Question

graph this function.
$f(x)=\

$$\begin{cases}-\\frac{1}{2}x - 3&\\text{if }x < 0\\7x - 3&\\text{if }x\\geq0\\end{cases}$$

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

Explanation:

Step1: Find points for $x < 0$

For $y =-\frac{1}{2}x - 3$, when $x=-2$, $y=-\frac{1}{2}\times(-2)-3=1 - 3=-2$. When $x = - 4$, $y=-\frac{1}{2}\times(-4)-3=2 - 3=-1$. Since $x<0$, the point at $x = 0$ for this part is open. When $x = 0$, $y=-\frac{1}{2}\times0-3=-3$.

Step2: Find points for $x\geq0$

For $y = 7x-3$, when $x = 0$, $y=7\times0-3=-3$. When $x = 1$, $y=7\times1-3=4$. When $x = 2$, $y=7\times2-3=11$. The point at $x = 0$ for this part is closed.

Step3: Plot the points

Plot the points $(-2,-2),(-4,-1)$ for $y =-\frac{1}{2}x - 3$ (open - circle at $(0, - 3)$ for this part) and $(0,-3),(1,4),(2,11)$ for $y = 7x-3$ (closed - circle at $(0,-3)$ for this part) and draw the lines for each part of the piece - wise function.

Answer:

Plot the points $(-2,-2),(-4,-1)$ with an open - circle at $(0,-3)$ for $y=-\frac{1}{2}x - 3$ and $(0,-3),(1,4),(2,11)$ with a closed - circle at $(0,-3)$ for $y = 7x-3$ and draw the appropriate lines.