To find the corresponding y-values (heights) for the given x-values (time in seconds) using the function \( h = -6t^2 + 33t + 6 \), we simply need to substitute each x-value into the equation and calculate the corresponding h-value.
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For \( t = 0 \): \[ h(0) = -6(0)^2 + 33(0) + 6 = 6 \]
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For \( t = 1 \): \[ h(1) = -6(1)^2 + 33(1) + 6 = -6 + 33 + 6 = 33 \]
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For \( t = 2 \): \[ h(2) = -6(2)^2 + 33(2) + 6 = -6(4) + 66 + 6 = -24 + 66 + 6 = 48 \]
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For \( t = 3 \): \[ h(3) = -6(3)^2 + 33(3) + 6 = -6(9) + 99 + 6 = -54 + 99 + 6 = 51 \]
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For \( t = 4 \): \[ h(4) = -6(4)^2 + 33(4) + 6 = -6(16) + 132 + 6 = -96 + 132 + 6 = 42 \]
Now, we can summarize our results:
- \( t = 0 \) → \( h = 6 \)
- \( t = 1 \) → \( h = 33 \)
- \( t = 2 \) → \( h = 48 \)
- \( t = 3 \) → \( h = 51 \)
- \( t = 4 \) → \( h = 42 \)
Thus, the corresponding y-values (heights) for the x-values (time in seconds) of \( 0, 1, 2, 3, \) and \( 4 \) are as follows:
- For \( t = 0 \): \( 6 \)
- For \( t = 1 \): \( 33 \)
- For \( t = 2 \): \( 48 \)
- For \( t = 3 \): \( 51 \)
- For \( t = 4 \): \( 42 \)