Question

Need help with Quadratic Formula!

The following equation is asked to be used to determine the amount of feet of where a ball is thrown.

y= -\frac(1)(16)x^2 + 4x + 3
( y= -1/16xsquared +4x +3)

It first asked what the height was when the ball was thrown from the kid's hand which was 3 feet.
It then asked what the highest point the ball was at which was the maximum part of the vertex, 67 feet.
The last question is where I'm having problems. It asks what the distance is in feet of where the ball lands away from the kid. I tried calculating this in quadratic formula but am stuck.

Answer 1

I ended up getting -4 + square root of 16.75 all divided by -1/8
which is incorrect.

Answer 2

if x defines distance, then what is x when y=0?

to which it appears you tried the quadratic formula

Answer 3

\[y= -\frac{1}{16}x^2 + 4x + 3\]
\[0= -\frac{1}{16}x^2 + 4x + 3\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[a=-1/16~:~b=4~:~c=3\]
\[x=\frac{-4\pm\sqrt{4^2-4(-1/16)(3)}}{-2/16}\]
\[x=\frac{-4\pm2\sqrt{4+(3/16)}}{-2/16}\]
\[x=\frac{-2\pm\sqrt{4+(3/16)}}{-1/16}\]
\[x=-16(-2\pm\sqrt{4+(3/16)}~)\]

Answer 4

\[x=32\pm4\sqrt{67}\]
if the information posted is correct