# Thread: SAT HELP: finding range

1. ## SAT HELP: finding range

Hello! I already mentioned this in a previous topic, but anyways, I bought an SAT practice book for getting ready for the big test. It was pretty well formated, except it only has the answers/explanations to random questions. I was just hoping that you guys could help me with a question here or there that I didn't know how to solve. Thanks!

Find the range of the following function:

y = (12x^2 + 48) / (27-3x^2)

2. Originally Posted by Mr_Green

Find the range of the following function:

y = (12x^2 + 48) / (27-3x^2)
The following method is my own, nobody seems to use it, thus if you do not understand I will try to find another one.

We need to find all $\displaystyle y$ such that,
$\displaystyle y=\frac{12x^2+48}{27-3x^2}$
Has a solution in the domain ($\displaystyle x\not = \pm 3$).
Multiply, by the denominator,
$\displaystyle y(27-3x^2)=12x^2+48$
$\displaystyle 27y-3x^2y=12x^2+48$
$\displaystyle (12x^2+3x^2y)=27y-48$
$\displaystyle 3x^2(4+y)=27y-48$
$\displaystyle x^2(4+y)=9y-16$
Note, $\displaystyle y\not = -4$ because otherwise,
$\displaystyle x^2(4-4)=9(-4)+16$---> Contradiction.
Thus, we can divide by it,
$\displaystyle x^2=\frac{9y-16}{y+4}$
In order to have a solution is $\displaystyle x$ we require that,
$\displaystyle \frac{9y-16}{y+4}\geq 0$
Thus, that happens when,
$\displaystyle 9y-16\geq 0 \mbox{ and }y+4>0$
Or,
$\displaystyle 9y-16\leq 0 \mbox{ and }y+4<0$
Solving the pair of inequalities,
$\displaystyle y\geq 16/9 \mbox{ and } y>-4$ (1)
Or,
$\displaystyle y\leq 16/9 \mbox{ and } y<-4$ (2)
The first inequality is really saying that,
$\displaystyle y\geq 16/9$
The second inequality is really saying that,
$\displaystyle y<-4$.
Thus, the range is,
$\displaystyle (-\infty,-4)\cup [16/9,\infty)$

3. ## .......

Thanks. I apreciate u going over it step by step.

I like your method as well.

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