The vertex on a parabola can be generalized trivially using differential calculus. See that [tex]f(x) = ax^2 + bx + c[/tex] is a parabola.
Now we use the formula [tex]\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}(x^m) = mx^{m-1}[/tex] and [tex]\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}(f(x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x} f(x) + \frac{\mathrm{d}}{\mathrm{d}x}g(x)[/tex] to solve this problem. We use the derivative of [tex]f[/tex]
[tex]\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} f(x) = \frac{\mathrm{d}}{\mathrm{d}x}(ax^2) + \frac{\mathrm{d}}{\mathrm{d}x}(bx) + \frac{\mathrm{d}}{\mathrm{d}x}(c) = 2ax + b + 0[/tex]
We find where [tex]\frac{\mathrm{d}}{\mathrm{d}x} f(x) = 0[/tex]. So this is when [tex]\displaystyle 2ax + b = 0 \Rightarrow x = \frac{-b}{2a}[/tex].
This gives vertex x location. To find y location you calculate ycoordinate using f(x).
[tex]f\left( \frac{-b}{2a} \right) = a\left( \frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c = c - \frac{b^2}{4a}[/tex]
The vertex is [tex]V\left( \frac{-b}{2a}, c - \frac{b^2}{4a} \right)[/tex]
