# How do you find the vertex and intercepts for f(x) = 2x^2 - 4x + 3?

Jun 14, 2018

$\text{see explanation}$

#### Explanation:

$\text{express the quadratic in "color(blue)"vertex form}$

•color(white)(x)y=a(x-h)^2+k

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

$\text{using the method of "color(blue)"completing the square}$

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

$\textcolor{w h i t e}{f \left(x\right)} = 2 \left({x}^{2} + 2 \left(- 1\right) x + 1 - 1 + \frac{3}{2}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = 2 {\left(x - 1\right)}^{2} + 1$

$\text{vertex } = \left(1 , 1\right)$

$\text{for y-intercept set x = 0}$

$f \left(0\right) = 3 \Rightarrow \left(0 , 9\right)$

$\text{for x-intercepts set y = 0}$

$2 {\left(x - 1\right)}^{2} + 1 = 0$

${\left(x - 1\right)}^{2} = - \frac{1}{2}$

$\text{this has no real solutions hence no x-intercepts}$
graph{2x^2-4x+3 [-10, 10, -5, 5]}