Question

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

answer attempt 1 out of 3

( f(x) = \begin{cases} square & \text{for } square \\ square & \text{for } square end{cases} )

Explanation:

Step1: Analyze the horizontal line segment

The left - hand part of the graph is a horizontal line. A horizontal line has a slope of 0. The y - value of this horizontal line is 4, and it is defined for \(x < - 1\) (since there is an open circle at \(x=-1\) for this segment). The equation of a horizontal line is \(y = k\), where \(k\) is the y - coordinate. So for \(x < - 1\), \(f(x)=4\).

Step2: Analyze the linear segment (right - hand part)

First, find the slope of the line. The open circle of this segment is at \((-1,6)\)? Wait, no, looking at the graph, the open circle of the right - hand segment is at \(x = 3\)? Wait, no, let's re - examine. Wait, the left segment: the open circle is at \(x=-1\), \(y = 4\). The right segment: let's find two points. The open circle is at \((3,6)\)? Wait, no, from the graph, the right - hand line: let's take two points. Let's say when \(x = 3\), \(y = 6\) (open circle) and when \(x=7\), \(y = 10\)? Wait, the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take the two points: the open circle is at \((3,6)\) and another point, say when \(x = 7\), \(y=10\). Then \(m=\frac{10 - 6}{7 - 3}=\frac{4}{4}=1\). Using the point - slope form \(y - y_1=m(x - x_1)\), with \((x_1,y_1)=(3,6)\) and \(m = 1\), we get \(y-6=1\times(x - 3)\), so \(y=x + 3\). Wait, when \(x = 3\), \(y=3 + 3=6\), which matches the open circle. And the domain for this segment is \(x>3\)? Wait, no, the open circle is at \(x = 3\)? Wait, the left segment is for \(x < - 1\)? Wait, maybe I misread the x - coordinate of the open circle. Wait, the left horizontal line: the open circle is at \(x=-1\), \(y = 4\). The right line: the open circle is at \(x = 3\), \(y = 6\). Let's recalculate the slope. Let's take two points on the right line. Suppose when \(x = 3\), \(y = 6\) (open circle) and when \(x=6\), \(y = 9\). Then slope \(m=\frac{9 - 6}{6 - 3}=\frac{3}{3}=1\). So the equation is \(y-6=1\times(x - 3)\), so \(y=x + 3\). The domain for the right line is \(x>3\)? Wait, no, the open circle is at \(x = 3\), so the domain is \(x>3\)? Wait, maybe the left segment is \(x < 3\) and the right is \(x>3\)? Wait, no, the left horizontal line: looking at the x - axis, the horizontal line goes to the left of \(x=-1\)? Wait, the x - axis has markings from - 10 to 10. The left horizontal line: the open circle is at \(x=-1\), so for \(x < - 1\), \(f(x)=4\). The right line: the open circle is at \(x = 3\), so for \(x>3\), let's find the equation. Let's take two points on the right line. Let's say the open circle is at \((3,6)\) and another point, say \((7,10)\). Slope \(m=\frac{10 - 6}{7 - 3}=1\). So the equation is \(y=x + 3\) (since \(y-6=1\times(x - 3)\Rightarrow y=x + 3\)).

Wait, maybe I made a mistake in the x - coordinate of the open circle for the left segment. Let's re - look at the graph. The left horizontal line: the open circle is at \(x=-1\), \(y = 4\). So for \(x < - 1\), \(f(x)=4\). The right line: the open circle is at \(x = 3\), \(y = 6\). Let's check the slope again. If we take \(x = 3\), \(y = 6\) and \(x = 4\), \(y = 7\), then slope \(m=\frac{7 - 6}{4 - 3}=1\). So the equation is \(y=x + 3\). And the domain for the right line is \(x>3\).

Answer:

\(f(x)=

$$\begin{cases}4&\text{for }x < - 1\\x + 3&\text{for }x>3\end{cases}$$

\)

Wait, maybe the open circle for the left is at \(x = 3\)? No, the left line is horizontal, going to the left, with open circle at \(x=-1\). Wait, perhaps the correct domain for the left is \(x < 3\) and right is \(x>3\)? Let's re - examine the graph. The left horizontal line: the open circle is at \(x = 3\), \(y = 4\)? No, the y - coordinate of the left horizontal line is 4, and the right line starts with an open circle at \(x = 3\), \(y = 6\). So:

For the left part (horizontal line):

• The line is horizontal, so \(y = 4\). The open circle is at \(x = 3\)? No, the x - axis: the left horizontal line is from \(x=-\infty\) to \(x = 3\) (open circle at \(x = 3\), \(y = 4\)), and the right line is from \(x>3\) with slope 1, passing through \((3,6)\). Then the slope between \((3,6)\) and \((4,7)\) is 1, so equation \(y=x + 3\).

So correct piecewise function:

\(f(x)=

$$\begin{cases}4&\text{for }x < 3\\x + 3&\text{for }x>3\end{cases}$$

\)

Yes, that makes sense. The left horizontal line has \(y = 4\) and is defined for \(x < 3\) (open circle at \(x = 3\)), and the right line has slope 1, passing through \((3,6)\) (open circle), so equation \(y=x + 3\) for \(x>3\).