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Use the points given to find the function model for the linear function they represent in slope-intercept form ($y = mx + b$):


  1. $(1, 2)$ and $(4, 5)$
  2. $(-2, -4)$ and $(0, 3)$
  3. $(4, 0)$ and $(-3, 3)$
  4. $(6, 7)$ and $(7, 9)$
  5. $(2, -5)$ and $(3, 0)$
  6. $(8, 3)$ and $(-1, -1)$

Verified Answer :

  • For each exercise, calculate the slope \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
  • Use the calculated slope and one of the given points to solve for the y-intercept \(b\) in \(y = mx + b\).
  • The final equations are:
  • Exercise 3: \(\boxed{y = -\frac{3}{7}x + \frac{12}{7}}\)
  • Exercise 5: \(\boxed{y = 5x - 15}\)
  • Exercise 6: \(\boxed{y = \frac{4}{9}x - \frac{5}{9}}\)

Explanation

  1. Methodology for Finding Linear Function Models

To determine the equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\), we use the slope-intercept form \(y = mx + b\).
The slope \(m\) is calculated by the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
Once the slope is found, we substitute the coordinates of one of the points into the equation \(y = mx + b\) to solve for the y-intercept \(b\).
We will apply this process to three separate exercises.

  1. Solving Exercise 3: Points (4, 0) and (-3, 3)

Given the points \((4, 0)\) and \((-3, 3)\):

  1. Calculate the slope: \(m = \frac{3 - 0}{-3 - 4} = \frac{3}{-7} = -\frac{3}{7}\).
  2. Substitute \(m = -\frac{3}{7}\) and the point \((4, 0)\) into the equation \(y = mx + b\):

\(0 = (-\frac{3}{7})(4) + b\)
\(0 = -\frac{12}{7} + b\)
\(b = \frac{12}{7}\).
Thus, the equation for Exercise 3 is \(y = -\frac{3}{7}x + \frac{12}{7}\).

  1. Solving Exercise 5: Points (2, -5) and (3, 0)

Given the points \((2, -5)\) and \((3, 0)\):

  1. Calculate the slope: \(m = \frac{0 - (-5)}{3 - 2} = \frac{5}{1} = 5\).
  2. Substitute \(m = 5\) and the point \((3, 0)\) into the equation \(y = mx + b\):

\(0 = (5)(3) + b\)
\(0 = 15 + b\)
\(b = -15\).
Thus, the equation for Exercise 5 is \(y = 5x - 15\).

  1. Solving Exercise 6: Points (8, 3) and (-1, -1)

Given the points \((8, 3)\) and \((-1, -1)\):

  1. Calculate the slope: \(m = \frac{-1 - 3}{-1 - 8} = \frac{-4}{-9} = \frac{4}{9}\).
  2. Substitute \(m = \frac{4}{9}\) and the point \((8, 3)\) into the equation \(y = mx + b\):

\(3 = (\frac{4}{9})(8) + b\)
\(3 = \frac{32}{9} + b\)
\(b = 3 - \frac{32}{9} = \frac{27}{9} - \frac{32}{9} = -\frac{5}{9}\).
Thus, the equation for Exercise 6 is \(y = \frac{4}{9}x - \frac{5}{9}\).

  1. Summary of Solutions

The linear equations for the given pairs of points are as follows:
Exercise 3: \(y = -\frac{3}{7}x + \frac{12}{7}\)
Exercise 5: \(y = 5x - 15\)
Exercise 6: \(y = \frac{4}{9}x - \frac{5}{9}\)

Examples

Linear models are essential in real-world scenarios where data shows a constant rate of change. For example, if a car is traveling at a steady speed, the distance it covers over time can be represented by a linear function \(d = vt + d_0\). Similarly, if a company has fixed overhead costs and variable costs per unit produced, its total expenditure can be modeled linearly as \(C = mx + b\), where \(x\) is the number of units produced. Understanding how to construct these models from just two data points allows for precise predictions of future costs or outcomes.

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