Hello, Mr_Green!

Another approach . . .

Find the range of the following function: .$\displaystyle y \:= \:\frac{12x^2 + 48}{27-3x^2}$

We see that the graph is symmetric to the y-axis.

So we need consider only $\displaystyle x \geq 0$ and "reflect" the graph.

There is a vertical asymptote at $\displaystyle x = 3$.

We see that: .$\displaystyle \lim_{x\to3^+}\frac{12(x^2 + 4)}{3(9 - x^2)} \:=\:-\infty$

. . .and that: .$\displaystyle \lim_{x\to3^-}\frac{12(x^2+4)}{3(9-x^2)} \:=\:+\infty$

There is a y-intercept $\displaystyle \left(0,\,\frac{16}{9}\right)$ and no x-intercepts.

Since $\displaystyle \lim_{x\to\infty}\frac{12(x^2 + 4)}{3(9 -x^2)} \;=\;\lim_{x\to\infty}\frac{4(1 + \frac{4}{x^2})}{\frac{9}{x^2} - 1} \:=\:\frac{4}{-1} \:=\:-4$

. . there is a horizontal asymptote: .$\displaystyle y\,=\,-4$

The graph looks something like this: Code:

: | :
:* | *:
: | :
: * | * :
: * | * :
: * | * :
: * :
: |16/9 :
: | :
------------------:-------+-------:--------------------
-3: | :3
: | :
- - - - - - - - - : - - - + - - - : - - - - - - - - -
* : -4| : *
* : | : *
* : | : *
* : | : *
* : | : *
: | :
*: | :*

The range appears to be below $\displaystyle \text{-}4$ and above $\displaystyle \frac{16}{9}$.

As ThePerfectHacker said: .$\displaystyle (\text{-}\infty,\,\text{-}4)\, \cup \left[\frac{16}{9},\,\infty\right) $