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 April 12th, 2017, 02:30 PM #1 Newbie   Joined: Mar 2017 From: leuven Posts: 18 Thanks: 0 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)
 April 12th, 2017, 02:49 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,106 Thanks: 578 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 $a,b,c$ Thanks from vlekje5
 April 12th, 2017, 03:03 PM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,425 Thanks: 1194 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 $a$ Thanks from vlekje5
April 13th, 2017, 01:22 AM   #4
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From: leuven

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Quote:
 Originally Posted by skeeter 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.

 April 13th, 2017, 02:21 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 2,425 Thanks: 1194 Put away the calculator. Recommend you leave the coefficient, $a$, in its exact form as a fraction.
April 13th, 2017, 07:31 AM   #6
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From: leuven

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Quote:
 Originally Posted by skeeter 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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