## Precalculus (6th Edition) Blitzer

The required value is$\left( 2,\ -5 \right)$
The quadratic function of the form, $f\left( x \right)=a{{x}^{2}}+bx+c$ , can be converted into its standard form as $f\left( x \right)=a{{\left( x+\frac{b}{2a} \right)}^{2}}+c-\frac{{{b}^{2}}}{4a}$ The quadratic function in its standard form can be written as $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k,\,\,\,\,\,\,\,\,a\ne 0$ Here, $a,\ h,\ k$ are constants and $x$ is a variable. The graph of $f\left( x \right)$ is a parabola which is symmetric about the line $x=h$. Now, the coordinates of the vertex of the parabola are $\left( h,k \right)$. Compare this with standard form to get $h=-\frac{b}{2a}$ and $k=c-\frac{{{b}^{2}}}{4a}$. The coordinates of the vertex of the parabola can also be written as $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)$. Let us consider the given function $f\left( x \right)=2{{x}^{2}}-8x+3$. Compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$ to get $a=2,\ b=-8$. The $x$ -coordinates of vertex are \begin{align} & h=-\frac{b}{2a} \\ & =-\left( \frac{-8}{4} \right) \\ & =2 \end{align} Calculate y-coordinate by determining $f\left( 2 \right)$: \begin{align} & f\left( 2 \right)=2{{\left( 2 \right)}^{2}}-8\left( 2 \right)+3 \\ & =-5 \end{align} Hence, the coordinates of the vertex are $\left( 2,\ -5 \right)$