Precalculus (6th Edition) Blitzer

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.
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.