Whats wrong?

The path of a soccer ball is modeled by the function h(x) = -0.005x2 + 0.25x, where h is the height in meters and x is the horizontal distance that the ball travels in meters.
What is the maximum height that the ball reaches? ___________ m. Round your answer to the nearest thousandth.

$$x=\frac{ -0.25\pm \sqrt{0.25^2-4\times -0.005 \times 0} }{ 2 \times -0.005 }$$
$$x=\frac{ -0.25\pm \sqrt{0.0625} }{ -0.01 }$$
$$x=\frac{ 0.25\pm 0.25 }{ -0.01 }$$
$$x=\frac{ 0.25+ 0.25 }{ -0.01 } =-50$$
$$x=\frac{ 0.25- 0.25 }{ -0.01 } =0$$

These answers mean that the ball either never leaves the ground or goes 50 meters underground... not practical...

Question

Whats wrong?

The path of a soccer ball is modeled by the function h(x) = -0.005x2 + 0.25x, where h is the height in meters and x is the horizontal distance that the ball travels in meters. 
What is the maximum height that the ball reaches?  ___________ m. Round your answer to the nearest thousandth.

$$x=\frac{ -0.25\pm \sqrt{0.25^2-4\times -0.005 \times 0} }{ 2 \times -0.005 }$$
$$x=\frac{ -0.25\pm \sqrt{0.0625} }{ -0.01 }$$
$$x=\frac{ 0.25\pm 0.25 }{ -0.01 }$$
$$x=\frac{ 0.25+ 0.25 }{ -0.01 } =-50$$
$$x=\frac{ 0.25- 0.25 }{ -0.01 } =0$$

These answers mean that the ball either never leaves the ground or goes 50 meters underground... not practical...

mathstudent55 · Answer

The path of the ball is an inverted parabola.
It is at height zero at position x = 0, and at position x = when it hits the ground.
Right in between, what I labeled xmax, is where it is at maximum height.
Find the 2 x-values where h(x) = 0. In other words, 
solve  -0.005x^2 + 0.25x = 0 for x.
Then add the two values of x and divide by 2. That gives you xmax.
Then find h(xmax), and that is the maximum height.
|dw:1388690294721:dw|

alyssajobug · Answer

-25? thats still underground tho... did I misunderstand? @mathstudent55

mathstudent55 · Answer

-0.005x2 + 0.25x, = 0

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

$x = \dfrac{-0.25 \pm \sqrt{(-0.25)^2 - 4(-0.005)(0)}}{2(-0.005)} $

$x = \dfrac{-0.25 \pm \sqrt{0.0625 - 0}}{-0.01} $

$x = \dfrac{-0.25 \pm \sqrt{0.0625}}{-0.01} $

$x = \dfrac{-0.25 \pm 0.25}{-0.01} $

$x = \dfrac{-0.25  + 0.25}{-0.01} $   or    $x = \dfrac{-0.25 - 0.25}{-0.01} $

$x = \dfrac{0}{-0.01} $   or    $x = \dfrac{-0.5 }{-0.01} $

$x = 0 $   or    $x = 50  $

mathstudent55 · Answer

On the step where you took the square root of 0.0625, you made the mistake of leaving out the negative sign of the -0.25 in the numerator. Then you carried that mistake forward. That's why you got -50 instead of positive 50.

alyssajobug · Answer

ok so it is positive 25 then?

mathstudent55 · Answer

BTW, there's one more thing you should be careful with. In this problem you are told x is the distance the ball travels. When you got your incorrect answer of -50, it didn't mean that the ball was 50 meters in the ground. It meant it was 50 meters in the wrong direction. That is impossible because if the ball is only traveling in one direction, it can show up on the other side of the starting point.

mathstudent55 · Answer

Yes, the x position for maximum height is x = 25.
Now plug in x = 25 into the h(x) equation to find the maximum height.

alyssajobug · Answer

oh, ok, I'll have to watch that... 

$$-0.005(25)^2+0.25(25)= 6.250625$$ or 6.25

alyssajobug · Answer

actually, it would be 6.251 because it says round to the thousanth

mathstudent55 · Answer

Not quite.

Be careful with your calculations.

Copy down the formula. Then replace variables with numbers. Then calculate each term. Then add the terms together.

$h(x) = -0.005x^2 + 0.25x$

$h(25) = -0.005(25)^2 + 0.25(25)$

$h(25) = -0.005(625) + 6.25$

$h(25) = -3.125 + 6.25$

$h(25) =3.125$

The answer is maximum height is 3.125 meters.