Range of values of f(x) using quadratic inequalities (need intuition) I'm working on an exercise from a book in the chapter on quadratic inequalities: "Find the set of possible values of the given function $\frac{x - 2}{(x + 2)(x - 3)}$". The answer in the book is "all values". I don't have much intuition on the calculations so would be grateful if someone could explain what happens at the various stages. Corrections, qualifications, etc. to my statements are very much welcomed!
Here's my workings:
Looking for the range of values of $y$, so $y = \frac{x - 2}{(x + 2)(x - 3)} \Rightarrow y = \frac{x - 2}{x^2 - x - 6}$
$\Rightarrow y(x^2 - x - 6) = x - 2$
$\Rightarrow yx^2 -(y + 1)x - 6y + 2 = 0$
As far as I can understand from my book, now that I've got a quadratic in $x$, I can use the discriminant to determine the $y$ range . With $a = y, b = -(y + 1), c = (-6y + 2)$, the discriminant ($b^2 - 4ac$) is:
$(-y-1)^2 - 4(y)(-6y + 2)$
$= 25y^2 - 6y + 1$
When $25y^2 - 6y + 1 \geqslant 0$, the roots of the quadratic in $x$ are real. At this stage, from what I understand, the values satisfying the inequality $25y^2 - 6y + 1 \geqslant 0$ are the range of $y$ values for all valid, real $x$ values plugged into the function $\frac{x - 2}{(x + 2)(x - 3)}$.
Solving the inequality using the quadratic formula with $a = 25, b = -6, c = 1$:
roots: $\frac{6\pm\sqrt{36 - (4)(25)(1)}}{50}$
$ = \frac{6\pm\sqrt{-64}}{50}$. Because the discriminant here $< 0$ then there are no real roots. I don't get how this is interpreted as "all values" (i.e. the answer in the book).
 A: Since there are no real roots, the graph of $25y^2−6y+1$ cannot cross the horizontal axis and has to be always negative or always positive. Since it is positive for $y=0$, it is always positive.
Since this is the expression for your original equation, this means that you can always find a solution $x$ which gives this value of $y$ (although you should check the trivial fact that this never gives a 0-denominator).
A: This answer is just to point out another way to find the range, since I believe your specific questions have been answered by user9325.
Consider $f(x)=\frac{x-2}{(x+2)(x-3)}$ on the interval $(-2,3)$, where it is everywhere defined and continuous.  Since $x+2$ is always positive on this interval and $x-3$ is always negative, the sign is opposite that of $x-2$.  
As $x$ approaches $3$ (from the left), $x-2$ will be positive, so $f(x)$ will be negative.  The denominator approaches $0$ while the numerator approaches $1$, so the absolute value of $f$ will go to $\infty$.  Thus $f(x)\to-\infty$ as $x\to 3$ from the left.
As $x$ approaches $-2$ (from the right), $x-2$ will be negative, so $f(x)$ will be positive.  The denominator approaches $0$ while the numerator approaches $-4$, so the absolute value of $f$ will go to $\infty$.  Thus $f(x)\to+\infty$ as $x\to-2$ from the right.
Because $f$ is continuous on $(-2,3)$ and takes on arbitrarily large positive and negative values on that interval, the range of $f$ is all real numbers by the Intermediate Value Theorem.
A: Not sure I can directly answer your final question, since I've forgotten what I've learn about how to apply discriminants in your case.  However...
It is clear from inspection of the factored form of the original "y = ..." equation, that the range of y is indeed all values of y from -infinity to +infinity, by observing the behavior of the function for x = -2 +/- dx, x = 2, and x = 3 +/- dx.
A: You can rewrite
$25y^2−6y+1$
as
$25((y - 3/25)^2 + 16/25^2)$
which is clearly > 0 for any value of y
