The height, h, in feet, of the rocket above the ground at time t seconds can be modeled by the equation h=−16t2+32t+4. What is the maximum height the rocket reaches?

Question

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Answer 1

h=−16t^2+32t+4

The quadratic equation is in the form of

ax^2 + bx +c

So

a = - 16; b = 32 and c = 4

t max of the vertex will have the formula of - b/2a

Substitute with a and b from the equation

t max = - 32/(2*(-16))

= - 32/(-32)

= 1

Maximum height of the rocket will be reached when t = 1. Put t =1 into the h - equation

h=−16t^2+32t+4

= - 16 * 1 ^2 + 32 * 1 +4

= - 16 + 32 + 4

= 20

The maximum height the rocket can reach is 20feet

Answer 2

h(t) =−16t^2+32t+4; h(t) is quadratic which means that it is shaped like a parabola. We understand that parabolas are symmetric around the vertical line passing through their vertices. We can understand that, because the leading coefficient in this quadratic expression is negative (-16) that the parabola is concave down or points down, and thus the vertex is at the maximum not the minimum. If we plug a = -16, b=32 and c = 4, the coefficients of the quadratic expression, into the quadratic formula, we wind up with the roots of this quadratic expression (the values of x where h = 0) are -(b/2a) +/- (something which doesn't matter) which tells us that the axis of symmetry is at x = (-b/2a) = -32/2(-16) = 1. h(1) is−16(1^2)+32(1)+4 = 36-16 = 20 FEET