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

The equation represents a quadratic function, hence its graph would be a vertical parabola, opening upwards, as the coefficient of ${x}^{2}$ is positive. To sketch the curve we can get the vertex and the axis of symmetry from the give equation as follows:
y= $3 \left({x}^{2} + 2 x\right) + 1$
= $3 \left({x}^{2} + 2 x + 1\right) + 1 - 3$
= $3 {\left(x + 1\right)}^{2} - 2$
This shows the vertex as (-1, -2) and the axis of symmetry x= -1. The x intercepts would be$- 1 \pm \frac{1}{3} \sqrt{6}$. The y intercept would be (0,1). With all these inputs, the curve can be easily sketched.