# Thread: Completing the square to convert to vertex form

1. ## Completing the square to convert to vertex form

I understand the operations involved in the process, but what I do not understand is this statement:

"In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "xh" is squared, so its value is always zero or greater; being squared, it can never be negative."

I found this on the purplemath website.

2. ## Re: Completing the square to convert to vertex form

Also, in the example they provide, the first-degree x term disappears at and they don't give any explanation for this. Here is the link

Completing the Square: Finding the Vertex

3. ## Re: Completing the square to convert to vertex form

Can you be more clear what you don't understand?
I guess you understand $\displaystyle (x-h)^2>0$ this statement is always true.
And I don't see where the first-degree x term disappears, may be can you give the example?

4. ## Re: Completing the square to convert to vertex form

Now that I look back at the example from the purplemath page I see it more clearly, I understand it now.Yes, I understand that (x-k)^2 > 0, but how come it having to be zero or positive makes it obvious that it is the vertex?

5. ## Re: Completing the square to convert to vertex form

Just to stick my oar in- strictly speaking, $\displaystyle (x- h)^2> 0$ is NOT true. What is true is that $\displaystyle (x- h)^2\ge 0$. Of course, x= 0 gives the vertex itself. For all other x x- h is not 0, and then $\displaystyle (x- h)^2> 0$ so that x= 0 gives the lowest point of the parabola.

6. ## Re: Completing the square to convert to vertex form

I understand the first portion of what you are saying, but could you clarify this part "For all other x x- h is not 0, and then (x- h)^2> 0 so that x= 0 gives the lowest point of the parabola."

Thank you