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

Jul 22, 2015

Plot the vertex, the intercepts (and possibly a few other points) then connect the points with a smooth curve.

#### Explanation:

$f \left(x\right) = \frac{2}{3} {\left(x - 3\right)}^{2} - 6$ is in vertex form
so as one of our points we have the vertex $\left(3 , - 6\right)$

The $f \left(x\right)$ (or $y$) intercept occurs where $x = 0$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(0\right) = \frac{2}{3} {\left(- 3\right)}^{2} - 6 = 0$
giving us another point: $\left(0 , 0\right)$

The x-intercepts occur where $f \left(x\right) = 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{2}{3} {\left(x - 3\right)}^{2} - 6 = 0$

$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - 3\right)}^{2} = 6 \cdot \left(\frac{3}{2}\right)$

$\textcolor{w h i t e}{\text{XXXX}}$$x - 3 = \pm 3$

$\textcolor{w h i t e}{\text{XXXX}}$$x = 0 \mathmr{and} x = 6$
Unfortunately this only gives us 1 more point (we already had $\left(0 , 0\right)$)
$\textcolor{w h i t e}{\text{XXXX}}$$\left(6 , 0\right)$

We could evaluate $f \left(x\right)$ for a few more values of $x$ ($x = \pm 6$ and $x = \pm 9$ would seem promising; but the 3 points we have will do for now:

graph{2/3(x-3)^2-6 [-4.39, 13.39, -6.404, 2.485]}