Question

graph this function.
$f(x) = \

$$\begin{cases} -\\frac{1}{3}x - 7 & \\text{if } x \\leq -3 \\\\ x & \\text{if } x > -3 \\end{cases}$$

$
select points on the graph to plot them. select \point fill\ to change a point from closed to open.

Explanation:

Step1: Analyze the first piece ($x \leq -3$)

The function is $f(x) = -\frac{1}{3}x - 7$ for $x \leq -3$. Let's find two points. When $x = -3$, $f(-3) = -\frac{1}{3}(-3) - 7 = 1 - 7 = -6$. So the point $(-3, -6)$ is a closed dot (since $x \leq -3$ includes $x = -3$). Let's take another $x$ value, say $x = -6$. Then $f(-6) = -\frac{1}{3}(-6) - 7 = 2 - 7 = -5$. So the point $(-6, -5)$ is on this line.

Step2: Analyze the second piece ($x > -3$)

The function is $f(x) = x$ for $x > -3$. When $x = -3$, $f(-3) = -3$, but since $x > -3$ does not include $x = -3$, this is an open dot at $(-3, -3)$. Let's take $x = 0$, then $f(0) = 0$, so the point $(0, 0)$ is on this line. Also, $x = 2$ gives $f(2) = 2$, so $(2, 2)$ is another point.

Step3: Plot the points and draw the lines

For the first piece ($x \leq -3$), draw a line through $(-3, -6)$ and $(-6, -5)$ (and other points if needed), with a closed dot at $(-3, -6)$. For the second piece ($x > -3$), draw a line through the open dot at $(-3, -3)$ and points like $(0, 0)$, $(2, 2)$, etc. The line for $f(x) = x$ has a slope of 1, and the line for $f(x) = -\frac{1}{3}x - 7$ has a slope of $-\frac{1}{3}$.

Answer:

To graph the piecewise function:

• For \( x \leq -3 \), plot the closed point \((-3, -6)\) and another point (e.g., \((-6, -5)\)) on \( y = -\frac{1}{3}x - 7 \), then draw the line.
• For \( x > -3 \), plot the open point \((-3, -3)\) and points (e.g., \((0, 0)\), \((2, 2)\)) on \( y = x \), then draw the line.

(Note: Since the question is about graphing, the final answer is the graphical representation as described, with the two lines and appropriate open/closed dots.)