# What is the vertex of  y = (1/6)(3x – 15)^2 - 31?

Feb 19, 2016

Vertex$\text{ "=" "(x,y)" "->" } \left(5 , - 31\right)$

#### Explanation:

There are three things we need to consider as a pre-amble before we start.

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$\textcolor{b l u e}{\text{Point 1}}$
Consider ${\left(3 x\right)}^{2}$ Inside the brackets the coefficient is presented as 3. Outside the bracket it has been squared so it will be 9 in that:

$9 \times {\left(x\right)}^{2} = {\left(3 x\right)}^{2}$ another example $\to \text{ } 16 \times {\left(x\right)}^{2} = {\left(4 x\right)}^{2}$

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$\textcolor{b l u e}{\text{Point 2}}$

$\frac{1}{3} \times {\left(3 x - 15\right)}^{2} = {\left(\frac{3 x}{\sqrt{3}} - \frac{15}{\sqrt{3}}\right)}^{2}$

so $\frac{1}{9} {\left(3 x - 15\right)}^{2} = {\left(\frac{3 x}{3} - \frac{15}{3}\right)}^{2}$
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$\textcolor{b l u e}{\text{Point 3}}$

To convert the given equation into vertex form we need to end up with the format of:

$y = a {\left(x - \frac{b}{2 a}\right)}^{2} + c \text{ }$ where $b$ can be positive or negative.
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$\textcolor{b l u e}{\text{Solving your question}}$

With the format of the given question you are already part way to building the vertex equation format of completing the square. So this is what I am going to do.

Given:$\text{ } y = \left(\frac{1}{6}\right) {\left(3 x - 15\right)}^{2} - 31$

To remove the coefficient of $x$ within the brackets multiply the bracketed part by 1, but in the form of $\textcolor{b l u e}{\frac{9}{9}}$

$y = \textcolor{b l u e}{\frac{9}{9}} \left(\frac{1}{6}\right) {\left(3 x - 15\right)}^{2} - 31$

$y = \frac{\textcolor{b l u e}{9}}{6} {\left(\frac{3 x}{\textcolor{b l u e}{3}} - \frac{15}{\textcolor{b l u e}{3}}\right)}^{2} - 31$

y=9/6(x-5)^2-31" "color(brown)("This is vertex form") Thus:

${x}_{\text{vertex}} = \left(- 1\right) \times \left(- 5\right) = 5$

${y}_{\text{vertex}} = - 31$ Notice that this is the value of the constant $c$

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Vertex$\text{ "=" "(x,y)" "->" } \left(5 , - 31\right)$