SOLUTION: How do you find the vertex and axis of symmetry in a quadratic equation?

Question 20867: How do you find the vertex and axis of symmetry in a quadratic equation?

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING 2 EXAMPLES TO KNOW THE METHOD

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LET Y=X^2+4X+5..
GRAPHICAL METHOD.....................
GIVE DIFFERENT VALUES TO X AND FIND CORRESPONDING Y VALUE TABULATE..TAKE A SUITABLE SCALE AND PLOT FOR GRAPH
X......0....1.......2....3....-1....-2....-3.....-4.......-5.........-6...........-7
Y.....5....10.....17...26...2.....1.....2......5......10.........17..........26
FROM TABLE/GRAPH YOU CAN SEE THAT THE VERTEX IS AT X=-2 AND Y=1.....,AXIS OF SYMMETRY IS X=-2.......THERE IS A MINIMUM OF Y=1 AT THE VERTEX GIVEN ABOVE..........
ANALYTICALLY YOU CAN DO LIKE THIS
Y=X^2+4X+4-4+5=(X+2)^2+1.....NOW (X+2)^2 BEING A SQUARE IS ALWAYS POSITIVE.SO ITS MINIMUM VALUE IS ZERO WHEN X=-2..AT THAT X=-2 ,MINIMUM VALUE OF Y OCCURS =0+1=1......NOW TAKING THIS AS VERTEX WE OBSERVE THAT ON EITHER SIDE OF X=-2,THAT IS SAY 5 UNITS ON EITHER SIDE OF X=-2...THAT IS X=-2+5=3 AND X=-2-5=-7......Y HAS SAME VALUE NAMELY (-7+2)^2+1=(-5)^2+1=26=(3+2)^2+1=5^2+1...SO X=-2 IS THE AXIS OF SYMMETRY....
I HOPE YOU CAN DO THE SECOND PROBLEM BY YOUR SELF NOW WITH THESE EXAMPLES GIVEN TO YOU.
VENUGOPAL
- Show quoted text -
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SEE RECENT ANSWERS GIVEN BY ME IN ALGEBRA.COM FOR THIS ...PROBLEM NO.20847,ANSWER NO.10010 DT 29TH.NOV.05....GIVEN BELOW
LET Y=
GIVE DIFFERENT VALUES TO X IN THE ABOVE EQUATION AND FIND CORRESPONDING VALUES OF Y..TABULATE AS FOLLOWS,CHOOSE A SUITABLE SCALE AND PLOT THEM ON A GRAPH AS SHOWN BELOW
X....0.....1......2.....3.......4......ETC..
Y....-1....2......3.....2.......-1

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