Posted by Ashley on .
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)?

Math (Help This Will Be on My Test Tomorrow!) 
Reiny,
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! 
Math (Help This Will Be on My Test Tomorrow!) 
Bosnian,
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 
Math (Help This Will Be on My Test Tomorrow!) 
Bosnian,
I make one mistake in typig.
2b= 1 Divide with 2 is mistake
2b= 10 Divide with 2 is correct