SOLUTION: ihave to find the quadratic model in standard form for each of the set values. The values are: (0,3) (1,10) (2,10). i have already figured the vertex it (0,3) but after that im los

Question 26614: ihave to find the quadratic model in standard form for each of the set values. The values are: (0,3) (1,10) (2,10). i have already figured the vertex it (0,3) but after that im lost. PLEASE HELP
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY.IF YOU STILL HAVE DIFFICULTY COME BACK
If given a graph, with a vertex of (-3,4) and points (-4,3) and (-5,0).. how do I write this in standard form?
IF YOU HAVE A GRAPH YOU KNOW THE VERTEX/CURVE POSITION IS ALONG X OR Y AXIS.THAT IS IT REPRESENTS MAXIMUM VALUE OF Y OR MINIMUM VALUE OF Y..IN ANY CASE IF THE VERTEX IS (-3,4),IT MEANS VERTEX OCCURS AT X=-3 AND ITS VALUE IS Y=4..SO THE EQN. IS
(Y-4)=K(X+3)^2.....OR....(X+3)=K(Y-4)^2..SINCE BOTH EQUATIONS SATISFY THE VERTEX COORDINATES
NOW WE TRY THE OTHER 2 POINTS TO FIND WHICH IS APPROPRIATE.
PUTTING (-4,3),WE GET
-1=K(-1)^2=K......................OR.............-1=K(-1)^2=K
PUTTING (-5,0)...WE GET
-4=K(-2)^2=4K..K=-1...............OR..............-2=K(-1)^2=K..WHICH IS NOT MATCHING WITH THE ABOVE..SO
Y-4=-(X+3)^2.... IS THE EQN.YOU CAN PLOT AND SEE AS BELOW

THIS IS FOR THE CORRECT ANSWER.ALL 3 POINTS ARE ON THE CURVE.THE CURVE IS ALONG Y AXIS.

THIS IS FOR THE WRONG ANSWER.ONLY 2 POINTS ARE ON THE CURVE.THE CURVE IS ALONG X AXIS.

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