# What are the vertex, focus and directrix of  y=x^2-x+19 ?

Feb 7, 2018

$\text{see explanation}$

#### Explanation:

$\text{given the equation of a parabola in standard form}$

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

$\text{then the x-coordinate of the vertex which is also}$
$\text{the axis of symmetry is}$

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

$y = {x}^{2} - x + 19 \text{ is in standard form}$

$\text{with "a=1,b=-1" and } c = 19$

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

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

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = {\left(\frac{1}{2}\right)}^{2} - \frac{1}{2} + 19 = \frac{75}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{2} , \frac{75}{4}\right)$

$\Rightarrow y = {\left(x - \frac{1}{2}\right)}^{2} + \frac{75}{4} \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\text{the translated form of a vertically opening parabola is}$

•color(white)(x)(x-h)^2=4p(y-k)

$\text{where "(h,k)" are the coordinates of the vertex and}$
$\text{p is the distance from the vertex to the focus/directrix}$

$\Rightarrow {\left(x - \frac{1}{2}\right)}^{2} = 1 \left(y - \frac{75}{4}\right) \leftarrow \textcolor{b l u e}{\text{translated form}}$

$\text{with } 4 p = 1 \Rightarrow p = \frac{1}{4}$

$\text{the focus lies on the axis of symmetry } x = \frac{1}{2}$

$\text{since "a>0" then parabola opens up } \bigcup$

$\text{hence the focus is "1/4" unit above the vertex and}$
$\text{the directrix "1/4" unit below the vertex}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{focus }} = \left(\frac{1}{2} , 19\right)$

$\text{and equation of directrix is } y = \frac{37}{2}$