Respuesta :
[tex]y=-\frac{1}{12}x^2+4x+5[/tex]
when the ball leaves the chil'd hand the distance is 0
so:
[tex]\begin{gathered} y=\frac{-(0)^2}{12}+4(0)+5 \\ y=\text{ 0+0+5} \\ y=5 \end{gathered}[/tex]the height is 5 feets
The maximun height happens when x=-b/(2a) because is the vertex
where b=4 and a=-1/12
[tex]\begin{gathered} x=\frac{-b}{2a} \\ \\ x=\frac{-(4)}{2(\frac{-1}{12})} \\ \\ x=\frac{-4}{-\frac{1}{6}}\text{ } \\ x=24 \end{gathered}[/tex]so x is the distance where maximun height happens, so replace on the function
[tex]\begin{gathered} y=-\frac{1}{12}x^2+4x+5 \\ \\ y=-\frac{1}{12}(24)^2+4(24)+5 \\ y=-\frac{1}{12}(576)+96+5 \\ y=-48+96+5 \\ y=53 \end{gathered}[/tex]the maximun height is 53 feets
the ball strike the ground when the height is 0 so we can replace y and fin x or the distance
[tex]\begin{gathered} 0=-\frac{1}{12}x^2+4x+5 \\ \end{gathered}[/tex]we need to factor and we can do it with this equation
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a= -1/12 , b=4 and c=5
[tex]\begin{gathered} x=\frac{-(4)\pm\sqrt[]{(4)^2-4(-\frac{1}{12})(5)}}{2(-\frac{1}{12})} \\ \\ x=\frac{-4\pm\sqrt[]{16+(\frac{5}{3})}}{-\frac{1}{6}} \\ x=-6(-4\pm\sqrt[]{\frac{53}{3}}) \\ \\ x=-6(-4\pm4.20) \end{gathered}[/tex]we have 2 solutions
1.
[tex]\begin{gathered} x_1=-6(-4+4.20) \\ x_1=-6(0.2) \\ x_1=-1.2 \end{gathered}[/tex]but its not a real solution because distance cant be negative
2.
[tex]\begin{gathered} x_2=-6(-4-4.20) \\ x_2=-6(-8.2) \\ x_2=49.2 \end{gathered}[/tex]so the distance when the ball strike the ground is 49.2 feets
