Question

graph this function.
f(x)=\begin{cases}-\frac{1}{2}x - 3 & \text{if }x < 0\\7x - 3 & \text{if }xgeq0end{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 \(x=-2\), \(f(-2)=-\frac{1}{2}\times(-2)-3=1 - 3=-2\).
For \(x = -4\), \(f(-4)=-\frac{1}{2}\times(-4)-3=2-3=-1\).

Step2: Find points for \(x\geq0\)

For \(x = 0\), \(f(0)=7\times0-3=-3\).
For \(x=1\), \(f(1)=7\times1 - 3=4\).
For \(x = 2\), \(f(2)=7\times2-3=14 - 3 = 11\).

Answer:

Plot the points \((-4,-1)\), \((-2,-2)\) (for \(x < 0\) part, with an open - circle at \(x = 0\) for this sub - function), and \((0,-3)\), \((1,4)\), \((2,11)\) (for \(x\geq0\) part). Then draw a line for \(y =-\frac{1}{2}x - 3\) (only for \(x < 0\)) and a line for \(y=7x - 3\) (only for \(x\geq0\)).