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What is the maximum value of h(t)=−16t^2+32t+64

What is the maximum value of h(t)=−16t^2+32t+64
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4 votes
What is the maximum value of h(t)=−16t^2+32t+64

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User Ozkanpakdil
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7.8k points

1 Answer

4 votes

Answer:

Explanation:

h(t)=-16t²+32t+64

=-16(t²-2t)+64

=-16(t²-2t+1-1)+64

=-16(t-1)²+16+64

=-16(t-1)²+80

maximum value=80 and is obtained at t=1

answered
User Mark Hughes
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8.2k points
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