Question

graph this function.
$f(x) = \

$$\begin{cases} -\\frac{2}{3}x + 1 & \\text{if } x < 0 \\\\ -8x + 1 & \\text{if } x \\geq 0 \\end{cases}$$

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

Explanation:

Step1: Analyze the piece for \( x < 0 \)

The function is \( f(x)=-\frac{2}{3}x + 1 \) when \( x < 0 \). Let's find a point: when \( x=-3 \), \( f(-3)=-\frac{2}{3}(-3)+1 = 2 + 1=3 \). So the point \((-3, 3)\) is on this line. Also, at \( x = 0 \), for \( x < 0 \), it's an open circle: \( f(0^-)=-\frac{2}{3}(0)+1 = 1 \), so open circle at \((0, 1)\).

Step2: Analyze the piece for \( x \geq 0 \)

The function is \( f(x)=-8x + 1 \) when \( x \geq 0 \). At \( x = 0 \), \( f(0)=-8(0)+1 = 1 \), so closed circle at \((0, 1)\). When \( x = 1 \), \( f(1)=-8(1)+1=-7 \), so point \((1, -7)\).

Step3: Plot the points and lines

For \( x < 0 \), use the line \( y = -\frac{2}{3}x + 1 \) with an open circle at \((0, 1)\) and other points like \((-3, 3)\), \((-6, 5)\) (since \( f(-6)=-\frac{2}{3}(-6)+1 = 4 + 1 = 5 \)). For \( x \geq 0 \), use the line \( y=-8x + 1 \) with a closed circle at \((0, 1)\) and other points like \((1, -7)\), \((0.5, -3)\) (since \( f(0.5)=-8(0.5)+1=-4 + 1=-3 \)).

Answer:

To graph \( f(x)=

$$\begin{cases}-\frac{2}{3}x + 1 & \text{if } x < 0 \\ -8x + 1 & \text{if } x \geq 0\end{cases}$$

\):

• For \( \boldsymbol{x < 0} \): Plot the line \( y = -\frac{2}{3}x + 1 \) (open circle at \( (0, 1) \), points like \( (-3, 3) \), \( (-6, 5) \), etc.).
• For \( \boldsymbol{x \geq 0} \): Plot the line \( y = -8x + 1 \) (closed circle at \( (0, 1) \), points like \( (1, -7) \), \( (0.5, -3) \), etc.).

(Graphing involves drawing two linear segments: a less - steep line for \( x < 0 \) and a very steep (negative slope) line for \( x \geq 0 \), meeting at \( (0, 1) \) with an open circle on the left segment and closed on the right.)