To find the vertex of the quadratic function given by \(h(t) = -6t^2 + 33t + 6\), we can use the formula for the t-coordinate of the vertex, which is given by:
\[ t = -\frac{b}{2a} \]
where \(a\) is the coefficient of \(t^2\) and \(b\) is the coefficient of \(t\).
In your function:
- \(a = -6\)
- \(b = 33\)
Plugging in the values:
\[ t = -\frac{33}{2(-6)} = \frac{33}{12} = 2.75 \]
Next, we need to find the height \(h\) at \(t = 2.75\).
Substituting \(t = 2.75\) back into the function:
\[ h(2.75) = -6(2.75)^2 + 33(2.75) + 6 \] Calculating \( (2.75)^2 \):
\[ (2.75)^2 = 7.5625 \] Now substituting back into the equation:
\[ h(2.75) = -6(7.5625) + 33(2.75) + 6 \] Calculate \(-6(7.5625)\):
\[ -6(7.5625) = -45.375 \] Calculate \(33(2.75)\):
\[ 33(2.75) = 90.75 \] Now combine these values:
\[ h(2.75) = -45.375 + 90.75 + 6 \] Calculating this gives:
\[ h(2.75) = 51.375 \]
Thus, the vertex of the function is at:
\[ (2.75, 51.375) \]
Therefore, the correct response is:
(2.75, 51.375)
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