# What is the vertex of  y= 6(x-3)^2-x^2-5x+3?

Aug 12, 2018

$\text{vertex } = \left(\frac{41}{10} , - \frac{541}{20}\right)$

#### Explanation:

$\text{expand and express in standard form}$

$y = 6 \left({x}^{2} - 6 x + 9\right) - {x}^{2} - 5 x + 3$

$\textcolor{w h i t e}{x} = 6 {x}^{2} - 36 x + 54 - {x}^{2} - 5 x + 3$

$\textcolor{w h i t e}{y} = 5 {x}^{2} - 41 x + 57 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{with "a=5,b=-41" and } c = 57$

$\text{given the equation in standard form then the x-coordinate}$
$\text{of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

${x}_{\text{vertex}} = - \frac{- 41}{10} = \frac{41}{10}$

$\text{substitute this value into the equation for y-coordinate}$

${y}_{\text{vertex}} = 5 {\left(\frac{41}{10}\right)}^{2} - 41 \left(\frac{41}{10}\right) + 57$

$\textcolor{w h i t e}{\times \times} = \frac{1681}{20} - \frac{3362}{20} + \frac{1140}{20} = - \frac{541}{20}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{41}{10} , - \frac{541}{20}\right)$