1Cademy - Finding the Zeros and Specific Heights for $$h(t) = -16t^2 + 32t + 128$$

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How to Solve a Quadratic Equation by Factoring

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Finding the Zeros and Specific Heights for $h(t) = -16t^2 + 32t + 128$

To analyze the height of an object thrown upward, modeled by the polynomial function $h(t) = -16t^2 + 32t + 128$ , we can find its zeros, the time it reaches a specific height, and its height at a specific time. First, to find the zeros which indicate when the object hits the ground, set $h(t) = 0$ . Factoring the equation $-16t^2 + 32t + 128 = 0$ gives $-16(t^2 - 2t - 8) = 0$ , which factors further to $-16(t - 4)(t + 2) = 0$ . This yields $t = 4$ and $t = -2$ . Discarding the negative time, the object hits the ground at 4 seconds. Second, to find when the object is at a specific height of 128 feet, set $h(t) = 128$ . Subtracting 128 from both sides gives $-16t^2 + 32t = 0$ . Factoring gives $-16t(t - 2) = 0$ , yielding $t = 0$ and $t = 2$ . The object is at 128 feet at 0 and 2 seconds. Third, to find the height at a specific time, such as $t = 1$ second, substitute 1 into the function. The result is $h(1) = -16(1)^2 + 32(1) + 128 = 144$ . The object is at 144 feet after 1 second, which is its highest point.

Updated 2026-06-03

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OpenStax Intermediate Algebra

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