the height y (in feet) of a ball thrown by a child is
[tex]y = \frac{1}{16} {x}^{2} + 4x + 3[/tex]
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

b).what is the maximum high of the ball

c).how far from the child does the ball strike the ground

I will assume you mistyped this question. For y = -1/16x^2 + 4x + 3, the answers to this question are
a) 3 feet
b) 67 feet
c) 64.741 feet

For a) we note that at x = 0, that is the instant where the ball leaves the hand. y(0) = 0.

For b), we find the vertex of y = -1/16x^2 + 4x + 3
                               y = -1/16x^2 + 4x + 3
                               y = -1/16(x^2 - 64x) + 3
                               y = -1/16(x^2 - 64x + 1024 - 1024) + 3
                               y = -1/16((x-32)^2 - 1024) + 3
                               y = -1/16(x-32)^2 + 64 + 3
                               y = -1/16(x-32)^2 + 67
The vertex is at (32,67) so 67 is the maximum height.

For c), we find the x-intercepts with the quadratic formula on
y = -1/16x^2 + 4x + 3=0:
                     x = [ -b ± √b^2 - 4ac ] / (2a)
                     x = [ -4 ± √4^2 - 4(-1/16)(3) ] / (2(-1/16))
   x = -0.741, 64.741
Only the positive solution, so 64.741 feet

erinna erinna · Answer 2 · 2019-10-20T08:25:10+02:00

Answer:

a) 3 ft

b) 67 ft

c) 64.741 ft.

Step-by-step explanation:

Let as consider the height y (in feet) of a ball thrown by a child is

[tex]y = -\frac{1}{16} {x}^{2} + 4x + 3[/tex]

where x is the horizontal distance in feet from the point at which the ball is thrown

.

(a)

We need to find the initial height of the ball.

Substitute x=0 in the given equation.

[tex]y = -\frac{1}{16} {(0)}^{2} + 4(0) + 3[/tex]

[tex]y =3[/tex]

Therefore, the height of ball is 3 ft when it leaves the child's hand.

(b)

If a quadratic equation is defined as

[tex]f(x)=ax^2+bx+c[/tex]

then the vertex of the function is

[tex]vertex=(-\frac{b}{2a},f(-\frac{b}{2a}))[/tex]

In the given equation [tex]a=-\frac{1}{16}[/tex] and b=4. So,

[tex]-\dfrac{b}{2a}=-\dfrac{4}{2(-\frac{1}{16})}=32[/tex]

Substitute x=32 in the given equation.

[tex]y = -\frac{1}{16} {32}^{2} + 4(32) + 3=67[/tex]

The vertex of the parabola is (32,67). So, the maximum height of the ball is 67 ft.

(c)

The height of the ball is zero when it touch the ground.

Substitute y=0 in the given function.

[tex]0= -\frac{1}{16} {x}^{2} + 4x + 3[/tex]

Using quadratic formula, we get

[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]x=\dfrac{-4\pm \sqrt{4^2-4( -\frac{1}{16})(3)}}{2( -\frac{1}{16})}[/tex]

[tex]x=-0.741,64.741[/tex]

The distance can not be negative. So the ball will touch the ground at the distance of 64.741 ft.