Originally Posted by

**2976math** ===============================================

**magentarita,**

I see where your problem is.

2x^2 + 8x = -9 ------ ok so far.

Before you take half of the secod term and square,

Make sure the coefficient of suared term 2x^2 to be 1.

Therefore, 2x^2 + 8x = -9

becomes (x^2) + 4x = -9/2 -- simply didived by 2.

(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get

(x^2) + 4x + 2^2 = (-9/2 ) + 2^2

**(x + 2)^2 = -1/2** now complete the square part is done at this point.

Rewrite original question to** y = [(x + 2)^2 ]+ 1/2**

Can you see **y = a ( x-h )^2 + k** --- standard form.

Vertex is at ( h, k ) = ( -2, 1/2) and a =1

Opens up vertically, therefor, mini. point is at vertex.