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: Graph \( y = 1\) for \(-6\leq x<-3\)

For the function \(y = 1\) (a horizontal line), when \(x=-6\), \(y = 1\) (closed - circle since \(x=-6\) is included in the domain \(-6\leq x<-3\)). When \(x=-3\), \(y = 1\) (open - circle since \(x = - 3\) is not included in the domain \(-6\leq x<-3\)).

Step2: Graph \(y=x + 4\) for \(-3

Find two points on the line \(y=x + 4\).
When \(x=-3\), \(y=-3 + 4=1\) (open - circle since \(x=-3\) is not included in the domain \(-3When \(x = 2\), \(y=2 + 4=6\) (closed - circle since \(x = 2\) is included in the domain \(-3Plot these points and draw a straight line passing through them for the interval \(-3

Answer:

Graph a horizontal line \(y = 1\) from \(x=-6\) (closed - circle) to \(x=-3\) (open - circle) and a line \(y=x + 4\) from \(x=-3\) (open - circle) to \(x = 2\) (closed - circle).