How do you find the vertex and axis of symmetry for f (x)= 2x² - 6x+ 3?

May 3, 2017

Given and equation of the form: $f \left(x\right) = a {x}^{2} + b x + c$
The equation for the axis of symmetry is: $x = - \frac{b}{2 a}$
The x coordinate for the vertex, h, is the same.
The y coordinate, $k = f \left(h\right)$

Explanation:

Given: $f \left(x\right) = 2 {x}^{2} - 6 x + 3$

$a = 2 , b = - 6 \mathmr{and} c = 3$

The equation for the axis of symmetry is:

$x = - \frac{- 6}{2 \left(2\right)}$

$x = \frac{3}{2}$

The x coordinate of the vertex is the same:

$h = \frac{3}{2}$

The y coordinate of the vertex is:

$k = f \left(h\right)$

$k = 2 {\left(h\right)}^{2} - 6 \left(h\right) + 3$

$k = 2 {\left(\frac{3}{2}\right)}^{2} - 6 \left(\frac{3}{2}\right) + 3$

$k = - \frac{3}{2}$

The vertex is $\left(\frac{3}{2} , - \frac{3}{2}\right)$

May 3, 2017

$\left(\frac{3}{2} , - \frac{3}{2}\right) , x = \frac{3}{2}$

Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where " (h,k)" are the coordinates of the vertex}$
$\text{and a is a constant}$

$\text{to express f(x) in this form "color(blue)"complete the square}$

$f \left(x\right) = 2 \left({x}^{2} - 3 x \textcolor{red}{+ \frac{9}{4}} \textcolor{red}{- \frac{9}{4}}\right) + 3$

$\textcolor{w h i t e}{f \left(x\right)} = 2 {\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{2} + 3$

$\textcolor{w h i t e}{f \left(x\right)} = 2 {\left(x - \frac{3}{2}\right)}^{2} - \frac{3}{2} \leftarrow \textcolor{red}{\text{ in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex}} = \left(\frac{3}{2} , - \frac{3}{2}\right)$

$\text{since " a>0" then minimum } \bigcup$

$\text{the axis of symmetry passes through the vertex}$ is vertical

$\text{with equation } x = \frac{3}{2}$
graph{(y-2x^2+6x-3)(y-1000x+1500)=0 [-10, 10, -5, 5]}