1. ## Please explain step by step

I've come to the conclusion that canadian and american math is taught differently. If you know how we do it here in Canada plz explain this question step by step.

The path of a thrown baseball can be modelled by the function h=-0.004d^2+0.14d+2 where h is the height of the ball, in meters, and d is the horizontal distance of the ball from the player, in meters.
a) What is the maximum height reached by the ball?
b) What is the horizontal distance of the ball from the player when it reaches its maximum height?
c) How far fromt he ground is the ball when the player releases it?

2. Originally Posted by Sculthorp
I've come to the conclusion that canadian and american math is taught differently. If you know how we do it here in Canada plz explain this question step by step.

The path of a thrown baseball can be modelled by the function h=-0.004d^2+0.14d+2 where h is the height of the ball, in meters, and d is the horizontal distance of the ball from the player, in meters.
a) What is the maximum height reached by the ball?
b) What is the horizontal distance of the ball from the player when it reaches its maximum height?
note that the height is modeled by a downward opening parabola. the maximum of this parabola gives the maximum height. all we need to do is find the vertex of the parabola (since this is where the max occurs). the d value (which corresponds to the x-coordinate kind of) will give the horizontal distance, the h-value, (which corresponds to the y-coordinate) will give the height.

there are two main ways to go about this. we can complete the square to get the function in the form: $h = a(d - x)^2 + y$, in which $(h,d) = (x,y)$ gives the vertex

or we can use the vertex formula: $d = \frac {-b}{2a}$, where b is the coefficient of d and a is the coefficient of d^2

do whichever you are comfortable with.

c) How far fromt he ground is the ball when the player releases it?
at the moment the player releases the ball, the horizontal distance from the player to the ball is zero. just plug in d = 0 and solve for h

3. How does that vertex formula work? I've been taught that formula, with added stuff to the top half of the equation to figure out the x-intercepts, but not to find out the vertex. And I tried to complete the square but the decimals were too big to work with.. :S The answer ends up being 3.0625 square units somehow

4. Originally Posted by Sculthorp
How does that vertex formula work? I've been taught that formula, with added stuff to the top half of the equation to figure out the x-intercepts, but not to find out the vertex. And I tried to complete the square but the decimals were too big to work with.. :S The answer ends up being 3.0625 square units somehow
yeah, completing the square will be a pain here. use the vertex formula (yes, it works). that will give you the value for d. to find the corresponding h to get the height, just plug in the value you got for d into the original equation

5. So that means D = 17.5? OHHHHHHHHHHH! I love you Jhevon.

6. Originally Posted by Sculthorp
So that means D = 17.5? OHHHHHHHHHHH!
yes, that's correct
I love you Jhevon.
let's try to keep this professional now