ok. So i have 3 points on a graph... These points are all points on a quadratic equation.

(32, 6)
(48, 14)
(64, 24)

Find the function.... How do I do that?
Oh and the function goes through (0, 0).

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

use the three points to find a,b and c.
Could you maybe elaborate abit?
y=a(d+x)²+e

32=x 48=x 64=x
6=y 14=y 24=y
0=d
0=e

the first:

6=a(0+32)²+0
6=a (0+32)(0+32)+0
6=a(1024) |:1024

i think it has to go like this
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6 = a*(32)^2 + b*(32) + c
14 = a*(48)^2 + b*(48) + c
24 = a*(64)^2 + b*(64) + c
0 = a*(0)^2 + b*(0) + c

solve for a,b, and c
If you have a graphing calculator, just put the points into matricies (sp?). Problem solved.
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...............
.006X^2+.0136X-.036 is as close as I can get. What class is this and based on previous work is this answer likely to be complex or very simple (whole numbers). Also are you sure it passes through (0,0)
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6 = a*(32)^2 + b*(32) + c
14 = a*(48)^2 + b*(48) + c
24 = a*(64)^2 + b*(64) + c
0 = a*(0)^2 + b*(0) + c

solve for a,b, and c

Well that shows you that c=0 so the rest should be just boring algebra.
Ok, i have the answer, :

0.006x^2... How did I get that?
Hmmm.... I don't know if this has anything to do with it, but this is the function of the line with the gradient. Even if you don't need it, it was good practice for my maths prelim next week .

Last edited by Eilidh at Jan 11, 2007,
6 = a*(32)^2 + b*(32) + c
14 = a*(48)^2 + b*(48) + c
24 = a*(64)^2 + b*(64) + c Both red eqations are the same.
0 = a*(0)^2 + b*(0) + c ====> c=0

So You get:
C=0

14 = a*(48)^2 + b*(48)
6 = a*(32)^2 + b*(32)
-----------------------------
14=2304*a+48*b
6=1024*a+32*b

a=5/768
b=-1/48
i wish i still had math problems like that.
plot it in excel and fit a curve to it (cheating)
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You can't get accurate results using excel plot, hell you don't even know if it'll draw you a parabula or not.