# How do you graph f(x)= 5/3(x-3)^2 - 6?

Aug 10, 2015

Calculate the vertex and the $x$- and $y$-intercepts and then sketch the graph.

#### Explanation:

$f \left(x\right) = \frac{5}{3} {\left(x - 3\right)}^{2} - 6$

Step 1. Your equation is in vertex form.

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$.

We see that $a = \frac{5}{3}$, $h = 3$, and $k = - 6$.

Step 2. Find the vertex.

The vertex is at ($h , k$) or ($3 , - 6$).

Step 3. Find the $y$-intercept.

Set $x = 0$ and solve for $y$.

$f \left(0\right) = \frac{5}{3} {\left(x - 3\right)}^{2} - 6 = \frac{5}{3} {\left(0 - 3\right)}^{2} - 6 = \frac{5}{3} {\left(- 3\right)}^{2} - 6 = \frac{5}{3} \left(9\right) - 6 = 9$

The $y$-intercept is at ($0 , 9$).

Step 4. Find the $x$-intercept(s).

Set $f \left(x\right) = 0$ and solve for $x$.

$0 = \frac{5}{3} {\left(x - 3\right)}^{2} - 6$

$\frac{5}{3} {\left(x - 3\right)}^{2} = 6$

(x-3)^2= 6×3/5 =18/5

x-3=±sqrt(18/5)= =±sqrt((9×2×5)/(5×5))=±3/5sqrt10

x=3±3/5sqrt10

x=3+3/5sqrt10≈4.9 and x=3-3/5sqrt10≈1.1

Step 5. Draw your axes and plot the four points. Step 6. Draw a smooth parabola that passes through the four points. 