The path of flight (trajectory) of a football thrown by a quarterback is described by the following function: $y ( x ) = - 0.002 x ^ { 2 } + 0.7 x + 7$ where $y =$ vertical position of football relative to the ground ( $\mathrm { ft } )$
$x =$ horizontal position of football relative to launch position $( \mathrm { ft } )$ How high above the ground is the football as it leaves the quarterback's hand?
A) 0.002 ft
B) 0.7 ft
C) 7 ft
D) 7.7 ft

Question

Quizplus · Accepted Answer

The football's height as it leaves the quarterback's hand is given by the value of $y$ when $x = 0$. Substituting $x = 0$ into the equation $y(x) = -0.002x^2 + 0.7x + 7$ gives $y(0) = 7$, indicating the football is 7 ft above the ground at launch.

The Path of Flight (Trajectory) of a Football Thrown by a Quarterback