To interpret the vertex of the function \( h(t) = 6t^2 - 24t + 30 \), we first need to determine the time \( t \) at which the height \( h \) is maximized or minimized. Given that this is a quadratic function that opens upwards (since the coefficient of \( t^2 \) is positive), the vertex will give us the minimum height.
The \( t \)-coordinate of the vertex of a quadratic function in the form \( at^2 + bt + c \) can be found using the formula:
\[ t = -\frac{b}{2a} \]
Here, \( a = 6 \) and \( b = -24 \). Plugging these values into the formula gives:
\[ t = -\frac{-24}{2 \cdot 6} = \frac{24}{12} = 2 \]
Now we can find the height at \( t = 2 \) by substituting \( t \) back into the function:
\[ h(2) = 6(2)^2 - 24(2) + 30 \] \[ = 6(4) - 48 + 30 \] \[ = 24 - 48 + 30 \] \[ = 6 \]
So the vertex is at the point \( (2, 6) \), which means at \( t = 2 \) seconds, the owl is at a minimum height of 6 feet.
Thus, the best interpretation of the vertex among the options you provided is:
(2, 6).