Question

graph the following function on the axes provided.
$f(x) = \

$$\begin{cases} 1 & \\text{for} & -6 \\leq x < -3 \\\\ x + 4 & \\text{for} & -3 < x \\leq 2 \\end{cases}$$

$
click and drag to make a line. click the line to delete it.
click on an endpoint of a line to change it.
graph axes with x from -10 to 10 and y from -10 to 10

Explanation:

Step1: Analyze the first piece of the function

For \( f(x)=1\) where \( - 6\leq x<-3\). This is a horizontal line segment. When \(x = - 6\), \(y = 1\) (closed - end point since \(x=-6\) is included in the domain). When \(x=-3\), \(y = 1\) (open - end point since \(x = - 3\) is not included in the domain for this piece).

Step2: Analyze the second piece of the function

For \(f(x)=x + 4\) where \(-3

• When \(x=-3\), \(y=-3 + 4=1\) (open - end point since \(x=-3\) is not included in the domain for this piece).
• When \(x = 2\), \(y=2 + 4=6\) (closed - end point since \(x = 2\) is included in the domain for this piece).

To graph \(y=x + 4\), we can also use the slope - intercept form \(y=mx + b\) (where \(m = 1\) and \(b = 4\)). We can find two points on the line. For example, when \(x=0\), \(y=4\).

Answer:

Graph a horizontal line segment from the point \((-6,1)\) (closed circle) to \((-3,1)\) (open circle) for the first part. Then graph the line \(y=x + 4\) from the point \((-3,1)\) (open circle) to \((2,6)\) (closed circle) for the second part.