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Anonymous
Simple? maths: parabola function 2016-11-07 19:31:45 Post No. 217076
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Simple? maths: parabola function 2016-11-07 19:31:45 Post No. 217076
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I need help with finding the correct function for the following problem:
I have a variable X which can have a minimum of 0 and a constant M that represents a given maximum.
- X and M are integers.
In addition I have a constant C that is given too.
C is also an integer.
I need a function that uses X and outputs a value (let's name it Y) between 0 < C > 0, where_
- Y will be 0 when X is 0
- Y will be C when X is half the size of M
- Y will be 0 again when X equals M
So X --> M maps Y from zero to C and back to zero.
The function is representing a parabola.
>>
Formally, you get your solution when you solve:
(-b (+-) sqrt(b^2-4*a*c))/(2*a) = {0, M}
and
a(M/2)^2+b(M/2)+c = C
for a, b, and c.
Practically, there's a ton of websites out there where you can play with parabolic functions by tweaking the variables.
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>>217076
You need to be 18, anon.
You know the zeros of the function Y, namely 0 and M, so
Y = A(X-0)(X-M) with a stretching factor A.
Now insert your other point and determine A.
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>>217091
>Y = A(X-0)(X-M)
C = A((M/2)-0)((M/2)-M)
A = C/((M/2)-0)((M/2)-M)
Y= C/((M/2)-0)((M/2)-M)(X-0)(X-M)
There you have it, OP.
>>
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messed up some parentheses
>tfw no reverse polish notation
Y= C/(((M/2)-0)((M/2)-M))(X-0)(X-M)
or maybe
Y= C(X-0)(X-M)/(((M/2)-0)((M/2)-M))
>>
>>217099
thanks a lot.
I guess I should have spend some time actually learning this shit in school.
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