# What is the vertex of y= 2x^2 - 4x + 3?

Nov 26, 2015

$\textcolor{p u r p \le}{\left({x}_{\text{vertex"), y_("vertex}}\right) = \left(- 1 , 9\right)}$

#### Explanation:

Given: $y = 2 {x}^{2} - 4 x + 3. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

Let the coordinates of the vertex be $\left({x}_{\text{vertex"), y_("vertex}}\right)$
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$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .} \textcolor{g r e e n}{\text{ Preamble}}$

There are two ways of doing this.

It looks as though the fashionable way at the moment is 'completing the square' alternatively known as a 'vertex equation'.

The other, which I am going to show you, is the basis upon which completing the square is built on.
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$\textcolor{B l u e}{\underline{S t e p \textcolor{w h i t e}{.} 1}}$

Write the given equation in the form of:
$y = 2 \left({x}^{2} - 2 x + \frac{3}{2}\right)$

I have made it such that I have a part that starts with ${x}^{2}$

We now look at the part inside the brackets of $- 2 x$

We then do this: $- \frac{1}{2} \times - 2 = - 1$ ignoring the variable $x$

We have now found $\textcolor{g r e e n}{{x}_{\text{vertex}} = - 1} \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

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$\textcolor{B l u e}{\underline{S t e p \textcolor{w h i t e}{.} 2}}$

Substitute (2) into (1) giving:

${y}_{\text{vertex}} = 2 {\left(- 1\right)}^{2} - 4 \left(- 1\right) + 3$

${y}_{\text{vertex}} = 2 + 4 + 3$

$\textcolor{g r e e n}{{y}_{\text{vertex}} = 9}$
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$\textcolor{p u r p \le}{\left({x}_{\text{vertex"), y_("vertex}}\right) = \left(- 1 , 9\right)}$ 