# What is the the vertex of y = x^2-x+16?

May 11, 2018

$\text{vertex } = \left(\frac{1}{2} , \frac{63}{4}\right)$

#### Explanation:

$\text{given a quadratic in standard form} \textcolor{w h i t e}{x} a {x}^{2} + b x + c$

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

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

$y = {x}^{2} - x + 16 \text{ is in standard form}$

$\text{with "a=1,b=-1" and } c = 16$

$\Rightarrow {x}_{\text{vertex}} = - \frac{- 1}{2} = \frac{1}{2}$

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

${y}_{\text{vertex}} = {\left(\frac{1}{2}\right)}^{2} - \frac{1}{2} + 16 = \frac{63}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{2} , \frac{63}{4}\right)$