Question

select the correct answer from each drop-down menu.
the graph represents the piecewise function
$f(x)=\\{\

$$\begin{array}{ll}\\square, & \\text{if } \\square; \\\\ \\square, & \\text{if } \\square.\\end{array}$$

$

Explanation:

Step1: Analyze the horizontal line

The horizontal line has \( y = 0 \) (since it's on the x - axis) and is defined for \( x\leq1 \) (as the other parts start at \( x = 1 \) or beyond). So one part of the piece - wise function is \( f(x)=0 \) when \( x\leq1 \).

Step2: Analyze the first non - horizontal line (lower left)

Let's find the slope of the line passing through points. Let's take two points on the lower left line. For example, when \( x=-1 \), \( y = - 3 \) and when \( x=-5 \), let's assume the line equation. The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points: from the graph, when \( x=-1 \), \( y=-3 \) and when \( x=-4 \), let's see, the line seems to have a slope. Wait, maybe a better way: the line on the lower left (for \( x < - 1 \)? Wait, no, let's re - examine. Wait, the horizontal line is \( y = 0 \) for \( x\leq1 \)? Wait, no, the point at \( x = 1 \) has \( y = 5 \). Wait, maybe I made a mistake. Let's look at the graph again. The horizontal line is from the left (with a closed dot) and goes to \( x = 1 \) (open or closed? Wait, the point at \( x = 1 \) for the horizontal line: no, the horizontal line is \( y = 0 \), and the other lines: one starts at \( x = 1 \) with \( y = 5 \) and goes up, and the other starts at \( x=-1 \) with \( y=-3 \) and goes down - left. Wait, maybe the piece - wise function has three parts? No, the problem shows two drop - downs for the function and two for the conditions. Wait, maybe the horizontal line is \( f(x)=0 \) for \( x\leq1 \), and the line with slope 2 (since from \( x = 1 \), \( y = 5 \); when \( x = 2 \), \( y = 6 \), so slope \( m=\frac{6 - 5}{2 - 1}=1 \)? Wait, no, \( 5\) to \( 6\) when \( x\) goes from \( 1\) to \( 2\), slope is \( 1 \). And the lower line: from \( x=-1 \), \( y=-3 \); when \( x=-2 \), let's see, if slope is 2, \( y=-3+( - 1)\times2=-5 \)? Wait, maybe the lower line has a slope of 2. Wait, let's correct.

Wait, the correct approach:

1. Horizontal segment: \( y = 0 \), defined for \( x\leq1 \) (since at \( x = 1 \), the other part starts with \( y = 5 \), so the horizontal line is \( f(x)=0 \) when \( x\leq1 \).

2. The line with positive slope: passes through \( (1,5) \) and \( (2,6) \), so slope \( m = 1 \), equation \( y-5=1\times(x - 1)\), so \( y=x + 4 \), defined for \( x\geq1 \).

3. The line with negative - positive slope (lower left): passes through \( (-1,-3) \) and \( (-2,-5) \), slope \( m=\frac{-5+3}{-2 + 1}=2 \), equation \( y+3=2(x + 1)\), \( y = 2x-1 \)? Wait, no, \( y+3=2(x + 1)\Rightarrow y=2x+2 - 3=2x - 1 \). When \( x=-1 \), \( y=2\times(-1)-1=-3 \), correct. When \( x=-2 \), \( y=2\times(-2)-1=-5 \), which matches the graph. So this line is defined for \( x < - 1 \). But the problem has two parts for the piece - wise function. Wait, maybe the horizontal line is \( f(x)=0 \) for \( - 1\leq x\leq1 \), and the other two lines. But the problem's drop - down has two functions. Maybe the intended functions are:

For \( x\leq1 \), \( f(x)=0 \); for \( x > 1 \), \( f(x)=x + 4 \) (since at \( x = 1 \), \( y = 5 \), \( 1+4 = 5 \), and at \( x = 2 \), \( 2 + 4=6 \), which matches the graph). And for \( x < - 1 \), \( f(x)=2x - 1 \). But the problem has two parts. Maybe the first part is \( f(x)=0 \) when \( x\leq1 \), and the second part is \( f(x)=x + 4 \) when \( x\geq1 \), and the third part is a mistake, but according to the drop - down, two functions. So let's proceed with the two main parts: horizontal line (\( y = 0 \)) for \( x\leq1 \) and the line \( y=x + 4 \) for \( x\geq1 \).

Step3: Confirm the functions and conditions

The horizontal line: \( f(x)=0 \), condition \( x\leq1 \).

The line with slope 1: \( f(x)=x + 4 \), condition \( x\geq1 \).

And the lower line: \( f(x)=2x-1 \), condition \( x < - 1 \). But since the problem has two drop - downs for functions, maybe the two main parts are \( f(x)=0 \) (for \( x\leq1 \)) and \( f(x)=x + 4 \) (for \( x\geq1 \)), and the lower line is a third part, but maybe the problem's drop - down is designed with two parts.

Wait, maybe I misread the graph. Let's re - examine:

- The horizontal line: \( y = 0 \), from \( x=-\infty \) to \( x = 1 \) (closed dot at \( x = 1 \)? No, the point at \( x = 1 \) for the horizontal line: no, the point at \( x = 1 \) is \( y = 5 \) (closed dot) for the upper line. So the horizontal line is \( y = 0 \) for \( x < 1 \), and the upper line is \( y=x + 4 \) for \( x\geq1 \) (since at \( x = 1 \), \( y=1 + 4=5 \), correct), and the lower line is \( y = 2x-1 \) for \( x < - 1 \). But the problem has two functions. Maybe the answer is:

First function: \( 0 \), condition: \( x\leq1 \)

Second function: \( x + 4 \), condition: \( x\geq1 \)

And the lower line: maybe a mistake in the problem, but based on the visible parts, the horizontal line is \( y = 0 \) and the upper line is \( y=x + 4 \).

Answer:

The piece - wise function is \( f(x)=

$$\begin{cases}0, & \text{if }x\leq1\\x + 4, & \text{if }x\geq1\end{cases}$$

\) (and the lower line \( 2x-1 \) for \( x < - 1 \) if considered, but based on the drop - down structure, the two main parts are \( 0\) for \( x\leq1 \) and \( x + 4\) for \( x\geq1 \))