# What is the vertex of y=3(x-3)^2-x^2+12x - 15?

Jan 19, 2016

$\text{Vertex } \to \left(x , y\right) \to \left(\frac{3}{2} , \frac{15}{2}\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Method:}}$
First simplify the equation so that it is in standard form of:
color(white)("xxxxxxxxxxx)y=ax^2+bx+c

Change this into the form:
color(white)("xxxxxxxxxxx)y=a(x^2+b/ax)+c This is NOT vertex form

Apply $- \frac{1}{2} \times \frac{b}{a} = {x}_{\text{vertex}}$

Substitute ${x}_{\text{vertex}}$ back into the standard form to determine
${y}_{\text{vertex}}$
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Given:$\textcolor{w h i t e}{\ldots . .} y = 3 {\left(x - 3\right)}^{2} - {x}^{2} + 12 x - 15$

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

$y = 3 \left({x}^{2} - 6 x + 9\right) - {x}^{2} + 12 x - 15$

$y = 3 {x}^{2} - 18 x + 27 - {x}^{2} + 12 x - 15$

$y = 2 {x}^{2} - 6 x + 12$ ...........................................(1)

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

Write as: $y = 2 \left({x}^{2} - 3 x\right) + 12$
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$\textcolor{b l u e}{\text{Step 3}}$

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 3\right) = + \frac{3}{2}}$.........................(2)

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$\textcolor{b l u e}{\text{Step 4}}$

Substitute value at (2) into equation (1) giving:

${y}_{\text{vertex}} = 2 {\left(\frac{3}{2}\right)}^{2} - 6 \left(\frac{3}{2}\right) + 12$

${y}_{\text{vertex}} = \frac{18}{4} - \frac{18}{2} + 12$

${y}_{\text{vertex}} = \frac{18}{4} - \frac{36}{4} + 12$

$\textcolor{g r e e n}{{y}_{\text{vertex}} = - \frac{9}{2} + 12 = \frac{15}{2}}$
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$\text{Vertex } \to \left(x , y\right) \to \left(\frac{3}{2} , \frac{15}{2}\right) \to \left(1 \frac{1}{2} , 7 \frac{1}{2}\right)$