Question

y=(x-2)^2

I need to put this in format -b/2a
Can anyone help me with this?

Answer 1

expand and complete the square...

Answer 2

Please don't expand and complete the square - unless you are really fond of driving around in circles.

IF you were to multiply this out, you would determine that you already have what you seek.

x = -b/2a is the axis of symmetry.

The axis of symmetry is x = 2, read directly from the present form.

Answer 3

So if I plug in 2 where x is I get 0

Answer 4

Okay. You have found the vertex. The vertex lives on the axis of symmetry.

You have not actually stated what the problem is. Are we just playing with parabolas?

Answer 5

yes parabolas so If i now place zero for x I can obtain a vertex of 0,0 correct?

Answer 6

Where did the 2 go? We already established that x = 2 is the axis of symmetry and (2,0) is the vertex. You still haven't suggested what it is you are supposed to be doing.

Answer 7

Ok yes vertex is 2,0, now I need to obtain some random points to find out if the parabola is going up or down, but my dilemma is that I am still not in the format of -b/2a so I can plug in random numbers.

Answer 8

nevermind I think I just figured it out

Answer 9

The form is of no consequence. Substitute any value you like in any form that it is.

Answer 10

Thank you tkhunny

Answer 11

If you REALLY want to multiply it out, go ahead. y = (x-2)^2 = x^2 - 4x + 4