## Precalculus: Mathematics for Calculus, 7th Edition

$h(t) = t^2 + 5$ $h(-3) = (-3)^2 +5 = (9) +5 = 14$ $h(6) = (6)^2 + 5 = (36) + 5 = 41$ $h(6) - h (-3) = 41 - 14 = 27$ Net change is an increase by 27