Question

graph the function.

1. $f(x)=\

$$\begin{cases}x + 4&\\text{if }-7\\leq x\\lt3\\\\-4&\\text{if }x = 3\\\\-x + 5&\\text{if }x\\gt3\\end{cases}$$

$

Explanation:

Step1: Graph \(y = x + 4\) for \(-7\leq x<3\)

• Find two points:
• When \(x=-7\), \(y=-7 + 4=-3\).
• When \(x = 3\) (but since \(x<3\) for this part, we use an open - circle at \(x = 3\)), \(y=3 + 4=7\).
• Plot the line segment from the point \((-7,-3)\) to the open - circle at \((3,7)\).

Step2: Plot the point for \(x = 3\)

• When \(x = 3\), \(f(x)=-4\). So, plot the point \((3,-4)\).

Step3: Graph \(y=-x + 5\) for \(x>3\)

• Find a point:
• When \(x = 4\), \(y=-4 + 5=1\).
• Since \(x>3\), we use an open - circle at \(x = 3\) (for this part, when \(x = 3\), \(y=-3 + 5=2\)).
• Plot the line starting from the open - circle at \((3,2)\) and passing through \((4,1)\) and extending to the right.

Answer:

The graph consists of a line segment \(y=x + 4\) from \((-7,-3)\) to an open - circle at \((3,7)\), a point \((3,-4)\), and a line \(y=-x + 5\) starting from an open - circle at \((3,2)\) and extending to the right.