# How do you find the VERTEX of a parabola y=x^2-3x-10?

May 14, 2018

$\text{vertex } = \left(\frac{3}{2} , - \frac{49}{4}\right)$

#### Explanation:

$\text{given a parabola in "color(blue)"standard form } \textcolor{w h i t e}{x} a {x}^{2} + b x + c$

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

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

$y = {x}^{2} - 3 x - 10 \text{ is in standard form}$

$\text{with "a=1,b=-3" and } c = - 10$

$\Rightarrow {x}_{\text{vertex}} = - \frac{- 3}{2} = \frac{3}{2}$

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

${y}_{\text{vertex}} = {\left(\frac{3}{2}\right)}^{2} - 3 \left(\frac{3}{2}\right) - 10 = - \frac{49}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{3}{2} , - \frac{49}{4}\right)$