To find the vertex of the quadratic function \( h(t) = -6t^2 + 33t + 6 \), we can use the vertex formula for a parabola given by \( t = -\frac{b}{2a} \), where \( h(t) = at^2 + bt + c \).
In this case:
- \( a = -6 \)
- \( b = 33 \)
- \( c = 6 \)
Now, we calculate the \( t \)-coordinate of the vertex:
\[ t = -\frac{b}{2a} = -\frac{33}{2 \times -6} = \frac{33}{12} = 2.75 \]
Next, we need to find the \( h \)-coordinate (height) at this time \( t = 2.75 \):
\[ h(2.75) = -6(2.75)^2 + 33(2.75) + 6 \]
Calculating \( (2.75)^2 \):
\[ (2.75)^2 = 7.5625 \] \[ h(2.75) = -6(7.5625) + 33(2.75) + 6 \] \[ = -45.375 + 90.75 + 6 \] \[ = 51.375 - 45.375 = 6.375 \]
Thus, the vertex is at the point \( (2.75, 51.375) \).
Looking at the options provided:
- (0,6)
- (51.375, 2.75)
- (2.75, 51.375)
- (6.676, 0)
The correct ordered pair representing the vertex is:
\[ (2.75, 51.375) \]