Question 33778: Write an equation in the form y=ax^2 + bx + c for the quadratic function whose graph passes through the points (8,0) (0,8) (-2,0). Thanks
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! AX^2+BX+C=Y...IS SATISFIED BY (8,0);(0,8);(-2,0)..SO SUBSTITUTING WE GET..
A*8^2+B*8+C=0...OR...64A+8B+C=0......................................I
A*0^2+B*0+C=8....................OR...........C=8.................II
A*(-2)^2+B*(-2)+C=0.....OR.......4A-2B+C=0...................III
SUBSTITUTING C=8 IN EQN.I AND EQN.III WEW GET
64A+8B+8=0....OR.....8A+B+1=0................................IV
4A-2B+8=0.......OR....2A-B+4=0............................V
EQN.IV+EQN.V...GIVES
10A+5=0...OR....A=-5/10=-1/2
SO 2*(-1/2)-B+4=0
-1-B+4=0
B=3
SO THE QUADRATIC IS
Y=-X^2/2+3X+8
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