# How do you find the vertex of a parabola y = x^2 + 3?

Jul 10, 2015

the vertex of $f \left(x\right)$ is $3$ when $x = 0$

#### Explanation:

Let $a , b , c$, 3 numbers with $a \ne 0$

Let $p$ a parabolic function such as $p \left(x\right) = a \cdot {x}^{2} + b \cdot x + c$

A parabola always admit a minimum or a maximum (= his vertex).

We have a formula to find easily the abscissa of a vertex of a parabola :

Abscissa of vertex of $p \left(x\right) = - \frac{b}{2 a}$



Let $f \left(x\right) = {x}^{2} + 3$
Then, the vertex of $f \left(x\right)$ is when $\frac{0}{2} = 0$

And $f \left(0\right) = 3$


Therefore the vertex of $f \left(x\right)$ is $3$ when $x = 0$

Because $a > 0$ here, the vertex is a minimum.

graph{x^2+3 [-5, 5, -0.34, 4.66]}
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