# What are the important points needed to graph y=3x^2+6x+1?

Nov 5, 2015

The vertex: $\left(- 1 , - 2\right)$
The y-intercept: $\left(0 , 1\right)$
The y-intercept reflected over axis of symmetry: $\left(- 2 , 1\right)$

#### Explanation:

$\frac{- b}{2 a} = \frac{- 6}{2 \cdot 3} = - 1$ This is the x-coordinate of the vertex.
$y = 3 {\left(- 1\right)}^{2} + 6 \left(- 1\right) + 1 = - 2$ This is the y-coordinate of the vertex.

The vertex: $\left(- 1 , - 2\right)$

Now plug in $0$ for x:
$y = 3 {\left(0\right)}^{2} + 6 \left(0\right) + 1 = 1$

The y-intercept: $\left(0 , 1\right)$

Now reflect that point over the axis of symmetry ($x = - 1$) to get
$\left(- 2 , 1\right)$
to get this, you take $- 1 - \left(0 - \left(- 1\right)\right)$