# How do you find the vertex of y= x^2 - 6x + 14?

Mar 19, 2016

Vertex$\text{ } \to \left(x , y\right) \to \left(3.5\right)$

#### Explanation:

The question does not define the method to be used so you have freedom of choice.

Let me show you a 'sort of' cheat method. Actually it forms part completing the square but does not take to the end of the process.

Consider standard form$\text{ } y = a {x}^{2} + b x + c$

Write this as:$\text{ } y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

Then you have $x - \left(\text{vertex}\right) = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

For this question:
$a = 1$
$\frac{b}{a} = \frac{- 6}{1} = - 6$

=>" "color(blue)(x_("vertex")=(-1/2)xx(-6) = +3)

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By substitution

$\textcolor{b r o w n}{y = {x}^{2} - 6 x + 14 \textcolor{b l u e}{\to {y}_{\text{vertex}} = {\left(3\right)}^{2} - 6 \left(3\right) + 14}}$

$\textcolor{b l u e}{{y}_{\text{vertex}} = 9 - 18 + 14 = 5}$

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