# How do you find the domain and range of F(x)= (x-3)^2 +2?

Mar 24, 2017

Domain: $\left\{x | x \in \mathbb{R}\right\}$
Range: $\left\{y | y > 2 , y \in \mathbb{R}\right\}$
The range depends on the direction of opening and the y-coordinate of the vertex. Since $a$ in $y = a {\left(x - p\right)}^{2} + q$ is positive, $f \left(x\right)$ opens up. Therefore, the vertex will be an absolute minimum. The vertex is given by $\left(p , q\right)$ in $y = a {\left(x - p\right)}^{2} + q$, so the y-coordinate of the vertex is $y = 2$. The range of $f \left(x\right)$ is $\left[2 , + \infty\right)$.