The height $y$ (in feet) of a ball thrown by a child is
$$y=-\frac{1}{12}{x}^2+2x+5$$
where $x$ is the horizontal distance in feet from the point at which the ball is thrown.
(a) How high is the ball when it leaves the child's hand? (Hint: Find $y$ when $x=0$.) feet
(b) What is the maximum height of the ball?
feet
(c) How far from the child does the ball strike the ground? feet

Step 1 ：The height of the ball when it leaves the child's hand is given by the value of $y$ when $x=0$. Substituting $x=0$ into the equation gives $y=-\frac{1}{12}\cdot 0^2+2\cdot 0+5$.

Step 2 ：Simplifying the above expression gives $y=5$.

Step 3 ：The maximum height of the ball can be found by finding the vertex of the parabola represented by the equation. The $x$-coordinate of the vertex is given by $-\frac{b}{2a}$, where $a$ and $b$ are the coefficients of $x^2$ and $x$ respectively. In this case, $a=-\frac{1}{12}$ and $b=2$, so the $x$-coordinate of the vertex is $-\frac{2}{2\cdot -\frac{1}{12}}$.

Step 4 ：Simplifying the above expression gives $x=12$.

Step 5 ：Substituting $x=12$ into the equation gives $y=-\frac{1}{12}\cdot {12}^2+2\cdot 12+5$.

Step 6 ：Simplifying the above expression gives $y=17$.

Step 7 ：The ball strikes the ground when $y=0$. Setting $y=0$ in the equation gives $0=-\frac{1}{12}{x}^2+2x+5$.

Step 8 ：Solving the above quadratic equation for $x$ gives $x=-5$ and $x=20$.

Step 9 ：Since $x$ represents the horizontal distance from the point at which the ball is thrown, we discard the negative solution. Therefore, the ball strikes the ground $20$ feet from the child.