Question 702058: A parabola has points at (-3,4) (1,10) and (3,8)
What is the standard form for this parabolic path?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A parabola has points at (-3,4) (1,10) and (3,8)
What is the standard form for this parabolic path?
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Standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
Set up 3 equations:y= Ax^2+Bx+c
(-3,4) 4=9A-3B+c
(1,10) 10=A+B+c
(3,8) 8=9A+3B+c
..
4=9A-3B+c
10=A+B+c
subtract
-6=8A-4B
..
4=9A-3B+c
8=9A+3B+c
subtract
-4=-6B
B=2/3
..
8A=4B-6=8/3-6
A=1/3-6/8=1/3-3/4=4/12-9/12=-5/12
..
c=10-A-B=10-(-5/12)-2/3=10+5/12-8/12=10-3/12=120/12-3/12=117/12
..
A=-5/12
B=2/3
c=117/12
..
y=Ax^2+Bx+c
y=(-5/12)x^2+(2/3)x+117/12
complete the square:
y=(-5/12)(x^2-(8/5)x+(64/100)+.8/3+117/12
y=(-5/12)(x-0.8)^2+10.016667
sorry, this problem has cumbersome numbers to work with
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