Question

How do you find the vertex and the intercepts for `y=2x^2 + 2x + 1`?

Answer 1

Vertex is at `(-1/2,1/2)`
y intercept is at `(0,1)`

Explanation:

To find the vertex , first complete the square:
`y=2x^2+2x+1`
`y=2[x^2+x+1/2]`
`y=2[(x+1/2)^2+1/4]`

`y=2(x+1/2)^2+1/2`

Now we know the horizontal, vertical, and dilation of `y = x^2`:
The vertex is at (-1/2,1/2).

Y-intercepts: Plug in zero for x, get 1

X-intercepts: solve for x using the vertex form by setting y equal to 0:
`0=2(x+1/2)^2+1/2`
`-1/2=2(x+1/2)^2`
`-1=(x+1/2)^2`

This function does not have any x intercepts. Its vertex lies above the axis and it is a positive parabola.
graph{2x^2+2x+1 [-8.23, 7.57, -1.61, 6.29]}