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Question 586402: find the equation in standard form for the parabola passing through the points (2,-8) (3,-8) (6,4)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the equation in standard form for the parabola passing through the points (2,-8) (3,-8) (6,4)
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y=ax^2+bx+c
1) For (2,-8): -8=4a+2b+c
2) For (3,-8): -8=9a+3b+c
3) For (6,4): 4=36a+6b+c
..
1) -8=4a+2b+c
2) -8=9a+3b+c
subtract eq2 from eq1
4) 0=-5a-b
..
1) -8=4a+2b+c
3) 4=36a+6b+c
subtract eq3 from eq1
5)-12=-32a-4b
4) 0= -5a-b
multiply eq4 by 4
6) 0=-20a-4b
subtract eq6 from eq5
7) -12=-12a
a=1
-5a-b=0
-5-b=0
b=-5
1) c=-8-4a-2b=-8-4+10=-2
2) c=-8-9a-3b=-8-9+15=-2
3) c=4-36a-6b=4-36+30=-2
..
a=1, b=-5, c=-2
..
equation of parabola:
y=ax^2+bx+c
y=x^2-5x-2
complete the square
y=(x^2-5x+25/4)-2-25/4
y=(x-5/2)^2-33/4
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