# How do you find the vertex of f(x)= 2x^2-4x+6?

Mar 21, 2017

Vertex$\to \left(x , y\right) = \left(1 , 4\right)$

#### Explanation:

Using part of the process of completing the square to determine the value of $x$. Then determining the value of $y$ by substitution.

Given:$\text{ } y = 2 {x}^{2} - 4 x + 6$

Write as:$\text{ } y = 2 \left({x}^{2} \textcolor{red}{- \frac{4}{2}} x\right) + 6$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- \frac{4}{2}}\right) = 1$

By substitution:

${y}_{\text{vertex}} = 2 {\left(1\right)}^{2} - 4 \left(1\right) + 6 = 4$

Vertex$\to \left(x , y\right) = \left(1 , 4\right)$

Compare to the standardised form of $y = a {x}^{2} + b x + c$

Notice that in the graph that ${y}_{\text{intercept}} = c = 6$