# What are the extrema of #f(x)=-8x^2+x# on #[-4,8]#?

##### 1 Answer

#### Answer:

Absolute minimum of

#### Explanation:

When finding the extrema on an interval, there are two locations they could be: at a critical value, or at one of the endpoints of the interval.

To find the critical values, find the function's derivative and set it equal to

Thus, our locations for potential maxima and minima are at

#f(-4)=-8(-4)^2-4=ul(-132)#

#f(1/16)=-8(1/16)^2+1/16=-1/32+1/16=ul(1/32)#

#f(8)=-8(8)^2+8=ul(-504)#

Since the highest value is

This is

graph{-8x^2+x [-4.1, 8.1, -550, 50]}