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: Analyze the function for \(x < 0\)

The function is \(y =-\frac{1}{2}x - 3\). When \(x=-2\), \(y=-\frac{1}{2}\times(-2)-3=1 - 3=-2\). When \(x = - 4\), \(y=-\frac{1}{2}\times(-4)-3=2 - 3=-1\). The \(y\) - intercept of \(y =-\frac{1}{2}x - 3\) is \(-3\). Since \(x<0\), the point at \(x = 0\) for this part of the function is open.

Step2: Analyze the function for \(x\geq0\)

The function is \(y = 7x-3\). When \(x = 0\), \(y=7\times0 - 3=-3\) (this is a closed - point as \(x\geq0\)). When \(x = 1\), \(y=7\times1-3=4\). When \(x = 2\), \(y=7\times2-3=11\).

Step3: Plot the points

For \(y =-\frac{1}{2}x - 3\) (\(x < 0\)): Plot points like \((-2,-2)\), \((-4,-1)\) with an open - point at \((0,-3)\). For \(y = 7x-3\) (\(x\geq0\)): Plot points like \((0,-3)\) (closed - point), \((1,4)\), \((2,11)\) and draw a line through them.

Answer:

To graph the function, plot points \((-2,-2)\), \((-4,-1)\) for \(y =-\frac{1}{2}x - 3\) (\(x < 0\)) with an open - circle at \((0,-3)\) and points \((0,-3)\) (closed - circle), \((1,4)\), \((2,11)\) for \(y = 7x-3\) (\(x\geq0\)) and draw straight - line segments through the respective sets of points.