Question

answer attempt 1 out of 2
$f(x) = \

$$\begin{cases} \\boxed{} \\text{ for } \\boxed{} \\\\ \\boxed{} \\text{ for } \\boxed{} \\end{cases}$$

$
(there are also buttons for <, >, ≤, ≥, =, ≠, add rule, remove rule, and submit answer, along with a graph of a piece - wise function above the answer section.)

Explanation:

Step1: Analyze the horizontal segment

The first part of the piecewise function is a horizontal line at \( y = 4 \) with an open circle at \( x = 2 \), so it's defined for \( x < 2 \). So the first rule is \( f(x)=4 \) for \( x < 2 \).

Step2: Analyze the linear segment

The second part is a line. Let's find its slope. The two points we can use (open circle at \( (5, 3) \) and intersection at \( (8, 0) \)). The slope \( m=\frac{0 - 3}{8 - 5}=\frac{-3}{3}=-1 \). Using point - slope form \( y - y_1=m(x - x_1) \), using \( (8,0) \), we get \( y-0=-1(x - 8) \), so \( y=-x + 8 \). The open circle is at \( x = 5 \), so this part is defined for \( 2 < x < 5 \)? Wait, no, wait the open circle on the horizontal line is at \( x = 2 \), and the open circle on the line is at \( x = 5 \). Wait, let's re - examine the graph. The horizontal line goes to the left from \( x = 2 \) (open circle), so domain \( x < 2 \). The line starts at \( x = 5 \) (open circle) and goes to the right? Wait, no, the line has an open circle at \( x = 5 \) (y = 3) and goes down to \( (8,0) \) and beyond. Wait, maybe I made a mistake. Wait, the horizontal segment: y = 4, open circle at x = 2, so \( f(x)=4 \) for \( x < 2 \). Then the line: let's find two points. The open circle is at (5, 3) and it crosses the x - axis at (8, 0). So the slope is \( \frac{0 - 3}{8 - 5}=-1 \). The equation of the line is \( y-3=-1(x - 5) \), which simplifies to \( y=-x + 5+3=-x + 8 \). So the line is \( y=-x + 8 \), and it's defined for \( 2 < x < 5 \)? Wait, no, the open circle on the horizontal line is at x = 2, and the open circle on the line is at x = 5. So the domain for the line is \( 2 < x < 5 \)? Wait, no, the horizontal line is from \( (-\infty, 2) \) with \( y = 4 \), and the line is from \( (2,5) \) with \( y=-x + 8 \)? Wait, no, the open circle on the horizontal line is at (2,4), and the open circle on the line is at (5,3). So the first piece: \( f(x)=4 \) for \( x < 2 \). The second piece: let's find the equation of the line. The two points are (5, 3) and (8, 0). Slope \( m=\frac{0 - 3}{8 - 5}=-1 \). Using point - slope form with (5,3): \( y - 3=-1(x - 5)\Rightarrow y=-x + 5 + 3=-x+8 \). So the second piece is \( f(x)=-x + 8 \) for \( 2 < x < 5 \)? Wait, no, the domain of the second piece: the open circle on the horizontal line is at x = 2, and the open circle on the line is at x = 5. So the second piece is for \( 2 < x < 5 \).

Answer:

\( f(x)=

$$\begin{cases}4 & \text{for } x < 2\\-x + 8 & \text{for } 2 < x < 5\end{cases}$$

\)