Answer
The x-coordinate of the vertex of the parabola whose equation is $y=a{{x}^{2}}+bx+c$ occurs when the derivative of the function is zero.
Work Step by Step
Consider the parabola $y=a{{x}^{2}}+bx+c$ ,
The derivative of a function is given by the formula, ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$
Find the derivative of the equation $y=a{{x}^{2}}+bx+c$ using the above formula,
$\begin{align}
& {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{a{{\left( x+h \right)}^{2}}+b\left( x+h \right)+c-\left( a{{x}^{2}}+bx+c \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{a\left( {{x}^{2}}+{{h}^{2}}+2xh \right)+bx+bh+c-a{{x}^{2}}-bx-c}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{a{{x}^{2}}+a{{h}^{2}}+2axh+bh-a{{x}^{2}}}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{a{{h}^{2}}+2axh+bh}{h}
\end{align}$
Further solve the above.
$\begin{align}
& {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( ah+2ax+b \right) \\
& =a\left( 0 \right)+2ax+b \\
& =2ax+b
\end{align}$
Thus, the derivative of the function is ${f}'\left( x \right)=2ax+b$.
Now, take ${f}'\left( x \right)=0$ and solve for x,
$\begin{align}
& {f}'\left( x \right)=2ax+b \\
& 0=2ax+b \\
& -2ax=b \\
& x=-\frac{b}{2a}
\end{align}$
Thus, the value of x for which the derivative of the function is zero is $x=-\frac{b}{2a}$
It is known that the x-coordinate of the vertex of a parabola $y=a{{x}^{2}}+bx+c$ is $-\frac{b}{2a}$ ,
Thus, the x-coordinate of the vertex of the parabola whose equation is $y=a{{x}^{2}}+bx+c$ occurs when the derivative of the function is zero.
