# How do you find the axis of symmetry for a quadratic equation y = x^2 + 6x + 13?

The axis of symmetry of a quadratic equation is the line parralel to the $O y$ axis passing through the vertex of the parabola. Therefore, we need the ${x}_{v}$ coordinate of the vertex. $V \left({x}_{v} , {y}_{v}\right)$
For a quadratic equation $f \left(x\right) = a {x}^{2} + b x + c$, we have the following formulae for the coordinates of the vertex:
${x}_{v} = - \frac{b}{2 a}$ and ${y}_{v} = - \frac{\Delta}{4 a}$, where $\Delta = {b}^{2} - 4 a c$
Therefore, ${x}_{v} = - \frac{6}{2 \cdot 1} = - 3$
So, the axis of symmetry of the equation $y = {x}^{2} + 6 x + 13$ is the line defined by the equation $x = - 3$.