Here's the question you clicked on:
burhan101
Determine the extreme values of this function on the given interval. f(x)= 1+ (x+3)² ; -2<x<6
Calc or pre-calc?
pre-cal @SithsAndGiggles
Ok, that method's a bit less work. So you're given a parabola in vertex form: \[f(x)=1+(x+3)^2\] \(f(x)\) has vertex (an extreme value point) at \((-3,1)\). \(-3\) is not in the interval \(-2<x<6\). So, you have something that looks like this: |dw:1371696732887:dw| This means that only the endpoints of this interval will contain the extreme values.
Which means you have only to check \(f(-2)\) and \(f(6)\).
where did you get (-3,1) from ? :S