# How do you find the axis of symmetry and vertex point of the function: y=2x^2+6x+4?

Find vertex of $y = 2 {x}^{2} + 6 x + 4$
$x = \left(- \frac{b}{2} a\right) = - \frac{6}{4} = - \frac{3}{2}$
$y = f \left(- \frac{3}{2}\right) = 2 \left(\frac{9}{4}\right) + 6 \left(- \frac{3}{2}\right) + 4 = \frac{18}{4} - \frac{18}{2} + 4 = - \frac{2}{4} = - \frac{1}{2}$