Question

How do you find the vertex and the intercepts for `f(x) = x^2 + 6x + 3`?

Answer 1

Vertex is at `(-3,-6)`.
The `x`-axis intercepts are `(-3-sqrt(6),0)` and `(-3+sqrt(6),0)`
The `y`-axis intercept is `(0,3)`.

Explanation:

To find the vertex, complete the square:
`f(x)=(x+6/2)^2-6`, which is
`(x+3)^2-6`
This is clearly a minimum value of `-6` when `x=-3`, so the vertex is at `(-3,-6)`, because `(-3)^2+6(-3)+3=-6`.

The intercept with the `y`-axis is when `x=0`, `y` as `0^2+6(0)+3`, giving the intercept as `(0,3)` (i.e., just read off the constant term).

The intercepts, if any, with the `x`-axis have `x`-coordinates which are the roots of the equation `f(x)=0`, which are the roots of `(x+3)^-6=0`. Clearly these are at `(x+3)^2=6`, that is `(x+3)=+-sqrt(6)` so `x=-3+-sqrt(6)`.

Alternatively, use the quadratic formula to get `x=(-(6)+-sqrt((6)^2-4xx(1)xx(3)))/(2xx(1))`
`x=-3+-sqrt(6)` because `sqrt(24)=2xxsqrt(6)`