Homework Help: pre-calc

Posted by steph on Thursday, November 10, 2011 at 8:10pm.

path of a frogs leap: a frog leaps from a box 3 feet high and lands 4 feet from the base of the stump. we can consider the initial position of the frog to be at (0,3) and its landing position to be at (4,0). it is determined that the height of the frog as a function of its horizontal distance x from the base of the box is given by: h(x)=-0.5x^2+1.25x+3 where x and h(x) are both in feet.

a)how high was the frog when its horizontal distances from the base of the box was 2 feet?
b)at what two horizontal distances from the base of the box was the frog 3.25 feet above the ground?
c)at what horizontal distance from the base of the box did the frog reach its highest point?
d) what was the maximum height reached by the frog?

pre-calc - Steve, Friday, November 11, 2011 at 10:26am
h(x)=-0.5x^2+1.25x+3

h(2) = 3.5

3.25 = -0.5x^2+1.25x+3
-0.5x^2+1.25x-.25 = 0
x = .22 (going up) or 2.28 (coming down)

a parabola reaches its vertex when x = -b/2a = -1.25/-1 = 1.25

h(1.25) = 3.781

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