# How do you find the minimum and maximum value of y=3(x-3)^2-3?

Minimum at $\left(3 , - 3\right)$
$y = a {\left(x - h\right)}^{2} + k$ => Eqution of parabola in vertex form, where $\left(h , k\right)$ is the vertex. For positive coefficient of squared term or a the parabola opens up, and if negative down. We have a minimum for upward and a maximum for down. So in this case:
$y = 3 {\left(x - 3\right)}^{2} - 3$
The parabola has a minimum value at vertex $\left(3 , - 3\right)$