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2 answers:
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Answer:
h(t) = -16(t - 1)² + 64
Step-by-step explanation:
The function that would best to use to identify the maximum height of the ball is the vertex form of the parabola.
The standard form of a quadratic function is
y = ax² + bx + c
The vertex form is
y = a(x - h)² + k
where (h, k) is the vertex of the parabola.
h = -b/(2a) and k = f(h)
In your equation, h(t) = -16t^2 + 32t + 48
a = -16; b = 32; c = 48
Calculate h
h = -32/[2(-16)]
h = (-32)/(-32)
h = 1
Calculate k
k = -16(1)² + 32×1 + 48
k = -16 + 32 + 48
k = 64
So, h = 1, k = 64, a = -16
The vertex form of the equation is h(t) = -16(t - 1)² + 64.
The graph below shows h(t) with the ball at its maximum height of 64 ft one second after being thrown.
