For anyone who knows how to solve a system of equations in order to find a quadratic function....
(0,15) (2,5) (3,6). Using elimination.

Question

For anyone who knows how to solve a system of equations in order to find a quadratic function....
(0,15) (2,5) (3,6). Using elimination.

anonymous · Answer

$$f(x)=ax^2+bx+c$$
$$f(0)=c=15$$ so 
$$f(x)=ax^2+bx+15$$ is a start

anonymous · Answer

$$f(2)=a\times (2)^2+\times 2+15=5\
4a+2b+15=5\
4a+2b=-10$$ is one equation

anonymous · Answer

repeat with $f(3)=6$ and you will have another equation in $a$ and $b$ 
then you can solve

anonymous · Answer

I got that far and have these three equations.  What do I do with the 15?
15 = C
5 = 4a + 2b + C
6 = 9a + 3b + C

anonymous · Answer

$c=15$

anonymous · Answer

i.e. replace $c$ by $15$ in your two equations

anonymous · Answer

so I just get rid of C from the equations by subtracting

anonymous · Answer

$$5 = 4a + 2b + C\
 6 = 9a + 3b + C$$
$$5 = 4a + 2b + 15\
 6 = 9a + 3b + 15$$

anonymous · Answer

yes, by subtracting $15$ to get 
$$-10=4a+2b\
-9=9a+3b$$

anonymous · Answer

Then elimination?

anonymous · Answer

-30 = 12a + 6b
-18 = 18a + 6b

anonymous · Answer

Does A = 2?

anonymous · Answer

B = -9? btw thanks for all the help

anonymous · Answer

hello?