Re: Standard to Vertex Form

Quote:

Originally Posted by

**michellederz** Alright, so I am trying to convert this formula to vertex form.

0.5x^{2}-8x+33

Here is what I have so far: (Am I doing this correctly)?

0.5(x^{2}-6x)+33

0.5(x^{2}-6x+9)+33

This is where I get stuck because I don't think I am completing the square correctly.. :c

You have the correct method, but where did the 6 come from in line 3?

$\displaystyle \frac{1}{2}x^2 - 8x + 33 = \frac{1}{2}(x^2 - 16x) + 33$

$\displaystyle = \frac{1}{2} (x^2 - 8x + 16 - 16) + 33$

Can you finish it?

-Dan