the given expression is
[tex]y=\frac{-1}{14}x^2+4x+3[/tex]here x is the horizontal distance and y is the vertical height.
(a) when the ball just leaves the child's hand the horizontal distance of the ball will be zero
so put x = 0
y = -1/14 (0)^2 + 4(0) + 3
y = 0 + 0 + 3
y = 3
the height of the ball is 3 when it leaves the child's hand.
(b)
the maximum height of the ball
the given equation is the equation of the parabola,
so the max. height of the ball at x = -2b/a
where b is the coefficient of x that is b = 4
a is the coefficient of x^2 that is a = -1/14
so
[tex]\begin{gathered} x=-\frac{2b}{a} \\ x=\frac{-2\times4}{-\frac{1}{14}} \\ x=14\times8=112 \end{gathered}[/tex]put x= 112 in the equation
[tex]\begin{gathered} y=-\frac{1}{14}(112)^2+4(112)+3 \\ y=-445 \end{gathered}[/tex]so the maximum height of the ball is 445 ft
(c) for the total horizontal distance of the ball
put y = 0 in the expression because the ball finally touched the ground and at ground y = 0.
so by putting y = 0
you will get a quadratic equation of x
by solving the quadratic equation you will get the value of x
