# How do you find the global extreme values for  y=x^2-6x-1 on [-2,2]?

Jul 10, 2015

${f}_{\min} \left(2\right) = - 9$

${f}_{\max} \left(- 2\right) = 15$

#### Explanation:

The possible extreme points of such a function are:
1) vertex of the parabola $V \left(p , q\right)$
2) end points of the interval (in this case $a = - 2 , b = 2$)

First we calculate the vertex:

$p = - \frac{b}{2 a} = \frac{6}{2} = 3$

3 does not belong to the interval mentioned in the task, so we don't have to calculate $q$, we know, that the extreme values are at the ends of the interval.

So we calculate:

$f \left(- 2\right) = 4 + 12 - 1 = 15$

$f \left(2\right) = 4 - 12 - 1 = - 9$

Now we can write the answer.

${f}_{\min} \left(2\right) = - 9$

${f}_{\max} \left(- 2\right) = 15$