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April 12th, 2017, 02:30 PM   #1
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Making a function with the domain and top.

5. An object is shot upwards vertically under influence of gravity it shall slow down, after 3 seconds it reaches a highest point and it will fall down with an ever increasing speed.
The maximum height is 20m

a) Find the equation of the parabola trajectory in a self picked
coordinate system.


So basicly what i got out of this is that i need to build up the function
or f(x) = based upon this data.

What i know is that the object is launched at (0,0) reaches a top at (3,20)

As it describes an object in the air that falls down i assume its a function of the second grade with the form:

ax²6bx+c

With a being <0

since the top is also the center of symetry in a second grade function it's easy to asume that it lands on the ground at (6,0)

So briefly put:
Code:
startpoint: (0.0) max(3,20) endpoint:( 0.6) 
 form: ax²+bx+c
with a<0
does anyone have any idea how i should make a function out of this,

I tried doing 20/3 since it takes 3 steps to reach 20 for a but simply doing that isn't enough i think.


(Also sorry if something is unclear, i had to transelate this to english, my textbook is in dutch. and i don't know alot of the english names for matimatical terms)
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April 12th, 2017, 02:49 PM   #2
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after a brilliant start you just fizzled right out

you have the form

$y = a x^2 + b x + c$

you have 3 points on this curve

$(0,0), ~(3,20),~(6,0)$

plug those points in and get a set of equations for $a,b,c$

solve the equations, they'll be linear in $a,b,c$
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April 12th, 2017, 03:03 PM   #3
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vertex form for the equation of a parabola ...

$y = a(x-h)^2 + k$, where the vertex is $(h,k)$



height is a function of time & you have the vertex coordinates, $(3,20)$ ...

$h = a(t - 3)^2 + 20$

use either $(0,0)$ or $(6,0)$ to determine the value of $a$
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April 13th, 2017, 01:22 AM   #4
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Quote:
Originally Posted by skeeter View Post
vertex form for the equation of a parabola ...

$y = a(x-h)^2 + k$, where the vertex is $(h,k)$



height is a function of time & you have the vertex coordinates, $(3,20)$ ...

$h = a(t - 3)^2 + 20$

use either $(0,0)$ or $(6,0)$ to determine the value of $a$

Woa i had never heard of that so called vertex equation of a parabola, Good to know that that excists for future excercises.

h= x(0-3)²+20
x*9+20=0
9x+20=0
9x=-20
x=-20/9= -2.2222....

so f(x)= -2.22222..(x-3)²+20



i tested this on my calculator and it is indeed correct!


always excited when i learn something new.
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April 13th, 2017, 02:21 AM   #5
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Put away the calculator.

Recommend you leave the coefficient, $a$, in its exact form as a fraction.
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April 13th, 2017, 07:31 AM   #6
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Quote:
Originally Posted by skeeter View Post
Put away the calculator.

Recommend you leave the coefficient, $a$, in its exact form as a fraction.
oh yeah i heard something about that before,
I'm so used to having to make a decimal number out of those, i guess thats a bad habit i picked up in school.
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