Problem
The function \( h(t)=4+64 t-16 t^{2} \) models the height \( h \), in feet, of a ball thrown in the air, after \( t \) seconds. Part A What is the vertex of the graph of the function, \( (t, h(t)) \) ?
Solution
Step 1 ：The vertex of a quadratic function in the form \( f(t)=a t^{2}+b t+c \) can be found by the formula \( t_{v}=\frac{-b}{2a} \).
Step 2 ：For \( h(t)=4+64 t-16 t^{2} \), we have \( a=-16 \), \( b=64 \), and \( c=4 \).
Step 3 ：Calculate \( t_{v} \): \( t_{v}=\frac{-64}{2(-16)} \), so \( t_{v}=2 \). Find \( h(t_{v}) \): \( h(2)=4+64(2)-16(2^{2}) \), so \( h(2)=68 \). Therefore, the vertex is \( (2, 68) \).
