Question

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 childs hand? feet (b) what is the maximum height of the ball? feet (c) how far from the child does the ball strike the ground? round your answers to the nearest 0.01. feet

Explanation:

Step1: Find the height when the ball leaves the hand

When the ball leaves the child's hand, the horizontal distance $x = 0$. Substitute $x = 0$ into the equation $y=-\frac{1}{12}x^{2}+2x + 5$.
$y=-\frac{1}{12}(0)^{2}+2(0)+5$
$y = 5$

Step2: Find the maximum - height of the ball

The equation $y=-\frac{1}{12}x^{2}+2x + 5$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-\frac{1}{12}$, $b = 2$, and $c = 5$. The $x$ - coordinate of the vertex of a quadratic function is given by $x=-\frac{b}{2a}$.
$x=-\frac{2}{2\times(-\frac{1}{12})}=-\frac{2}{-\frac{1}{6}} = 12$
Substitute $x = 12$ into the equation $y=-\frac{1}{12}x^{2}+2x + 5$.
$y=-\frac{1}{12}(12)^{2}+2(12)+5$
$y=- 12 + 24+5$
$y = 17$

Step3: Find the distance when the ball strikes the ground

When the ball strikes the ground, $y = 0$. So we need to solve the quadratic equation $-\frac{1}{12}x^{2}+2x + 5=0$. Multiply through by - 12 to get $x^{2}-24x - 60 = 0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $x^{2}-24x - 60 = 0$, where $a = 1$, $b=-24$, and $c=-60$.
$x=\frac{24\pm\sqrt{(-24)^{2}-4\times1\times(-60)}}{2\times1}=\frac{24\pm\sqrt{576 + 240}}{2}=\frac{24\pm\sqrt{816}}{2}=\frac{24\pm2\sqrt{204}}{2}=12\pm\sqrt{204}$
We take the positive root since distance cannot be negative. $x=12+\sqrt{204}\approx12 + 14.28=26.28$

Answer:

(a) 5
(b) 17
(c) 26.28