SOLUTION: I am working with parabolas. For this problem I need to Complete the square Give the Vertex Give the Axis Give the x-intercepts Give the y-intercepts Give a point symmetric

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Question 29860: I am working with parabolas. For this problem I need to
Complete the square
Give the Vertex
Give the Axis
Give the x-intercepts
Give the y-intercepts
Give a point symmetric to the y-intercept
Draw the graph
The problem is y=x^2-2x-3.
Can anyone help me solve this. I am working on it and would like to have something to check my answer with.

Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE WHICH IS ALMOST SAME AS YOUR PROBLEM AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK.
Y=X^2-2X-3=X^2-2*X*1+1^2-1-3=(X-1)^2-1-3=(X-1)^2-4=0
X INTERCEPT IS OBTAINED BY PUTTING Y=0....WE GET
X^2-2X-3=0=X^2-3X+X-3=X(X-3)+1(X-3)=(X-3)(X+1)=0...SO..X=3 OR -1...
Y INTERCEPT IS GOT BY PUTTING X =0...WE GET
Y=0-0-3=-3
SEE THE GRAPH BELOW ......

THE AXIS IS X=1 AS YOU CAN SEE FROM THE GRAPH
ONE POINT SYMMETRIC TO Y INTERCEPT??NO...SYMMETRIC TO AXIS IT IS......(-1,3)
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I have to put this equation y=x^2-2x-15 into this form y=a(x-h)^2+k.
MAKE A PERFECT SQUARE USING X^2 AND X TERMS.ADD AND SUBTRACT THE REQUIRED CONSTANT FOR THE PURPOSE.
Y=(X-1)^2-1-15=(X-1)^2-16...COMPARING WITH THE ABOVE
y=a(x-h)^2+k.
WE GET A=1 AND K=-16
I have to find the line of symmetry.
X-1=0 IS THE LINE OF SYMMETRY SINCE ON EITHER SIDE OF X=1,WE GET SYMMETRIC/SAME VALUES FOR Y..AT X=1+2=3..Y IS -12 AND AT X=1-2=-1 ALSO WE GET Y=-12
(h,k)=vertex I think you use complete the square technique.
YA ..THE VERTEX AS YOU SHOULD KNOW NOW IS AT X=1 AND AT X=1 ,Y=-16 .SO (1,-16) IS THE VERTEX
THE GRAPH WILL LOOK LIKE THIS

Please help. thanks

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