=4`, which also gives you the position of your vertical asymptote!

To find the horizontal asymptote(s) you do the limit when `x` becomes very large:
`lim_(x->+oo)[g(x)]=0`
and
`lim_(x->-oo)[g(x)]=0`

So the x axis (equation: `y=0`) is your horizontal asymptote!

Summarizing:
Range: all real values of `x` except `x=4` which is a point of discontinuity of your function;
Vertical asymptote: at `x=4`
Horizontal asymptote: `y=0`

Answer 2

`g(x)=(3x+9)/(x^2-x-12)`

Domain:
All reals except solutions to `x^2-x-12=0`

`x^2-x-12=(x-4)(x+3)`

So the domain is all real numbers except `-3` and `4`. In interval notation: `(-oo, -3)uu(-1, 4)uu(4, oo)`

Horizontal asymptotes:
`lim_(xrarroo)(3x+9)/(x^2-x-12) = lim_(xrarroo)(x^2(3/x+9/x^2))/(x^2(1-1/x-12/x^2)) = lim_(xrarroo)(3/x+9/x^2)/(1-1/x-12/x^2) = 0`

So the line `y=0` is a horizontal asymptote on the right.

The same reasoning show that `lim_(xrarr-oo)(3x+9)/(x^2-x-12) =0`

So the line `y=0` is a horizontal asymptote on the left as well.

Vertical asymptotes
`g(x)=(3x+9)/(x^2-x-12) = (3(x+3)) / ((x-4)(x+3))`

`lim_(xrarr-3)g(x)=lim_(xrarr-3)(3(x+3)) / ((x-4)(x+3))=3/(3-4)=-3 !=oo`

So there is no vertical asymptote at -3.

`lim_(xrarr4^+)g(x)=lim_(xrarr4^+)3 / (x-4)=oo`

So the line `x=4` is a vertical asymptote

(And `lim_(xrarr4^-)g(x)=lim_(xrarr4^+)3 / (x-4)= -oo`)

Discontinuities

There is a discontinuity at `-3`. (Removable because the limit exists.)

And one at `4` non-removable because the limit does not exist. (In fact it is an infinite discontinuity.)