# How do you graph f(x)= 1/6(x+3)^2 - 6?

Aug 30, 2015

The axis of symmetry is $h = - 3$, and the vertex is $\left(- 3 , - 6\right)$. Plot the vertex and some points on either side of the axis of symmetry. Sketch the graph as a curve. Do not connect the dots.

#### Explanation:

Substitute $y$ for $f \left(x\right)$.

$y = \frac{1}{6} {\left(x + 3\right)}^{2} - 6$ is in the vertex form for a parabola, $y = a {\left(x - h\right)}^{2} - 6$, where $h = - 3 , k = - 6 , \mathmr{and} \left(h , k\right) = \left(- 3 , - 6\right)$ is the vertex. The axis of symmetry is $h = - 3$.

The vertex is the minimum point on the parabola and the axis of symmetry is the imaginary line dividing the parabola into two equal but opposite halves.

Determine some points on the graph where the $x$ values fall on both sides of the axis of symmetry.

$x = - 5 ,$ $y = - \frac{16}{3}$
$x = - 4 ,$ $y = - \frac{35}{6}$
$x = - 3 ,$ $y = - 6$
$x = - 2 ,$ $y = - \frac{35}{6}$
$x = - 1 ,$ $y = - \frac{16}{3}$

Plot the vertex and the other points. Sketch a curve through the points. Don't connect the dots.

graph{y=1/6(x+3)^2-6 [-10.72, 9.28, -9.64, 0.36]}