Question

What is the vertex of `y=3x^2 + 6x + 1`?

Answer 1

`(-1, -2)`

Explanation:

Derivate the function and calculate `y'(0)` to find where the the slope is equal to `0`.

`y = 3x^2 + 6x + 1`
`y' = 2 * 3x^(2-1) + 1 * 6x^(1-0)`
`y' = 6x + 6`

Calculate `y'(0)`:
`y'(0) = 0`
`6x + 6 = 0`
`6x = -6`
`x = -1`

Put this `x` value into the original function to find the y-value.
NOTE: Put it in `y`, not `y'`.

`y = 3*(-1)^2 + 6*(-1) + 1`
`y = 3 * 1 - 6 + 1`
`y = 3 - 6 + 1 = -2`

The vertex is at `(-1, -2)`