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
y=2 so 2=a+2-1
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 ?