I do not understand this at all. Click image to enlarge Attachment 25278. I keep getting the answer , which means that the vertex of this equation is NOT . I even used h: to plug it into the entire equation .

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- October 18th 2012, 07:06 PMEraser147Completing the square
I do not understand this at all. Click image to enlarge Attachment 25278. I keep getting the answer , which means that the vertex of this equation is NOT . I even used h: to plug it into the entire equation .

- October 18th 2012, 07:19 PMtopsquarkRe: Completing the square
Your completing the square work is correct. (The answer in the "Click to enlarge" panel, that is.)

-Dan - October 18th 2012, 07:38 PMEraser147Re: Completing the square
Why? Why is the k value not 1/5?

- October 18th 2012, 07:38 PMEraser147Re: Completing the square
Can you show me the steps for the solution so I can look at the mechanics of it?

- October 18th 2012, 08:14 PMProve ItRe: Completing the square
- October 18th 2012, 08:42 PMEraser147Re: Completing the square
It wouldn't make sense because if you plugin 8/5 into the original equation, 1/5 IS the answer for the k value.

- October 18th 2012, 09:06 PMProve ItRe: Completing the square
- October 18th 2012, 09:55 PMEraser147Re: Completing the square
No no no no, just no. That image is what the answer the solution manual has. THAT's NOT my answer. My answer is with the 1/5. I know how to expand, but no matter how I calculate it, it won't show the k value to have 1/5. Again, it is easily testable by jut plugging in k into the original polynomial above.