# How do you find the axis of symmetry and vertex point of the function:  y=x^2+8x+12?

Oct 1, 2015

Use the Method of Completing the Square to find that the axis of symmetry is $x = - 4$ and the vertex is at $\left(x , y\right) = \left(- 4 , - 4\right)$. Details to follow.

#### Explanation:

Complete the square:

$y = f \left(x\right) = {x}^{2} + 8 x + 12 = \left({x}^{2} + 8 x + 16\right) + \left(12 - 16\right) = {\left(x + 4\right)}^{2} - 4$

Since ${\left(x + 4\right)}^{2}$ is minimized at $x = - 4$, the vertex (low point) for this upward-pointing parabola is at $\left(x , y\right) = \left(- 4 , f \left(- 4\right)\right) = \left(- 4 , 16 - 32 + 12\right) = \left(- 4 , - 4\right)$.

The vertical line at $x = - 4$ is also the axis of symmetry of this parabola. For any given $c > 0$, the values of $f \left(- 4 + c\right)$ and $f \left(- 4 - c\right)$ are equal (this is the algebraic form of the symmetry about the vertical line $x = - 4$). We can check this: $f \left(- 4 + c\right) = {\left(- 4 + c + 4\right)}^{2} - 4 = {c}^{2} - 4$ and $f \left(- 4 - c\right) = {\left(- 4 - c + 4\right)}^{2} - 4 = {\left(- c\right)}^{2} - 4 = {c}^{2} - 4$.