Math (Help This Will Be on My Test Tomorrow!)
posted by Ashley .
y= ax^2+bx4. How would i find the equation of this parabola with the points (0,4) (1,0) (2,2) (4,0) and 6,10)?

Use the first few given points
for (0,4) : 4 = 0 + 0  4 , nothing new learned here
for (1,0) : 0 = a + b  4 > a+b+4 or b = 4a
for (2,2) : 2 = 4a + 2b  4  2a + b = 3
use substitution
2a + (4a) = 3
a = 1
then b = 4(=1) = 5
y = x^2 + 5x  4
test if the other points satisfy, they do! 
You must write equation for any pair of points.
y=ax^2+bx4
x=0 y= 4
4=a*0^2+b*0+c
0+0+c= 4
c= 4
x=1 y=0
0=a*1^2+b*14
a*1+b*14=0
a+b4=0
a=4b
x=2 y=2
y=ax^2+bx4
2=(4b)*2^2+b*24
(4b)*4+2b4=2
164b+2b=2+4
2b=616
2b= 1 Divide with 2
b=5
a=4b
a=45
a= 1
So:
a= 1 b=5 c= 4
y=ax^2+bx+c= x^2+5x4
y=x^2+5x4
Checking of results:
For x=4
y=ax^2+bx4
y= 4^2+5*44
y= 16+204
y=44
y=0
Correct value
For x=6
y=ax^2+bx4
y= 6^2+6*54
y= 36+304
y= 64
y= 10
Correct value 
I make one mistake in typig.
2b= 1 Divide with 2 is mistake
2b= 10 Divide with 2 is correct