Question

express the function graphed on the axes below as a piecewise function.
answer
attempt 1 out of 2
$f(x)=\

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

$

Explanation:

Step1: Find first piece (left segment)

Points: open at \((-5, 2)\), closed at \((-1, -3)\). Slope \(m = \frac{-3 - 2}{-1 - (-5)} = \frac{-5}{4}=-\frac{5}{4}\)? Wait, no, wait: \((-5,2)\) to \((-1,-3)\): \(x\) from -5 to -1 (change +4), \(y\) from 2 to -3 (change -5). So slope \(m=\frac{-3 - 2}{-1 - (-5)}=\frac{-5}{4}\)? Wait, no, wait, maybe I misread the points. Wait the left segment: open circle at \((-5, 2)\), then goes to closed circle at \((-1, -3)\)? Wait no, looking at the graph: the left part: open circle at \(x=-5\), \(y=2\); then the line goes to closed circle at \(x=-1\), \(y=-3\)? Wait no, maybe the two segments: first segment (left) from \((-5, 2)\) (open) to \((-1, -3)\) (closed)? Wait no, the right segment: from \((-1, -1)\) (open) to \((4, -5)\) (open)? Wait no, let's re-examine the graph:

- Left segment: open circle at \((-5, 2)\), then a line to closed circle at \((-1, -3)\). So domain for left segment: \(-5 < x \leq -1\) (since open at -5, closed at -1).

Slope for left segment: \(m = \frac{-3 - 2}{-1 - (-5)} = \frac{-5}{4}\)? Wait, no, \(y_2 - y_1 = -3 - 2 = -5\), \(x_2 - x_1 = -1 - (-5) = 4\), so \(m = -5/4\). Equation: using point-slope form, \(y - 2 = -\frac{5}{4}(x + 5)\). Simplify: \(y = -\frac{5}{4}x - \frac{25}{4} + 2 = -\frac{5}{4}x - \frac{25}{4} + \frac{8}{4} = -\frac{5}{4}x - \frac{17}{4}\). Wait, but maybe I made a mistake. Wait, alternatively, maybe the left segment is from \((-5, 2)\) to \((-1, -3)\), but let's check the right segment: open at \((-1, -1)\), then line to open at \((4, -5)\). So slope for right segment: \(\frac{-5 - (-1)}{4 - (-1)} = \frac{-4}{5} = -\frac{4}{5}\)? Wait no, \((-1, -1)\) to \((4, -5)\): \(x\) from -1 to 4 (change +5), \(y\) from -1 to -5 (change -4). So slope \(m = -4/5\). Equation: \(y - (-1) = -\frac{4}{5}(x - (-1))\), so \(y + 1 = -\frac{4}{5}(x + 1)\), \(y = -\frac{4}{5}x - \frac{4}{5} - 1 = -\frac{4}{5}x - \frac{9}{5}\). But wait, the closed circle is at \((-1, -3)\) for the left segment? Wait no, maybe I mixed up the segments. Let's correct:

Wait, the graph has two segments:

1. Left segment: open circle at \((-5, 2)\), closed circle at \((-1, -3)\). So domain: \(-5 < x \leq -1\).

Slope: \(m = \frac{-3 - 2}{-1 - (-5)} = \frac{-5}{4}\). Equation: \(y = -\frac{5}{4}x + b\). Plug in \((-1, -3)\): \(-3 = -\frac{5}{4}(-1) + b\) → \(-3 = \frac{5}{4} + b\) → \(b = -3 - \frac{5}{4} = -\frac{17}{4}\). So \(y = -\frac{5}{4}x - \frac{17}{4}\) for \(-5 < x \leq -1\).

2. Right segment: open circle at \((-1, -1)\), open circle at \((4, -5)\). Wait, no, the closed circle is at \((-1, -3)\) for left, and the right segment starts at open circle \((-1, -1)\)? Wait, no, looking at the graph: the vertical line at \(x=-1\): left segment has closed circle at \(x=-1\) (so includes \(x=-1\)), right segment has open circle at \(x=-1\) (so excludes \(x=-1\)). So right segment: domain \(-1 < x < 4\) (since open at both ends? Wait, the right end is open at \(x=4\), \(y=-5\). So right segment: from \((-1, -1)\) (open) to \((4, -5)\) (open). Slope: \(\frac{-5 - (-1)}{4 - (-1)} = \frac{-4}{5}\). Equation: \(y - (-1) = -\frac{4}{5}(x - (-1))\) → \(y + 1 = -\frac{4}{5}(x + 1)\) → \(y = -\frac{4}{5}x - \frac{4}{5} - 1 = -\frac{4}{5}x - \frac{9}{5}\) for \(-1 < x < 4\).

Wait, but the left segment: open at \(x=-5\), closed at \(x=-1\): so \(f(x) = -\frac{5}{4}x - \frac{17}{4}\) for \(-5 < x \leq -1\).

Right segment: open at \(x=-1\), open at \(x=4\): \(f(x) = -\frac{4}{5}x - \frac{9}{5}\) for \(-1 < x < 4\).

Wait, but let's check the points:

For left segment: at \(x=-5\), open, so \(f(-5)\) is not defined there. At \(x=-1\), closed, so \(f(-1) = -3\). Let's plug \(x=-1\) into left segment equation: \( -\frac{5}{4}(-1) - \frac{17}{4} = \frac{5}{4} - \frac{17}{4} = -\frac{12}{4} = -3\). Correct.

For right segment: at \(x=-1\), open, so \(f(-1)\) is not defined there (matches open circle). At \(x=4\), open, so \(f(4)\) is not defined. Plug \(x=0\) into right segment: \( -\frac{4}{5}(0) - \frac{9}{5} = -\frac{9}{5} = -1.8\), which is between -1 and -5, makes sense.

Wait, but maybe I made a mistake in the slope for the left segment. Let's recheck the left segment points: open at \((-5, 2)\), closed at \((-1, -3)\). So \(x\) from -5 to -1, \(y\) from 2 to -3. So the change in \(x\) is \( -1 - (-5) = 4\), change in \(y\) is \( -3 - 2 = -5\). So slope is \(-5/4\), correct.

So first piece: \(f(x) = -\frac{5}{4}x - \frac{17}{4}\) for \(-5 < x \leq -1\).

Second piece: \(f(x) = -\frac{4}{5}x - \frac{9}{5}\) for \(-1 < x < 4\).

Wait, but let's check the right segment's open circle at \(x=4\), \(y=-5\). Plug \(x=4\) into second piece: \( -\frac{4}{5}(4) - \frac{9}{5} = -\frac{16}{5} - \frac{9}{5} = -\frac{25}{5} = -5\), which matches the open circle at \((4, -5)\). Good.

And the left segment's open circle at \((-5, 2)\): plug \(x=-5\) into first piece: \( -\frac{5}{4}(-5) - \frac{17}{4} = \frac{25}{4} - \frac{17}{4} = \frac{8}{4} = 2\), which matches the open circle. Perfect.

Step2: Find second piece (right segment)

As above, slope \(-4/5\), equation \(y = -\frac{4}{5}x - \frac{9}{5}\), domain \(-1 < x < 4\).

Answer:

$$ f(x) =
LATEXBLOCK0
$$