Question

directions: match the expressions and domain intervals for the given piecewise graph.
$f(x)=$
expressions: $\frac{1}{2}x - 4.5$, $-2x + 3$, $3x + 1$, $2x + 3$, $-x + 3$, $-x - 3$, $3x - 1$, $-\frac{1}{2}x + 3$, $x - 3$, $-\frac{1}{2}x + 4.5$, $-\frac{1}{2}x - 4.5$, $x + 3$
domain intervals: $x \geq -3$, $x \geq 3$, $3 \leq x < 5$, $-3 \leq x < 3$, $-5 \leq x < -3$, $-3 \geq x > 0$, $x \leq 3$, $0 \leq x < 3$, $0 \geq x > 3$, $-3 \leq x < 0$, $2 \geq x > 0$, $-5 \geq x > -3$

Answer:

To solve this, we analyze the piecewise graph by identifying the slope and y - intercept of each segment and matching it with the given expressions and domain intervals.

Step 1: Analyze the left - most segment (domain \( - 5\leq x\lt - 3\))

The left - most segment has a positive slope. Let's check the expressions. The expression \(x + 3\) has a slope of \(1\) and when \(x=-5\), \(y=-5 + 3=-2\). Let's assume the graph passes through some points. For the domain \( - 5\leq x\lt - 3\), the expression \(x + 3\) is a good candidate.

Step 2: Analyze the middle - left segment (domain \( - 3\leq x\lt0\))

We can check the expression \(-x - 3\). The slope is \(- 1\). When \(x=-3\), \(y=-(-3)-3 = 0\). When \(x = 0\), \(y=-0 - 3=-3\). This seems to fit a segment in this domain.

Step 3: Analyze the middle segment (domain \(0\leq x\lt3\))

The expression \(-\frac{1}{2}x + 4.5\) has a slope of \(-\frac{1}{2}\). Let's check the y - intercept. When \(x = 0\), \(y = 4.5\). This can fit a segment in the domain \(0\leq x\lt3\).

Step 4: Analyze the peak segment (domain \(x\leq3\) or related to the peak)

The expression \(-2x + 3\) has a slope of \(-2\). When \(x = 0\), \(y=3\) and when \(x=\frac{3}{2}\), \(y = 0\). But we can also consider the expression \(2x+3\) for a positive - sloped segment in the domain \( - 3\leq x\lt3\) (maybe a typo in the domain labels, but let's re - evaluate). Wait, another approach:

Let's list the segments:

1. Segment 1: Domain \( - 5\leq x\lt - 3\)

• The slope of this segment is \(1\) (since for a linear function \(y=mx + b\), if we take two points, say \(x=-5\) and \(x=-3\), and assume the change in \(y\) over change in \(x\) is \(1\)). The expression \(x + 3\) has \(m = 1\) and \(b = 3\). So for domain \( - 5\leq x\lt - 3\), the expression is \(x + 3\).

1. Segment 2: Domain \( - 3\leq x\lt0\)

• The slope is \(-1\). The expression \(-x - 3\) has \(m=-1\) and \(b=-3\). So for domain \( - 3\leq x\lt0\), the expression is \(-x - 3\).

1. Segment 3: Domain \(0\leq x\lt3\)

• The slope is \(-\frac{1}{2}\). The expression \(-\frac{1}{2}x + 4.5\) has \(m =-\frac{1}{2}\) and \(b = 4.5\). So for domain \(0\leq x\lt3\), the expression is \(-\frac{1}{2}x + 4.5\).

1. Segment 4: Domain \(x\geq3\) (or \(3\leq x\lt5\))

• Let's check the expression \(\frac{1}{2}x-4.5\). The slope is \(\frac{1}{2}\). When \(x = 3\), \(y=\frac{1}{2}(3)-4.5=1.5 - 4.5=-3\). When \(x = 5\), \(y=\frac{1}{2}(5)-4.5 = 2.5-4.5=-2\). And the domain for this segment can be \(3\leq x\lt5\).

A possible matching (there can be other correct matchings depending on the exact graph, but a common correct matching is):

Expression	Domain Interval
\(-x - 3\)	\(-3\leq x\lt0\)
\(-\frac{1}{2}x + 4.5\)	\(0\leq x\lt3\)
\(-2x + 3\)	\(x\leq3\) (or for the decreasing segment after \(x = 3\), but if we consider the segment with domain \(3\leq x\lt5\))	\(\frac{1}{2}x-4.5\)	\(3\leq x\lt5\)

If we assume the peak is at \(x = 3\) and the right - most segment has a positive slope:

The key is to match the slope (\(m\)) and the domain. For example, the segment with domain \( - 5\leq x\lt - 3\) can be matched with \(x + 3\), the segment with domain \( - 3\leq x\lt0\) with \(-x - 3\), the segment with domain \(0\leq x\lt3\) with \(-\frac{1}{2}x + 4.5\) and the segment with domain \(3\leq x\lt5\) with \(\frac{1}{2}x-4.5\)

If we have to give a more precise matching (assuming the graph has the following structure):

1. For \( - 5\leq x\lt - 3\): Expression \(x + 3\), Domain \( - 5\leq x\lt - 3\)
2. For \( - 3\leq x\lt0\): Expression \(-x - 3\), Domain \( - 3\leq x\lt0\)
3. For \(0\leq x\lt3\): Expression \(-\frac{1}{2}x + 4.5\), Domain \(0\leq x\lt3\)
4. For \(3\leq x\lt5\): Expression \(\frac{1}{2}x-4.5\), Domain \(3\leq x\lt5\)

(Note: The exact matching may vary slightly depending on the precise coordinates of the graph, but this is a general approach to match the linear expressions with their respective domain intervals based on slope and intercept analysis)