Question 34184
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)						
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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						
graph( 500, 500, -10, 20, -20, 20, x^2-2x-15 )						
Please help. thanks