Question

express the function graphed on the axes below as a piecewise function.

Explanation:

Step1: Analyze the left segment

The left segment has a slope. Let's find two points. One point is \((-4, -9)\) (black dot) and another is \((0, -1)\)? Wait, no, wait. Wait, the line passes through \((0, -1)\)? Wait, no, looking at the graph, the left line: let's take two points. Let's see, the left segment: when \(x = -4\), \(y = -9\); when \(x = 3\), there's an open circle at \(y = 5\)? Wait, no, maybe I misread. Wait, the graph: the left part (from \(x = -4\) to \(x < 3\))? Wait, no, let's check the points. The left segment has a black dot at \((-4, -9)\) and goes through \((0, -1)\)? Wait, no, when \(x = 0\), \(y = -1\)? Wait, no, the line crosses the y-axis at \((0, -1)\)? Wait, no, looking at the grid, the line from \((-4, -9)\) to \((3, 5)\) (open circle) and then from \((3, 5)\) (open circle) to \((6, 7)\) (black dot). Wait, let's calculate the slope for the left part. Let's take two points: \((-4, -9)\) and \((3, 5)\). The slope \(m = \frac{5 - (-9)}{3 - (-4)}=\frac{14}{7}=2\). Wait, no, wait, when \(x = 0\), what's \(y\)? Let's use point-slope form. Let's take point \((-4, -9)\) and slope \(m = 2\). So the equation is \(y - (-9)=2(x - (-4))\), so \(y + 9 = 2(x + 4)\), so \(y = 2x + 8 - 9 = 2x - 1\). Wait, when \(x = -4\), \(y = 2(-4) -1 = -9\), correct. When \(x = 3\), \(y = 2(3) -1 = 5\), which matches the open circle at \((3, 5)\). Then the right segment: from \(x > 3\) to \(x = 6\). Let's take two points: \((3, 5)\) (open circle) and \((6, 7)\) (black dot). The slope \(m=\frac{7 - 5}{6 - 3}=\frac{2}{3}\)? Wait, no, wait, \(7 - 5 = 2\), \(6 - 3 = 3\), so slope is \(\frac{2}{3}\)? Wait, no, wait, \(7 - 5 = 2\), \(6 - 3 = 3\), so \(y - 5=\frac{2}{3}(x - 3)\), so \(y=\frac{2}{3}x - 2 + 5=\frac{2}{3}x + 3\)? Wait, no, when \(x = 6\), \(y=\frac{2}{3}(6)+3 = 4 + 3 = 7\), correct. Wait, but maybe I made a mistake. Wait, let's re-examine. Wait, the left segment: from \(x \leq -4\)? No, the black dot is at \((-4, -9)\), so the left segment is from \(x \geq -4\) to \(x < 3\), and the right segment is from \(x > 3\) to \(x \leq 6\). Wait, no, the left segment: the black dot is at \((-4, -9)\), so the domain for the left part is \(x \geq -4\) (since it's a black dot, inclusive) up to \(x < 3\) (open circle, exclusive). The right part: open circle at \(x = 3\), black dot at \(x = 6\), so domain \(x > 3\) (exclusive) to \(x \leq 6\) (inclusive). Now, let's confirm the left function. Let's take two points: \((-4, -9)\) and \((3, 5)\). Slope \(m = \frac{5 - (-9)}{3 - (-4)}=\frac{14}{7}=2\). So equation: \(y = 2x + b\). Plug in \((-4, -9)\): \(-9 = 2(-4) + b\), so \(-9 = -8 + b\), so \(b = -1\). So \(y = 2x - 1\) for \(-4 \leq x < 3\). Then the right function: points \((3, 5)\) (open) and \((6, 7)\). Slope \(m=\frac{7 - 5}{6 - 3}=\frac{2}{3}\)? Wait, no, \(7 - 5 = 2\), \(6 - 3 = 3\), so slope is \(\frac{2}{3}\). Equation: \(y - 5=\frac{2}{3}(x - 3)\), so \(y=\frac{2}{3}x - 2 + 5=\frac{2}{3}x + 3\)? Wait, no, when \(x = 3\), \(y=\frac{2}{3}(3)+3 = 2 + 3 = 5\), correct. When \(x = 6\), \(y=\frac{2}{3}(6)+3 = 4 + 3 = 7\), correct. Wait, but maybe I messed up the left segment. Wait, let's check \(x = 0\): \(y = 2(0) -1 = -1\), which is on the line. Yes, the line passes through \((0, -1)\). So the left function is \(f(x) = 2x - 1\) for \(-4 \leq x < 3\). The right function: let's check the slope again. Wait, from \((3, 5)\) to \((6, 7)\), the change in \(y\) is \(2\), change in \(x\) is \(3\), so slope is \(\frac{2}{3}\). Wait, but maybe I made a mistake. Wait, the right segment: from \(x = 3\) (open) to \(x = 6\) (closed). So the equation is \(y - 5=\frac{2}{3}(x - 3)\), so \(y=\frac{2}{3}x + 3\)? Wait, no, \(y = \frac{2}{3}x - 2 + 5 = \frac{2}{3}x + 3\). Wait, but when \(x = 6\), \(y = 7\), which is correct. So the piecewise function is:

For \( -4 \leq x < 3 \), \(f(x) = 2x - 1\)

For \( 3 < x \leq 6 \), \(f(x) = \frac{2}{3}x + 3\)

Wait, but let's verify the left segment again. When \(x = -4\), \(f(-4) = 2(-4) -1 = -9\), which matches the black dot. When \(x = 3\), \(f(3) = 2(3) -1 = 5\), which matches the open circle (so \(x < 3\) for the left part). Then the right part: when \(x = 6\), \(f(6) = \frac{2}{3}(6) + 3 = 4 + 3 = 7\), which matches the black dot. When \(x = 3\), the open circle is at \(5\), so \(x > 3\) for the right part.

Step2: Write the piecewise function

So the first piece is \(2x - 1\) for \( -4 \leq x < 3 \), and the second piece is \(\frac{2}{3}x + 3\) for \( 3 < x \leq 6 \).

Answer:

\( f(x) =

$$\begin{cases} 2x - 1 & \text{for } -4 \leq x < 3 \\ \frac{2}{3}x + 3 & \text{for } 3 < x \leq 6 \end{cases}$$

\)