# What is the the vertex of y = 2x^2 - 14x-5?

Jan 14, 2016

$\left({x}_{\text{vertex") , y_("vertex}}\right) \to \left(3 \frac{1}{2} , - 29 \frac{1}{2}\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Method 1}}$

Given that the standard form for a quadratic equation is:

$a {x}^{2} + b x + c = 0$

and: $\textcolor{w h i t e}{\ldots .} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Then you could use this to find the $x$ intercepts and that ${x}_{\text{vertex}}$ is half way between them. That is $\textcolor{b l u e}{- \frac{b}{2 a}}$
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$\textcolor{b l u e}{\text{Method 2}}$

$\textcolor{b r o w n}{\text{Use something that is similar to completing the square:}}$
$\textcolor{g r e e n}{\text{When you think about this, it is the same thing as method 1!}}$

Write as: $y = 2 \left({x}^{2} - \frac{14}{2} x\right) - 5$

Now consider just the brackets

$\textcolor{b l u e}{{x}_{\text{vertex}} =} \left(- \frac{1}{2}\right) \times \left(- \frac{14}{2}\right) = + \frac{14}{4} = \textcolor{b l u e}{+ 3 \frac{1}{2}}$
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Having found ${x}_{\text{vertex}}$ we can find the value of ${y}_{\text{vertex}}$ by substituting for $x$ in the original equation.

${y}_{\text{vertex}} = 2 {x}^{2} - 14 x - 5$

${y}_{\text{vertex}} = 2 {\left(\frac{7}{2}\right)}^{2} - 14 \left(\frac{7}{2}\right) - 5$

$\textcolor{b l u e}{{y}_{\text{vertex}} =} \frac{49}{2} - 49 - 5 = \textcolor{b l u e}{- 29 \frac{1}{2}}$
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$\left({x}_{\text{vertex") , y_("vertex}}\right) \to \left(3 \frac{1}{2} , - 29 \frac{1}{2}\right)$