DELTA MATH HELP

Question

Given the function h(x)=−x^2−5x+8, determine the average rate of change of the function over the interval −4≤x≤−2.

Answer 1

h(-2)=14

h(-4)=12

(14-12)/(-2+4)= 2/2= 1

= secant line slope

= average rate of change

between (-4, 12) and (-2,14)

-x^2-5x+8

-(x+5x+25/4)+8+25/4

=-(x+5/2)^2+ 57/4 is a downward opening parabola

x=-2.5 is axis of smmetry

(-2.5, 18.25) = vertex=max