Relationship between apex and derivative

Does somebody know if there's relationship between apex and derivative ?
If there's formula y=ax^2+bx+c then the x apex is -b/2a and y apex is y(x). But can I get the same result by using function's derivative ?

I tried to determine a and b according to the condition that parabola's y=a+bx-x^2 apex point would be (1;2)
I assumed that at the apex the slop is 0 and so did some substitutions:
firstly first derivative is: y'=0+b-2x
0=b(1)-2
b=2
y=2 so 2=a+2-1
2=a+1
a=1

It appears to be correct but the way I am used to use for finding apexes don't work with the formula y=a+bx-x^2

Any ideas ?

So, what is your problem? What don't you understand? You got it correctly [except "0 = b(1) -2". That should have been 0 = b -2(1)], so ....?

You used derivatives--yes, the derivative is correct, and yes, at the "apex" or vertex, the slope of the tangent line, or the 1st derivative, is zero.
So, what is wrong?

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"...but the way I am used to use for finding apexes don't work with the formula y=a+bx-x^2."

And what is that way?

The "x = -b/2a" way?

y = a +bx -x^2
y' = b -2x
y' = 0
0 = b -2x
2x = b
x = b/2 --------the x-coordinate of the apex/vertex of y = a +bx -x^2.

Is that wrong?
Is this the reason why it does not work with y = a +bx -x^2 ?

y = a +bx -x^2
Rewriting that,
y = -x^2 +bx +a ----(1)
At the vertex,
x = "-b/2a"
Here, in (1), "b" is +b; and "a" is -1, so,
x = -(+b)/[2(-1)]
x = -b/-2
x = b/2 ----------is that the same as the x = b/2 above?

Quote:

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Originally Posted by ticbol

y = a +bx -x^2
y' = b -2x
y' = 0
0 = b -2x
2x = b
x = b/2 --------the x-coordinate of the apex/vertex of y = a +bx -x^2.

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That's the one I wanted to know. Thanks for that.
You shouldn't write so long explanation. The one which is above on quota is enough to understand.

You have a problem with me then.
I always give long explanations.