# How do you find the minimum and maximum value of y=(x-6)^2+3?

$x = 6$ is the minimum and we don't have a maximum
First we have a square and the plus three. If x is a real number, its square must be non-negative. So, taking this and the fact that three is constant, the minimum value happens when $x - 6 = 0 \implies x = 6$