State the range of this

Question

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.

Answer 1

In $$\color{red}{-4}x^2+\color{blue}8x+\color{green} 0$$

we have: $$A=\color{red}{-4},B=\color{blue}8, C=\color{green} 0$$

Answer 2

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.

Answer 3

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 $f$?