Answer
$\left(-\frac{b}{2a},-\frac{b^2-4ac}{4a}\right)$
Work Step by Step
The $x$ coordinate of the vertex of $y=ax^2+bx+c$ is
$$x=-\frac{b}{2a}.$$
The $y$ coordinate can be found by substituting obtained $x$-coordinate into the function and computing $y$:
$$y=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=-\frac{b^2-4ac}{4a}$$
The vertex is:
$$\left(-\frac{b}{2a},-\frac{b^2-4ac}{4a}\right)$$
