How do you find the vertex of y=2x^2-10x+3?

Aug 3, 2018

$\text{vertex } = \left(\frac{5}{2} , - \frac{19}{2}\right)$

Explanation:

$\text{given a parabola in standard form}$

$y = a {x}^{2} + b x + c \textcolor{w h i t e}{x} a \ne 0$

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

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

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

$\text{with "a=2,b=-10" and } c = 3$

${x}_{\text{vertex}} = - \frac{- 10}{4} = \frac{5}{2}$

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

${y}_{\text{vertex}} = 2 {\left(\frac{5}{2}\right)}^{2} - 10 \left(\frac{5}{2}\right) + 3 = - \frac{19}{2}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{5}{2} , - \frac{19}{2}\right)$
graph{2x^2-10x+3 [-20, 20, -10, 10]}