# How do you find the domain and range for y=5/4(x+2)^2 -1 ?

Jul 11, 2018

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

#### Explanation:

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

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

$\left(- \infty , \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$

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

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

•color(white)(x)y=a(x-h)^2+k

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

$y = \frac{5}{4} {\left(x + 2\right)}^{2} - 1 \text{ is in this form}$

$\text{with "(h,k)=(-2,-1)larrcolor(red)"vertex}$

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

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