What is the vertex form of y=3x^2+4x+1?

Jun 12, 2018

$y = 3 {\left(x + \frac{2}{3}\right)}^{2} + \frac{7}{3}$

Explanation:

$y = 3 {x}^{2} + 4 x + 1$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{4}{6} = - \frac{2}{3}$
y-coordinate of vertex:
$y \left(- \frac{2}{3}\right) = \frac{12}{9} - \frac{8}{3} + 1 = \frac{4}{3} - \frac{8}{3} + 1 = \frac{7}{3}$
$V e r t e x \left(- \frac{2}{3} , \frac{7}{3}\right)$
Vertex form of y:
$y = 3 {\left(x + \frac{2}{3}\right)}^{2} + \frac{7}{3}$