What is the axis of symmetry and vertex for the graph y = x^2 + 6x + 13?

May 3, 2016

Axis of symmetry -> x = -3

Vertex -> (x,y)-> (-3, 4 )

Explanation:

Consider the general form $y = a {x}^{2} + b x + c$

Write the general form as $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

In your case $a = 1$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a} \to \left(- \frac{1}{2}\right) \times 6 = - 3}$

$\textcolor{b l u e}{\text{axis of symmetry } \to x = - 3}$

To find ${y}_{\text{vertex}}$ substitute $x = - 3$ in the original equation.

$\implies {y}_{\text{vertex}} = {\left(- 3\right)}^{2} + 6 \left(- 3\right) + 13$

$\textcolor{b l u e}{\implies {y}_{\text{vertex}} = + 4}$

$\textcolor{b r o w n}{\text{Vertex} \to \left(x , y\right) \to \left(- 3 , 4\right)}$