# How do you find the domain and range of f(x)= x^2- 6x + 8?

Feb 21, 2018

$x \in \mathbb{R} , y \in \left[- 1 , + \infty\right)$

#### Explanation:

$f \left(x\right) \text{ is defined for all real values of x}$

$\Rightarrow \text{domain is } x \in \mathbb{R}$

$\text{to determine the range express "f(x)" in "color(blue)"vertex form}$

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

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

$\text{using the method of "color(blue)"completing the square}$

$f \left(x\right) = {x}^{2} + 2 \left(- 3\right) x \textcolor{red}{+ 9} \textcolor{red}{- 9} + 8$

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

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , - 1\right)$

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

• " if "a>0" then vertex is minimum "uuu

• " if "a<0" then vertex is maximum "nnn

$\text{here "a=1>0rArr" vertex is a minimum}$

$\Rightarrow \text{range is } \left[- 1 , + \infty\right)$
graph{x^2-6x+8 [-10, 10, -5, 5]}