SOLUTION: 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^2-5

Question 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
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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.

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