Question

graph the function.

$f(x)=\left\\{\

$$\begin{array}{c}2 x+1 \\text { if } x<-2 \\\\-3 x-1 \\text { if } x \\geq-2\\end{array}$$

\
ight.$

Explanation:

Step1: Analyze \(y = 2x+1\) for \(x < - 2\)

• Find two points:
• When \(x=-3\), \(y=2\times(-3)+1=-6 + 1=-5\).
• The line \(y = 2x + 1\) has a slope \(m = 2\) and a \(y\)-intercept \(b = 1\). But since \(x<-2\), we use the point \((-3,-5)\) and the direction of the line (with slope \(2\)). The point \((-2,2\times(-2)+1=-3)\) is not included (open - circle at \(x = - 2\) for this part of the piece - wise function).

Step2: Analyze \(y=-3x - 1\) for \(x\geq-2\)

• Find two points:
• When \(x=-2\), \(y=-3\times(-2)-1=6 - 1 = 5\) (closed - circle at \(x=-2\) for this part of the piece - wise function).
• When \(x = 0\), \(y=-3\times0-1=-1\). The line \(y=-3x - 1\) has a slope \(m=-3\) and a \(y\)-intercept \(b=-1\).

To graph the function:

• Plot the open - circle at the point \((-2,-3)\) for the part \(y = 2x + 1\) (since \(x=-2\) is not in the domain \(x < - 2\) for this sub - function).
• Plot the closed - circle at the point \((-2,5)\) for the part \(y=-3x - 1\) (since \(x=-2\) is in the domain \(x\geq-2\) for this sub - function).
• For \(y = 2x+1\) with \(x < - 2\), use the point \((-3,-5)\) and draw a line with slope \(2\) going to the left.
• For \(y=-3x - 1\) with \(x\geq-2\), use the points \((-2,5)\) and \((0,-1)\) and draw a line with slope \(-3\) going to the right.

Answer:

The graph of the piece - wise function \(f(x)=

$$\begin{cases}2x + 1, &x < - 2\\-3x - 1, &x\geq-2\end{cases}$$

\) is composed of two line segments: one for \(x < - 2\) (open - circle at \(x=-2\)) with the equation \(y = 2x + 1\) and one for \(x\geq-2\) (closed - circle at \(x=-2\)) with the equation \(y=-3x - 1\).