To interpret the vertex of the function \( h = 6t^2 - 24t + 30 \), we first need to find the vertex of the quadratic equation. The vertex of a parabola represented by the equation \( h = at^2 + bt + c \) can be calculated using the formula for the time \( t \) at the vertex, given by:
\[ t = -\frac{b}{2a} \]
For our equation:
- \( a = 6 \)
- \( b = -24 \)
Now, substituting the values into the formula:
\[ t = -\frac{-24}{2 \cdot 6} = \frac{24}{12} = 2 \]
Now we will find the height \( h \) at \( t = 2 \) by substituting \( t = 2 \) back into the original equation:
\[ h = 6(2)^2 - 24(2) + 30 \] \[ h = 6(4) - 48 + 30 \] \[ h = 24 - 48 + 30 \] \[ h = 6 \]
Thus, the vertex of the parabola occurs at the point \( (2, 6) \).
In the context of the problem, this means that at \( t = 2 \) seconds, the owl is at its highest point of 6 feet above the ground. Therefore, the best interpretation of the vertex is:
(2,6)