SOLUTION: Consider the parabola: y=x^2-6x+5 Complete the Square Give the Vertex Give the x- and y-intercepts Graph the parabola Could someone please help me with this problem? Thank

Question 34184: Consider the parabola: y=x^2-6x+5
Complete the Square
Give the Vertex
Give the x- and y-intercepts
Graph the parabola
Could someone please help me with this problem? Thank you so very much.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY

Equations/30056: I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS.
1.problem
y=x^2-5x+6

2.problem
y=-x^2+2x-1
1 solutions
Answer 16805 by venugopalramana(1167) About Me on 2006-03-13 10:41:52 (Show Source):
I need to determine the following for these two problems :HOW MANY X-INTERCEPTS THE PARABOLA HAS, and WHETER ITS VERTEX LIES ABOVE OR BELOW OR ON THE X-AXIS.
1.problem
y=x^2-5x+6
PUT Y=0 AND SOLVE FOR X TO GET X INTERCEPTS.
X^2-5X+6=0=X^2-2X-3X+6=0=X(X-2)-3(X-2)=0=(X-2)(X-3)=0....X=2 AND 3...
SO THE X INTERCEPTS ARE AT X = 2 AND X = 3
Y=X^2-5X+6={X^2-2*X*5/2+(5/2)^2}-(5/2)^2+6 =(X-2.5)^2 - 0.25
SO THE VERTEX IS AT X=2.5 AND Y=-0.25...THAT IS BELOW X AXIS
2.problem
y=-x^2+2x-1
DOING THE SAME WAY WE GET
Y=-(X-1)^2=0 AND HENCE
X INTERCEPTS ARE X=1
AND VERTEX IS AT X=1 AND Y=0 SO THE VERTEX IS ON THE X AXIS.

Coordinate-system/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.
1 solutions
Answer 16624 by venugopalramana(1167) About Me on 2006-03-11 07:58:25 (Show Source):
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 ......
graph( 500, 500, -10, 20, -20, 20,x^2-2x-3 )
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)
******************************************************************************
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
graph( 500, 500, -10, 20, -20, 20, x^2-2x-15 )
Please help. thanks
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