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
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 "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative."
I found this on the purplemath website.
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
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?
Just to stick my oar in- strictly speaking, is NOT true. What is true is that . Of course, x= 0 gives the vertex itself. For all other x x- h is not 0, and then so that x= 0 gives the lowest point of the parabola.
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