Question in my text book
Solve for range of the function, $$y=\frac{x^2+4x-1}{3x^2+12x +20}$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x \in \mathbb{R}$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$
Best Answer
Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$ 2x^2+8x+19=0,$$ which has roots $$x=\frac{-8\pm \sqrt{64-152}}{4}$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.
The method suggested in the book, is exactly this, with a general $y$ in place of $1$.