# What is the range of the function f(x) = -3x^2 + 3x - 2?

May 14, 2018

$\left(- \infty , - \frac{5}{4}\right]$

#### Explanation:

$\text{we require to find the vertex and it's nature, that is}$
$\text{maximum or minimum}$

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to obtain this form use "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

$\text{factor out } - 3$

$y = - 3 \left({x}^{2} - x + \frac{2}{3}\right)$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - x$

$y = - 3 \left({x}^{2} + 2 \left(- \frac{1}{2}\right) x \textcolor{red}{+ \frac{1}{4}} \textcolor{red}{- \frac{1}{4}} + \frac{2}{3}\right)$

$\textcolor{w h i t e}{y} = - 3 {\left(x - \frac{1}{2}\right)}^{2} - 3 \left(- \frac{1}{4} + \frac{2}{3}\right)$

$\textcolor{w h i t e}{y} = - 3 {\left(x - \frac{1}{2}\right)}^{2} - \frac{5}{4} \leftarrow \textcolor{red}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{2} , - \frac{5}{4}\right)$

$\text{to determine if vertex is max/min}$

• " if "a>0" then minimum "uuu

• " if "a<0" then maximum "nnn

$\text{here "a=-3<0" hence maximum}$

$\text{range } y \in \left(- \infty , - \frac{5}{4}\right]$
graph{-3x^2+3x-2 [-8.89, 8.89, -4.444, 4.445]}