The function \( h = 6t^2 - 24t + 30 \) is a quadratic equation in standard form \( h(t) = at^2 + bt + c \), where \( a = 6 \), \( b = -24 \), and \( c = 30 \).
To find the vertex of the parabola represented by this function, we can use the vertex formula for a quadratic function, which gives us the \( t \)-coordinate of the vertex as:
\[ t = -\frac{b}{2a} \]
Substituting in the values of \( a \) and \( b \):
\[ t = -\frac{-24}{2 \times 6} = \frac{24}{12} = 2 \]
Next, we substitute \( t = 2 \) back into the height function \( h(t) \):
\[ h(2) = 6(2)^2 - 24(2) + 30 \] \[ h(2) = 6(4) - 48 + 30 \] \[ h(2) = 24 - 48 + 30 = 6 \]
Thus, the vertex of the function is at the point \( (2, 6) \).
Interpreting this in the context of the problem, the vertex \( (2, 6) \) indicates that after 2 seconds, the owl reaches its maximum height of 6 feet before descending.
Therefore, the option that best interprets the vertex is \( (2, 6) \).
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