Question

match the following representations of linear functions with the correct equation
{(-3, -13), (-1, -9), (3, -1), (6, 5)}

x	-3	-1	2	4	7
y	13	5	-7	-15	-27

the monthly cost of renting a trumpet is modeled in the graph
(graph: x-axis: number of months, y-axis: cost of rental ($); line from (0,45) to (7, 185))
the water level in the lake during the spring was 1.5 feet below the spill level. the unusually hot and dry summer caused the water level to drop 0.6 foot each week.
equations: y = -1.5x - 0.6; y = -4x + 1; y = 2x - 7; y = 20x + 45; y = (4/3)x - 7; y = -0.6x + 1.5

Explanation:

To solve this, we analyze each representation and find the corresponding linear equation.

1. Set of Points \(\{(-3, -13), (-1, -9), (3, -1), (6, 5)\}\)

• Step 1: Calculate the slope (\(m\))

Using two points, e.g., \((-3, -13)\) and \((-1, -9)\):
\(m = \frac{-9 - (-13)}{-1 - (-3)} = \frac{4}{2} = 2\).

• Step 2: Find the equation

Using point-slope form \(y - y_1 = m(x - x_1)\) with \((-3, -13)\):
\(y - (-13) = 2(x - (-3)) \implies y + 13 = 2x + 6 \implies y = 2x - 7\).

2. Table (\(x: -3, -1, 2, 4, 7\); \(y: 13, 5, -7, -15, -27\))

• Step 1: Calculate the slope (\(m\))

Using \(x = -3, y = 13\) and \(x = -1, y = 5\):
\(m = \frac{5 - 13}{-1 - (-3)} = \frac{-8}{2} = -4\).

• Step 2: Find the equation

Using point-slope form with \((-3, 13)\):
\(y - 13 = -4(x + 3) \implies y - 13 = -4x - 12 \implies y = -4x + 1\).

3. Trumpet Rental Graph

• The graph has a \(y\)-intercept (\(b\)) of \(45\) (when \(x = 0\), \(y = 45\)) and slope \(m = 20\) (e.g., from \(x = 0\) to \(x = 1\), \(y\) increases by \(20\)).
• Equation: \(y = 20x + 45\).

4. Lake Water Level

• Initial level: \(1.5\) feet (but below spill level, so the equation models the drop: slope \(m = -0.6\) (dropping \(0.6\) ft/week), \(y\)-intercept \(b = 1.5\) (initial level relative to spill).
• Equation: \(y = -0.6x + 1.5\).

Final Matches:

• \(\{(-3, -13), (-1, -9), (3, -1), (6, 5)\}\) → \(y = 2x - 7\)
• Table → \(y = -4x + 1\)
• Trumpet Rental Graph → \(y = 20x + 45\)
• Lake Water Level → \(y = -0.6x + 1.5\)

(Note: The remaining equation \(y = -1.5x - 0.6\) and \(y = \frac{4}{3}x - 7\) do not match any representation here.)

Answer:

To solve this, we analyze each representation and find the corresponding linear equation.

1. Set of Points \(\{(-3, -13), (-1, -9), (3, -1), (6, 5)\}\)

• Step 1: Calculate the slope (\(m\))

Using two points, e.g., \((-3, -13)\) and \((-1, -9)\):
\(m = \frac{-9 - (-13)}{-1 - (-3)} = \frac{4}{2} = 2\).

• Step 2: Find the equation

Using point-slope form \(y - y_1 = m(x - x_1)\) with \((-3, -13)\):
\(y - (-13) = 2(x - (-3)) \implies y + 13 = 2x + 6 \implies y = 2x - 7\).

2. Table (\(x: -3, -1, 2, 4, 7\); \(y: 13, 5, -7, -15, -27\))

• Step 1: Calculate the slope (\(m\))

Using \(x = -3, y = 13\) and \(x = -1, y = 5\):
\(m = \frac{5 - 13}{-1 - (-3)} = \frac{-8}{2} = -4\).

• Step 2: Find the equation

Using point-slope form with \((-3, 13)\):
\(y - 13 = -4(x + 3) \implies y - 13 = -4x - 12 \implies y = -4x + 1\).

3. Trumpet Rental Graph

• The graph has a \(y\)-intercept (\(b\)) of \(45\) (when \(x = 0\), \(y = 45\)) and slope \(m = 20\) (e.g., from \(x = 0\) to \(x = 1\), \(y\) increases by \(20\)).
• Equation: \(y = 20x + 45\).

4. Lake Water Level

• Initial level: \(1.5\) feet (but below spill level, so the equation models the drop: slope \(m = -0.6\) (dropping \(0.6\) ft/week), \(y\)-intercept \(b = 1.5\) (initial level relative to spill).
• Equation: \(y = -0.6x + 1.5\).

Final Matches:

• \(\{(-3, -13), (-1, -9), (3, -1), (6, 5)\}\) → \(y = 2x - 7\)
• Table → \(y = -4x + 1\)
• Trumpet Rental Graph → \(y = 20x + 45\)
• Lake Water Level → \(y = -0.6x + 1.5\)

(Note: The remaining equation \(y = -1.5x - 0.6\) and \(y = \frac{4}{3}x - 7\) do not match any representation here.)