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

Apr 25, 2016

$\left(0 , 3\right)$

#### Explanation:

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

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

where $\left(h , k\right)$ is the vertex and $a \ne 0$ a constant multiplier.

In our case we find:

$f \left(x\right) = 1 {\left(x - 0\right)}^{2} + 3$

which means that $\left(h , k\right) = \left(0 , 3\right)$ and $a = 1$

So the vertex is $\left(0 , 3\right)$

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Alternatively, note that ${x}^{2} \ge 0$ with minimum value $0$ only when $x = 0$. Hence the minimum value of $f \left(x\right)$ is also when $x = 0$. So the $x$ coordinate of the vertex must be $0$.

Substituting $x = 0$ in the formula for $f \left(x\right)$ gives us $f \left(0\right) = 3$.

Hence the vertex is at $\left(0 , 3\right)$