Question

write a piecewise function for the given graph.
the piecewise function for the given graph is f(x) =
\

$$\begin{cases} \\square & \\text{if } x \\geq \\square \\\\ \\square & \\text{if } x < \\square \\end{cases}$$

(simplify your answers.)

Explanation:

Step1: Determine the domain split

The graph has a hole at \( x = -1 \) and different lines for \( x < -1 \) and \( x \geq -1 \). So the split is at \( x = -1 \).

Step2: Find the equation for \( x < -1 \)

Points on the left line: \((-3, 3)\) and let's check the slope. Let's take two points: \((-3, 3)\) and another point. Wait, the left line: let's use \((-3, 3)\) and \((-1, 1)\) (but there's a hole at \((-1,1)\)). The slope \( m=\frac{3 - 1}{-3 - (-1)}=\frac{2}{-2}=-1\)? Wait no, wait the left line: let's take \((-3, 3)\) and another point. Wait, actually, let's use two points on the left line (for \( x < -1 \)): say \((-3, 3)\) and a point like \((-5, 5)\) (extending the line). Wait, slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3 - 5}{-3 - (-5)}=\frac{-2}{2}=-1\)? Wait no, that's not right. Wait, looking at the graph, the left line (for \( x < -1 \)): let's take points \((-3, 3)\) and \((-1, 1)\) (hole). So slope \( m=\frac{3 - 1}{-3 - (-1)}=\frac{2}{-2}=-1\)? Wait, no, wait the line for \( x < -1 \): let's check the points. Wait, the left line has a point \((-3, 3)\) and when \( x = -1 \), there's a hole at \((-1, 1)\). Wait, no, maybe I made a mistake. Wait, let's take two points on the left line: \((-3, 3)\) and \((-5, 5)\) (if we extend). Wait, slope \( m=\frac{5 - 3}{-5 - (-3)}=\frac{2}{-2}=-1\)? No, that would be \( y - 3 = -1(x + 3) \), so \( y = -x \). Let's check: when \( x = -3 \), \( y = 3 \), correct. When \( x = -1 \), \( y = 1 \), which is the hole, correct. So for \( x < -1 \), the equation is \( y = x \)? Wait, wait \( y = x \): when \( x = -3 \), \( y = -3 \)? No, that's wrong. Wait, I messed up. Let's recalculate. Let's take two points on the left line: \((-3, 3)\) and \((-1, 1)\) (hole). Slope \( m=\frac{1 - 3}{-1 - (-3)}=\frac{-2}{2}=-1\). So equation: \( y - 3 = -1(x + 3) \) → \( y - 3 = -x - 3 \) → \( y = -x \). Wait, when \( x = -3 \), \( y = 3 \), correct. When \( x = -1 \), \( y = 1 \), correct. So for \( x < -1 \), \( f(x) = x \)? Wait, no, \( y = -x \)? Wait, \( -x \) when \( x = -3 \) is 3, yes. So \( f(x) = x \)? Wait, no, \( y = x \) would be \( x = -3 \), \( y = -3 \), no. Wait, I think I flipped. Let's do it again. Slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3 - 1}{-3 - (-1)}=\frac{2}{-2}=-1 \). So point-slope form: \( y - y_1 = m(x - x_1) \). Using \((-3, 3)\): \( y - 3 = -1(x + 3) \) → \( y = -x - 3 + 3 \) → \( y = -x \). Wait, no, \( -x - 3 + 3 = -x \). So \( y = -x \)? Wait, when \( x = -3 \), \( y = 3 \), correct. When \( x = -1 \), \( y = 1 \), correct. So for \( x < -1 \), \( f(x) = x \)? No, \( y = -x \) is \( f(x) = -x \)? Wait, no, \( -x \) when \( x = -3 \) is 3, yes. So \( f(x) = x \) is wrong. Wait, \( y = x \) would be \( x = -3 \), \( y = -3 \), no. So it's \( f(x) = x \)? Wait, no, I'm confused. Wait, let's take another point. The left line: when \( x = -5 \), what's \( y \)? If \( f(x) = x \), \( y = -5 \), but the line goes up, so slope should be positive. Oh! I messed up the slope calculation. Let's take two points on the left line: \((-3, 3)\) and \((-5, 1)\) (extending). Slope \( m=\frac{3 - 1}{-3 - (-5)}=\frac{2}{2}=1 \). Ah! There we go. I took the wrong points. So slope is 1. So equation: \( y - 3 = 1(x + 3) \) → \( y = x + 6 \)? Wait, no, \( y - 3 = x + 3 \) → \( y = x + 6 \). Wait, when \( x = -3 \), \( y = 3 \), correct. When \( x = -1 \), \( y = -1 + 6 = 5 \)? No, that's not the hole. Wait, the hole is at \((-1, 1)\). So my mistake. The left line: the points are \((-3, 3)\) and the hole is at \((-1, 1)\). Wait, slope between \((-3, 3)\) and \((-1, 1)\) is \( \frac{1 - 3}{-1 - (-3)}=\frac{-2}{2}=-1 \), but the line is going up, so maybe I got the points wrong. Wait, the graph: the left line (for \( x < -1 \)) has a point \((-3, 3)\) and a hole at \((-1, 1)\), and the right line (for \( x \geq -1 \)) has a point \((5, 2)\) and \((-1, -4)\). Let's check the right line. Points \((-1, -4)\) and \((5, 2)\). Slope \( m=\frac{2 - (-4)}{5 - (-1)}=\frac{6}{6}=1 \). So equation for right line: \( y - (-4) = 1(x + 1) \) → \( y + 4 = x + 1 \) → \( y = x - 3 \). Let's check \( x = 5 \): \( y = 5 - 3 = 2 \), correct. Now the left line: points \((-3, 3)\) and hole at \((-1, 1)\). Slope \( m=\frac{1 - 3}{-1 - (-3)}=\frac{-2}{2}=-1 \)? No, but the line is going up, so slope should be positive. Wait, maybe the hole is at \((-1, 1)\) for the left line, and the left line has slope 1. Let's recalculate: if slope is 1, then \( y - 3 = 1(x + 3) \) → \( y = x + 6 \). At \( x = -1 \), \( y = -1 + 6 = 5 \), but the hole is at \((-1, 1)\), so that's not it. Wait, the right line: \( y = x - 3 \). At \( x = -1 \), \( y = -1 - 3 = -4 \), which matches the point \((-1, -4)\). So the right line is \( y = x - 3 \) for \( x \geq -1 \). Now the left line: let's take two points on the left line (for \( x < -1 \)): \((-3, 3)\) and another point. Wait, the left line has a hole at \((-1, 1)\), so the equation for the left line should pass through \((-3, 3)\) and have the same slope as the left segment. Wait, maybe the left line is \( y = x \)? Let's check: when \( x = -3 \), \( y = -3 \), no. Wait, the left line: the points are \((-3, 3)\) and the hole is at \((-1, 1)\). So the slope is \( \frac{1 - 3}{-1 - (-3)} = -1 \), but the line is increasing, so I must have the points wrong. Wait, maybe the left line is \( y = x \), and the hole is at \((-1, -1)\), but the graph shows a hole at \((-1, 1)\). I think I need to re-express. Let's start over.

For \( x \geq -1 \): the line passes through \((-1, -4)\) and \((5, 2)\). Slope \( m = \frac{2 - (-4)}{5 - (-1)} = \frac{6}{6} = 1 \). Equation: \( y - (-4) = 1(x - (-1)) \) → \( y + 4 = x + 1 \) → \( y = x - 3 \). So for \( x \geq -1 \), \( f(x) = x - 3 \).

For \( x < -1 \): the line passes through \((-3, 3)\) and has a hole at \((-1, 1)\). Let's find the slope between \((-3, 3)\) and \((-1, 1)\): \( m = \frac{1 - 3}{-1 - (-3)} = \frac{-2}{2} = -1 \). Wait, but the line is going up, so maybe it's \( m = 1 \). Wait, if we take two points on the left line: \((-3, 3)\) and \((-5, 1)\), slope is \( \frac{3 - 1}{-3 - (-5)} = 1 \). So equation: \( y - 3 = 1(x + 3) \) → \( y = x + 6 \). But at \( x = -1 \), \( y = -1 + 6 = 5 \), not 1. So the hole is at \((-1, 1)\), so the equation should be \( y = x + 2 \)? Wait, no. Wait, the left line: when \( x = -3 \), \( y = 3 \); when \( x = -1 \), \( y = 1 \) (hole). So the equation is \( y = -x \)? Wait, \( -(-3) = 3 \), \( -(-1) = 1 \), yes! So \( y = -x \). So for \( x < -1 \), \( f(x) = -x \)? Wait, \( x = -3 \), \( y = 3 \) (correct), \( x = -1 \), \( y = 1 \) (hole, correct). And for \( x \geq -1 \), \( f(x) = x - 3 \) (since at \( x = -1 \), \( y = -1 - 3 = -4 \), correct, and at \( x = 5 \), \( y = 5 - 3 = 2 \), correct).

So the piecewise function is:
\[ f(x) = \begin{cases} -x & \text{if } x < -1 \\ x - 3 & \text{if } x \geq -1 \end{cases} \]

Answer:

The piecewise function is \( f(x) =

$$\begin{cases} \boldsymbol{-x} & \text{if } x < -1 \\ \boldsymbol{x - 3} & \text{if } x \geq -1 \end{cases}$$

\)