let the function by y = ax^2 + bx + c

Question

and sub in (1,2),(2,7), and (3,15)

this gives you 3 equations in 3 unknowns which solve quite nicely to
a = 3/2
b = 1/2
c = 0

so the function is y = (3/2)x^2 + x/2
or
y = (3x^2 + x)/2

can somone please explain to me how this person got a/b/c when they have x/y?

Answer 1

don't know what you mean by "...how this person got a/b/c when they have x/y"

who is "they" ?
a/b/c is ambiguous, do you mean a/(b/c), which is ac/b, or (a/b)/c, which would be a/(bc)

BTW, your equation is correct, all 3 points satisfy your equation.

Answer 2

obviously the equation is solved... thanks to u :)

i wanted to know how you found what a,b, and c were with the x & y

Answer 3

Put the three points in and get three equations
2 = a (1)^2 + b(1) + c
7 = a (2)^2 + b(2) + c
15 = a (3)^2 + b(3) + c
or
2 = a + b + c
7 = 4a + 2b + c
15 = 9a + 3b + c

Answer 4

thank you Reiny/Damon
sorry for all the trouble

Answer 5

subtract equation 1 from eqn 2
(eqn3) 5 = 3 a + b
subtract equation 1 from equation 3
(eqn4)13 = 8 a + 2 b
now multiply equation 3 by 2
(eqn5) 10 = 6 a + 2 b
subtract eqn 5 from eqn 4
3 = 2 a
so
a = 3/2 lo and behold
now work your way back up to get b and c

Answer 6

LOL, didn't even realize that you had posted my reply from two days ago.

ok, let's take the first given point (1,2)
It must satisfy the equation
y = ax^2 + bx + c

so

2 = a(1^2) + b(1) + c ---> a+b+c = 2 (#1)

same thing for (2,7)

7 = 4a + 2b + c (#2), etc

my third equation was 9a + 3b + c = 15 (#3)

I then did #2 - #1 to get 3a+b=5
and #3 - #2 to get 5a+b=8

I then subtracted those two to get a=3/2
subbed that back into 3a+b=5 to get b=1/2
and finally subbed those two values back into the first equation to get c=0