Question

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

$$\begin{cases} -6 & \\text{for} & x < 1 \\\\ x - 6 & \\text{for} & x > 4 \\end{cases}$$

$
click and drag to make a line. click the line to delete it.
click on an endpoint of a line to change it.

Explanation:

Step 1: Analyze \( y = - 6\) for \( x<1\)

This is a horizontal line. For \( x < 1\), the \( y\) - value is always \(-6\). We can plot points such as \((0,-6)\), \((- 1,-6)\). Since \(x = 1\) is not included in this part of the domain (\(x<1\)), we use an open - circle at the point \((1,-6)\).

Step 2: Analyze \( y=x - 6\) for \(x>4\)

Find two points on the line \(y=x - 6\). When \(x = 5\), \(y=5 - 6=-1\); when \(x = 6\), \(y=6 - 6 = 0\). Since \(x = 4\) is not included in this part of the domain (\(x>4\)), we use an open - circle at the point \((4,4 - 6=-2)\).

To graph \(y=-6\) for \(x < 1\):

• Plot several points with \(y=-6\) and \(x\) values less than \(1\) (e.g., \((-2,-6)\), \((-0.5,-6)\)).
• Draw a horizontal line through these points with an open circle at \(x = 1\).

To graph \(y=x - 6\) for \(x>4\):

• Plot the points \((5,-1)\) and \((6,0)\) (and others if needed).
• Draw a straight line through these points with an open circle at \(x = 4\).

Answer:

Graph a horizontal line \(y = - 6\) for \(x<1\) (open - circle at \((1,-6)\)) and a line \(y=x - 6\) for \(x>4\) (open - circle at \((4,-2)\)) on the coordinate plane.