SOLUTION: I am trying to identify the axis of symmetry and creating an x and y table for the graph. My numbers are wrong because it doesn't resemble a parabola... y=x^2-5x+3 and y=-x^2+6x-2.

Question 50343This question is from textbook
: I am trying to identify the axis of symmetry and creating an x and y table for the graph. My numbers are wrong because it doesn't resemble a parabola... y=x^2-5x+3 and y=-x^2+6x-2. I am kinda stuck if you can assit please. This question is from textbook

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
y=x^2-5x+3
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Complete the square to put into vertex form, as follows:
x^2-5x+(5/2)^2=y-3+(5/2)^2
(x-(5/2))^2 = y-12/4+25/4
(x-(5/2))^2 = y+ 13/4
The vertex is (5/2,-13/4)
The axis of symmetry is x=5/2
You can create a x,y table for the graph by substituting
values for x in the original equation and solving for y.
------------------

and y=-x^2+6x-2
--------
-(x^2-6x+3^2)=y+2-3^2
-(x-3)^2=y-7
Axis of symmetry is x=3
Cheers,
Stan H.
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