# What is the range of f(x) = 2x^2 + 1?

Jul 29, 2018

$y \in \left[1 , \infty\right)$

#### Explanation:

$\text{We require the vertex and whether it is a max/min}$
$\text{turning point}$

$y = 2 {x}^{2} + 1$

$\text{is symmetrical about the y-axis and the coordinates of}$
$\text{it's vertex are}$

$\text{vertex } = \left(0 , 1\right)$

$\text{since "a>0" then it has a minimum turning point } \bigcup$

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