# How do you find the vertex of the parabola y=3x^2+5x+8?

$\left(- \frac{5}{6} , \frac{71}{12}\right)$

#### Explanation:

Given parabola:

$y = 3 {x}^{2} + 5 x + 8$

$y = 3 \left({x}^{2} + \frac{5}{3} x\right) + 8$

$y = 3 \left({x}^{2} + 2 \left(\frac{5}{6}\right) x + {\left(\frac{5}{6}\right)}^{2}\right) - 3 {\left(\frac{5}{6}\right)}^{2} + 8$

$y = 3 {\left(x + \frac{5}{6}\right)}^{2} - \setminus \frac{25}{12} + 8$

$y = 3 {\left(x + \frac{5}{6}\right)}^{2} + \setminus \frac{71}{12}$

$3 {\left(x + \frac{5}{6}\right)}^{2} = y - \frac{71}{12}$

${\left(x + \frac{5}{6}\right)}^{2} = \frac{1}{3} \left(y - \frac{71}{12}\right)$

The above equation is in standard formula of upward parabola: ${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right) \setminus \setminus \forall \setminus \setminus a > 0$ which has

Vertex: $\left(x - {x}_{1} = 0 , y - {y}_{1} = 0\right)$

$\left(x + \frac{5}{6} = 0 , y - \frac{71}{12} = 0\right)$

$\setminus \equiv \left(- \frac{5}{6} , \frac{71}{12}\right)$

Jul 21, 2018

$\text{vertex } = \left(- \frac{5}{6} , \frac{71}{12}\right)$

#### Explanation:

$\text{given a parabola in "color(blue)"standard form}$

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

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

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

$y = 3 {x}^{2} + 5 x + 8 \text{ is in standard form}$

$\text{with "a=3,b=5" and } c = 8$

${x}_{\text{vertex}} = - \frac{5}{6}$

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

${y}_{\text{vertex}} = 3 {\left(- \frac{5}{6}\right)}^{2} + 5 \left(- \frac{5}{6}\right) + 8$

$\textcolor{w h i t e}{\times \times} = \frac{25}{12} - \frac{50}{12} + \frac{96}{12} = \frac{71}{12}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{5}{6} , \frac{71}{12}\right)$