# How do you graph y=3x^2+6x-1?

y= $3 \left({x}^{2} + 2 x\right) - 1$
= $3 \left({x}^{2} + 2 x + 1\right) + 1 - 3$
= $3 {\left(x + 1\right)}^{2} - 2$
Since coefficient of${x}^{2}$ is positive, the parabola would open up, its vertex would be at (-1,-2), the axis of symmetry would be line x= -1. X-intercepts would be $- 1 \pm \frac{2}{3} \sqrt{3}$