# What is the vertex of  y= 2(x-5)^2-x^2-3x-1?

Jan 29, 2018

$\left(\frac{23}{2} , - \frac{333}{4}\right)$

#### Explanation:

$\text{require to express y in standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\Rightarrow y = 2 \left({x}^{2} - 10 x + 25\right) - {x}^{2} - 3 x - 1$

$\textcolor{w h i t e}{\Rightarrow y} = 2 {x}^{2} - 20 x + 50 - {x}^{2} - 3 x - 1$

$\textcolor{w h i t e}{\Rightarrow y} = {x}^{2} - 23 x + 49 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{with "a=1,b=-23" and } c = 49$

$\text{then the x-coordinate of the vertex is}$

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

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 23}{2} = \frac{23}{2}$

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

${y}_{\textcolor{red}{\text{vertex}}} = {\left(\frac{23}{2}\right)}^{2} - 23 \left(\frac{23}{2}\right) + 49$

$\textcolor{w h i t e}{\times \times} = \frac{529}{4} - \frac{529}{2} + 49 = - \frac{333}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{23}{2} , - \frac{333}{4}\right)$