State the range of this How do I state the range of the following equation:
$$8x-4x^2$$ using $$c-\frac{b^2}{4a}$$
I'm again unsure of what $a$, $b$, and $c$ are.
 A: In $$\color{red}{-4}x^2+\color{blue}8x+\color{green} 0$$   
we have: $$A=\color{red}{-4},B=\color{blue}8, C=\color{green} 0$$
A: You should know how the formula has been derived. It is done assuming that the quadratic equation is given by
$$ax^2+bx+c=0$$ where $a\ne0$
So, whenever the formula is used it must be remembered that:
$$a = \text{coefficient of } x^2$$
$$b = \text{coefficient of } x$$
$$c = \text{constant term } $$
Here, write your equation as
$$8x-4x^2=-4x^2+8x+0=(-4)x^2+(8)x+(0)$$
which makes the values of $a,b,c$ very clear.
A: Generally, $f(x)=ax^2+bx+c$ has its extreme value at $f'(x)=0$, i.e.
$$2ax+b=0\implies x=-\frac{b}{2a}.$$
Substituting $x=-\frac{b}{2a}$ into $f(x)$ gives:
$$f\left(-\frac{b}{2a}\right)=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+\frac{2ac}{2a}$$
$$=\frac{ab^2}{4a^2}-\frac{2ab^2}{4a^2}+\frac{4a^2c}{4a^2}=\frac{4a^2c-ab^2}{4a^2}=c-\frac{b^2}{4a}.$$
So $f$ has its extremum at the point $\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right).$
Given the extreme value, do you know how to continue in order to find the whole range of $f$? To begin with, do you know what $f$ looks like? What kind of function is $f$?
