Question

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

$$\begin{cases}2x + 9&\\text{for}&x < -3\\\\2x - 2&\\text{for}&x > 4\\end{cases}$$

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

Explanation:

Step1: Analyze \( f(x) = 2x + 9 \) ( \( x < -3 \) )

This is a linear function with slope \( m = 2 \) and y - intercept \( b = 9 \). But since \( x < - 3 \), we find a point on this line. Let's take \( x=-4 \) (which is less than - 3). Then \( f(-4)=2\times(-4)+9=-8 + 9 = 1 \). Another point: when \( x=-3 \), \( f(-3)=2\times(-3)+9=-6 + 9 = 3 \), but since \( x < - 3 \), the point \( (-3,3) \) is not included (open circle). We can also use the slope. Starting from a point, say when \( x=-5 \), \( f(-5)=2\times(-5)+9=-10 + 9=-1 \). So we can plot the line for \( x < - 3 \) using points like \( (-5,-1) \), \( (-4,1) \) with an open circle at \( (-3,3) \).

Step2: Analyze \( f(x)=2x - 2 \) ( \( x > 4 \) )

This is a linear function with slope \( m = 2 \) and y - intercept \( b=-2 \). Since \( x > 4 \), we find a point on this line. Let's take \( x = 5 \) (which is greater than 4). Then \( f(5)=2\times5-2=10 - 2 = 8 \). When \( x = 4 \), \( f(4)=2\times4-2=8 - 2 = 6 \), but since \( x > 4 \), the point \( (4,6) \) is not included (open circle). Using the slope, when \( x = 6 \), \( f(6)=2\times6-2=12 - 2 = 10 \). So we can plot the line for \( x > 4 \) using points like \( (5,8) \), \( (6,10) \) with an open circle at \( (4,6) \).

To graph:

• For \( y = 2x+9, x < - 3 \): Draw a line with slope 2, passing through points like \( (-5,-1) \), \( (-4,1) \), with an open circle at \( (-3,3) \).
• For \( y = 2x - 2, x > 4 \): Draw a line with slope 2, passing through points like \( (5,8) \), \( (6,10) \), with an open circle at \( (4,6) \).

(Note: Since this is a graphing problem, the final answer is the graphical representation as described above. If we were to describe the key features: two linear segments, one for \( x < - 3 \) with slope 2, passing through points with an open end at \( (-3,3) \), and one for \( x > 4 \) with slope 2, passing through points with an open end at \( (4,6) \))

Answer:

The graph consists of two linear segments:

1. For \( y = 2x + 9, x < - 3 \): A line with slope 2, passing through points (e.g., \((-5, -1)\), \((-4, 1)\)) with an open circle at \((-3, 3)\).
2. For \( y = 2x - 2, x > 4 \): A line with slope 2, passing through points (e.g., \((5, 8)\), \((6, 10)\)) with an open circle at \((4, 6)\).