Question

How do you find the vertex of `f(x) = x^2 + 3`?

Answer 1

`(0, 3)`

Explanation:

The equation of a parabola (with vertical axis) in vertex form can be written:

`f(x) = a(x-h)^2 + k`

where `(h, k)` is the vertex and `a != 0` a constant multiplier.

In our case we find:

`f(x) = 1(x-0)^2+3`

which means that `(h, k) = (0, 3)` and `a=1`

So the vertex is `(0, 3)`

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Alternatively, note that `x^2 >= 0` with minimum value `0` only when `x=0`. Hence the minimum value of `f(x)` is also when `x=0`. So the `x` coordinate of the vertex must be `0`.

Substituting `x=0` in the formula for `f(x)` gives us `f(0) = 3`.

Hence the vertex is at `(0, 3)`