Shane Bechler is about to punt the ball. Let "t" represent the # of seconds from the moment his foot constacts the football. The height of the ball when Lechler makes contact is 4 feet. After one seconds, the ball has reached a height of 44 feet and after two seconds, the ball has reached a height of 52 feet. Hint: h(t) = -16t^2 + vt + h should not be used.

a. Write a quadratic function in standard form for the information given.
b.After how many seconds does the football reach its maximum height and what is the height?

P.S. I tried to work this problem using my TI 84 Calculator by finding the Quadratic regression but it is in the same form as the hint, but I can't use that form. So help me out...I'm really confused. How would I work this problem out?

2. Originally Posted by chrozer
Shane Bechler is about to punt the ball. Let "t" represent the # of seconds from the moment his foot constacts the football. The height of the ball when Lechler makes contact is 4 feet. After one seconds, the ball has reached a height of 44 feet and after two seconds, the ball has reached a height of 52 feet. Hint: h(t) = -16t^2 + vt + h should not be used.

a. Write a quadratic function in standard form for the information given.
b.After how many seconds does the football reach its maximum height and what is the height?

P.S. I tried to work this problem using my TI 84 Calculator by finding the Quadratic regression but it is in the same form as the hint, but I can't use that form. So help me out...I'm really confused. How would I work this problem out?
You don't have any data to do a quadratic regression.

Standard form for a quadratic is not $\displaystyle h = at^2 + bt + c$, it is $\displaystyle h = a(t - h)^2 + k$

So
$\displaystyle 4 = a(0 - h)^2 + k$
and
$\displaystyle 44 = a(1 - h)^2 + k$
and
$\displaystyle 52 = a(2 - h)^2 + k$

Solve for a, h, and k.

-Dan

3. Yeah I know I didn't have alot of data but I had 3 points, (0,4);(1,44);(2,52).

I get the equation of
$\displaystyle y = 16x^2 + 56x + 4$

But I doubt that's answer since the question said not to use

$\displaystyle h(t) = -16t^2 + vt + h$

And how would you go by solving the equations you set up?

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4. Originally Posted by topsquark

$\displaystyle 4 = a(0 - h)^2 + k$
and
$\displaystyle 44 = a(1 - h)^2 + k$
and
$\displaystyle 52 = a(2 - h)^2 + k$

Solve for a, h, and k.
The first equation says:
$\displaystyle 4 = ah^2 + k$

Thus $\displaystyle k = 4 - ah^2$

Insert this value for k into the other two equations:
$\displaystyle 44 = a(1 - h)^2 + 4 - ah^2$
and
$\displaystyle 52 = a(2 - h)^2 + 4 - ah^2$

Giving us
$\displaystyle 40 = a(1 - h)^2 - ah^2$
and
$\displaystyle 48 = a(2 - h)^2 - ah^2$

Take the first one:
$\displaystyle 40 = a(1 - h)^2 - ah^2$

$\displaystyle 40 = a - 2ah + ah^2 - ah^2$

$\displaystyle 40 = a - 2ah$

$\displaystyle a = \frac{40}{1 - 2h}$

Insert this value of a into the last equation:
$\displaystyle 48 = \left ( \frac{40}{1 - 2h} \right ) (2 - h)^2 - \left ( \frac{40}{1 - 2h} \right ) h^2$

Solve for h:
$\displaystyle 48(1 - 2h) = 40(2 - h)^2 - 40h^2$

$\displaystyle 48 - 96h = 160 - 160h + 40h^2 - 40h^2$

$\displaystyle 48 - 96h = 160 - 160h$

$\displaystyle 64h = 112$

$\displaystyle h = \frac{7}{4}$

Giving us
$\displaystyle a = \frac{40}{1 - 2 \cdot \frac{3}{16}} = -16$
and
$\displaystyle k = 4 - (-16) \cdot \left ( \frac{7}{4} \right )^2 = 53$

-Dan

5. I see.... Thnx for the help.