# How do you find the vertex of the parabola y=2x^2-8x+7?

Feb 9, 2015

You can find the coordinates of your vertex either using a mnemonic or the derivative:

1) Mnemonic: the coordinates of the vertex of a parabola in the form:
$y = a {x}^{2} + b x + c$ are:
${x}_{v} = - \frac{b}{2 a}$
${y}_{v} = - \frac{\Delta}{4 a}$ where $\Delta = {b}^{2} - 4 a c$
${x}_{v} = 2$
${y}_{v} = - 1$

2) Derivative: determine the derivative of your function and set it equal to zero:
$y ' = 4 x - 8$
$4 x - 8 = 0$ gived $x = {x}_{v} = 2$
$y = {y}_{v} = 2 \cdot 4 - 8 \cdot 2 + 7 = - 1$

Graphically:
graph{2x^2-8x+7 [-4.385, 4.386, -2.19, 2.193]}

Mar 5, 2016

A slight variant on method

color(blue)("Vertex" ->(x,y)->(2,-1)

#### Explanation:

Given:$\text{ } 2 {x}^{2} - 8 x + 7$

Write as:$\text{ } 2 \left({x}^{2} - \textcolor{red}{\frac{8}{2}} x\right) + 7$

Consider the $\textcolor{red}{- \frac{8}{2}} \text{ from } - \frac{8}{2} x$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- \frac{8}{2}}\right) = + \frac{8}{4} = 2}$

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Substitute $x = \textcolor{m a \ge n t a}{2}$ in the original equation to find ${y}_{\text{vertex}}$

y=2x^2-8x+7" "->" "y_("vertex")=2(color(magenta)(2))^2-8(color(magenta)(2))+7

$\textcolor{b l u e}{{y}_{\text{vertex}} = 8 - 16 + 7 = - 1}$

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color(blue)("Vertex" ->(x,y)->(2,-1)