# How do you find the domain and range of y=2x^2-4x+3?

Jul 2, 2018

Range: $y \ge 1$
Domain: $x \in \mathbb{R}$ (or "all x")

#### Explanation:

the domain is "all x" ($x \in \mathbb{R}$) because it doesn't matter which x you there, you will get "some y".

in order to find the range of the function, giving that it is "a smiling parabola" ($a > 0$), you just need to find the min (${x}_{\min} = - \frac{b}{2 a}$):

${x}_{\min} = - \frac{b}{2 a} = \frac{4}{2 \cdot 2} = 1$

now find ${y}_{\min} = 2 \cdot {1}^{2} - 4 \cdot 1 + 3 = 2 - 4 + 3 = - 2 + 3 = 1$

so the min is $\left(1 , 1\right)$, so the range is every y equal or above ($y \ge 1$)

Jul 2, 2018

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

#### Explanation:

$\text{this is a polynomial of degree 2 and is defined for all real}$
$\text{values of } x$

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

$\text{to find the range we require the vertex and whether it is}$
$\text{a maximum or minimum turning point}$

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

•color(white)(x)y=a(x-h)^2+k

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

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

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

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

$\textcolor{w h i t e}{y} = 2 {\left(x - 1\right)}^{2} + 1 \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

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

$\text{Since "a>0" then minimum turning point } \bigcup$

$\text{range is } y \in \left[1 , \infty\right)$
graph{2x^2-4x+3 [-10, 10, -5, 5]}