# How do you find the domain and range of f(x)=8x^2-5x+2?

Apr 7, 2017

$x \in \mathbb{R}$
$y \in \mathbb{R} , y \ge \frac{39}{32}$

#### Explanation:

f(x) has no $\textcolor{b l u e}{\text{excluded values}} .$That is values that make the function $\textcolor{b l u e}{\text{undefined}}$ and so all real values of x can be accepted by f(x)

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

To find the values of y in the range we require to find the vertex.

$8 {x}^{2} - 5 x + 2$

$\text{has " a=8,b=-5" and } c = 2$

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a} = - \frac{- 5}{16} = \frac{5}{16}$

To find the corresponding y-coordinate substitute this value into the equation.

${y}_{\textcolor{red}{\text{vertex}}} = 8 {\left(\frac{5}{16}\right)}^{2} - 5 \left(\frac{5}{16}\right) + 2$

$\textcolor{w h i t e}{\times \times} = \frac{25}{32} - \frac{50}{32} + \frac{64}{32} = \frac{39}{32}$

$\Rightarrow \text{vertex } = \left(\frac{5}{16} , \frac{39}{32}\right)$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ge \frac{39}{32}$
graph{8x^2-5x+2 [-12.66, 12.65, -6.33, 6.33]}