Question

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

Answer 1

The vertex: `(-1,-2)`
The y-intercept: `(0,1)`
The y-intercept reflected over axis of symmetry: `(-2,1)`

Explanation:

`(-b)/(2a) = (-6)/(2*3) = -1` This is the x-coordinate of the vertex.
`y=3(-1)^2+6(-1)+1=-2` This is the y-coordinate of the vertex.

The vertex: `(-1,-2)`

Now plug in `0` for x:
`y = 3(0)^2 + 6(0) + 1 = 1`

The y-intercept: `(0,1)`

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