Finding the vertex of a quadratic

[Chasney913]

I'd like some pointers on the following question:

Given that y=(x-r)²+(x-s)²
Prove that this equation is at its minimum (i.e. vertex) when x=(r+s)/2

First I tried substituting for x, and came out with something like y=(r-s)(s-r), so I wasn't really sure if that was right. It didn't seem to be.

Then I rearranged the equation into transformational form, without substituting, and came up with the following:

.5(y-.5r²+sr-.5s²)=x-((r+s)/2)

Now, I can see that that transformations are evident.

The generic form is:
1/a(y-k)=(x-q) where a is the vertical stretch, k is the vertical translation, and q is the horizontal translation.

So, the vertex is given by (q,k), and I can see that k does equal (r+s)/2. However, I'm not sure if this constitutes a proof.

Thanks in advance.

[DarkerLine]

You have what seems to be a valid proof. You could also observe that the transformation that maps x to r+s-x, which is a reflection about the line x=(r+s)/2, doesn't change the equation: ((r+s-x)-r)²+((r+s-x)-s)²=(x-r)²+(x-s)². Since the reflection doesn't change the graph, the vertex must actually be on the line x=(r+s)/2.

You could also use calculus, set y'=0, and get 2(x-r)+2(x-s)=0, which comes out to x=(r+s)/2.

[stranger]

Solve both equations for a variable, then substitute. From there, I cannot help you. Darkerline, can you use some simpler words, we aren't all in collegiate calculus.

[DarkerLine]

[Chasney913]

That's okay.

Apparently, my solution is the correct one, so I am pleased.
I really can't use the calculus part, primarily because I don't know it, and also because we haven't been taught it yet, so I really shouldn't know it anyways.

Thanks for confirming that my solution constitutes a proof. I'm kind of new to that aspect of math.