Answer
$$
y=a x^{2}+b x+c
$$
The vertex of this parabola is
$$
(\frac{-b}{2a} ,\frac{4 a c-b^{2}}{4 a}).
$$
Work Step by Step
$$
y=a x^{2}+b x+c
$$
To find the vertex of this parabola. First, we find $y^{\prime}(x) $,
$$
\begin{aligned}
y^{\prime}(x) &=2 a x+b\\
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $y^{\prime} =0 $. Now, solve the equation $y^{\prime} =0 $ to get
$$
\begin{aligned}
y^{\prime}(x) =2 a x+b &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2 a x &=-b\\
\Rightarrow\quad\quad\quad\quad\quad\\
x & =\frac{-b}{2a} \\
\end{aligned}
$$
The only critical number is : $x =\frac{-b}{2a} .$
When $x=\frac{-b}{2a} $
$$
\begin{aligned}
y &=a x^{2}+b x+c \\
&= a\left(\frac{-b}{2 a}\right)^{2}+b\left(\frac{-b}{2 a}\right)+c \\
&=\frac{a b^{2}}{4 a^{2}}-\frac{b^{2}}{2 a}+c \\
&=\frac{b^{2}}{4 a}-\frac{2 b^{2}}{4 a}+\frac{4 a c}{4 a} \\
&=\frac{4 a c-b^{2}}{4 a}
\end{aligned}
$$
So, the vertex of this parabola is $(\frac{-b}{2a} ,\frac{4 a c-b^{2}}{4 a})$.
