Linear Functions: Practice Problems with Step-by-Step Solutions

Linear functions appear throughout mathematics and real-world modeling. These practice problems emphasize the meaning of the slope (rate of change) and the \(y\)-intercept (initial value). Try each problem first, then open the solution to check your reasoning.

Review: Key Facts About Linear Functions

  • Slope-intercept form: \(f(x) = mx + b\)
  • Point-slope form: \(y - y_1 = m(x - x_1)\)
  • Slope: \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)
  • Slope meaning: the rate of change of the output per unit change in the input.
  • \(y\)-intercept meaning: the output when the input is zero (initial value).

Practice Problems and Solutions

Problem 1: Finding a linear function from a table

Question: \(f\) is a linear function. Values of \(x\) and \(f(x)\) are given below; complete the table.

\(x\)\(f(x)\)
-317
0--
--1
4-18
7--
---30

Solution:

Since \(f\) is linear, write \(f(x) = ax + b\). Two ordered pairs are given: \((-3, 17)\) and \((4, -18)\). Substitute into \(f(x) = ax + b\):

\[ 17 = -3a + b \] \[ -18 = 4a + b \]

Subtract the first from the second:

\[ -35 = 7a \implies a = -5 \]

Substitute \(a = -5\) into the first equation:

\[ 17 = -3(-5) + b \implies b = 2 \]

Thus, \(f(x) = -5x + 2\). Now fill in the table:

  • \(x = 0\): \(f(0) = 2\)
  • \(f(x) = 1\): \(1 = -5x + 2 \implies x = \dfrac{1}{5}\)
  • \(x = 7\): \(f(7) = -33\)
  • \(f(x) = -30\): \(-30 = -5x + 2 \implies x = \dfrac{32}{5}\)
Problem 2: Families of linear functions

Question: A family of linear functions is given by \(f(x) = mx + (3 - 2m)\), where \(m\) is a constant.

  1. Graph \(f\) for \(m = 0, 1, 2, -3,\) and \(-5\).
  2. What do all the graphs in part (a) have in common?
  3. Justify your answer to part (b) analytically.
  4. Write the equation of the family of functions whose graphs pass through \((-2, -4)\).

Solution:

  1. The graphs of the functions for the listed values of \(m\) are shown below:

    Family of linear functions for several values of m, all passing through the point (2, 3)
  2. All the graphs intersect and pass through the same point \((2, 3)\).
  3. To prove this, evaluate \(f(2)\): \[ f(2) = m(2) + (3 - 2m) = 2m + 3 - 2m = 3 \] This result is independent of \(m\), so every line in the family passes through \((2, 3)\).
  4. Using the point-slope form for a line through \((-2, -4)\): \[ y - y_1 = m(x - x_1) \] \[ y + 4 = m(x + 2) \] \[ y = mx + 2m - 4 \] \[ y = mx + (2m - 4) \]
Problem 3: Population growth word problem

Question: A high school had 1200 students enrolled in 2003 and 1500 students in 2006. If the student population \(P\) grows as a linear function of time \(t\), where \(t\) is the number of years after 2003:

  1. How many students were enrolled in 2010?
  2. Find a linear function that relates the student population to the time \(t\).

Solution:

  1. Represent the data as ordered pairs \((t, P)\). The year 2003 corresponds to \(t = 0\), and 2006 corresponds to \(t = 3\): \[ (0, 1200) \quad \text{and} \quad (3, 1500) \]

    Compute the slope (rate of growth):

    \[ m = \frac{1500 - 1200}{3 - 0} = 100 \ \text{students/year} \]

    The year 2010 corresponds to \(t = 7\):

    \[ P(2010) = 1200 + 7(100) = 1900 \ \text{students} \]
  2. With slope \(m = 100\) and \(y\)-intercept \(b = 1200\): \[ P(t) = 100t + 1200 \]
Problem 4: Interpreting a linear graph

Question: The graph below is a linear function that relates the value \(V\) (in dollars) of a car to its age \(t\), where \(t\) is the number of years after 2000.

Graph of a linear function showing the decreasing value of a car over time
  1. Find the slope and interpret it.
  2. What will be the value of the car in 2010?

Solution:

  1. Read two points from the graph: \((0, 16000)\) and \((7, 7600)\). \[ m = \frac{7600 - 16000}{7 - 0} = \frac{-8400}{7} = -1200 \ \$/\text{year} \]

    A slope of \(-1200 \ \$/\text{year}\) means the car's value decreases by \$1200 each year.

  2. In 2010, \(t = 10\). Starting from \((7, 7600)\) and moving 3 more years: \[ V(10) = 7600 - 3(1200) = 7600 - 3600 = 4000 \ \$ \]
Problem 5: Cost function

Question: The cost of producing \(x\) tools by a company is \(C(x) = 1200x + 5500\) (in dollars).

  1. What is the cost of 100 tools?
  2. What is the cost of 101 tools?
  3. Find the difference between the cost of 101 and 100 tools.
  4. Find the slope of the graph of \(C(x)\).
  5. Interpret the slope.

Solution:

  1. \(C(100) = 1200(100) + 5500 = 125{,}500 \ \$\)
  2. \(C(101) = 1200(101) + 5500 = 126{,}700 \ \$\)
  3. \(C(101) - C(100) = 1200 \ \$\)
  4. The slope is the coefficient of \(x\): \(m = 1200\).
  5. The slope is the marginal cost — the extra cost when one additional tool is produced.
Problem 6: Draining a tank

Question: A 500-liter tank full of oil is being drained at a constant rate of 20 liters per minute.

  1. Write a linear function \(V(t)\) for the volume in the tank after \(t\) minutes (drainage starts at \(t = 0\)).
  2. Find the \(V\)- and \(t\)-intercepts and interpret them.
  3. How many liters are in the tank after 11 minutes and 45 seconds?

Solution:

  1. Starting at 500 L and decreasing by 20 L per minute: \[ V(t) = 500 - 20t \]
  2. \(V\)-intercept (\(t = 0\)): \(V(0) = 500\) L — the initial oil volume.
    \(t\)-intercept (\(V = 0\)): \[ 0 = 500 - 20t \implies t = 25 \ \text{minutes} \] This is the total time to empty the tank.
  3. Convert 11 min 45 s to a decimal: \[ t = 11 + \frac{45}{60} = 11.75 \ \text{min} \] \[ V(11.75) = 500 - 20(11.75) = 265 \ \text{liters} \]
Problem 7: Perimeter of an expanding shape

Question: A 50 m by 70 m rectangular garden is surrounded by a walkway of constant width \(x\) meters.

Rectangular garden surrounded by a walkway of constant width x meters
  1. Write the outside perimeter \(P\) in terms of \(x\).
  2. Find the slope of the graph of \(P\).
  3. What is the meaning of the slope found in (b)?

Solution:

  1. Outside length: \(70 + 2x\); outside width: \(50 + 2x\). \[ P(x) = 2(70 + 2x) + 2(50 + 2x) \] \[ P(x) = 140 + 4x + 100 + 4x = 8x + 240 \]
  2. The slope is \(8\).
  3. Each 1 m increase in walkway width increases the outside perimeter by 8 m.
Problem 8: Fuel consumption

Question: A driver starts with 25 gallons in the tank. The car burns 5 gallons per 100 miles. Assuming the fuel decreases linearly:

  1. Write a linear function \(G(x)\) for gallons left after \(x\) miles.
  2. What is the value and meaning of the slope?
  3. What is the value and meaning of the \(x\)-intercept?

Solution:

  1. Consumption rate: \(\dfrac{5}{100} = 0.05\) gallons per mile. \[ G(x) = 25 - 0.05x \]
  2. The slope is \(-0.05\) gal/mi. The tank loses 0.05 gallons for every mile driven.
  3. Set \(G(x) = 0\): \[ 25 - 0.05x = 0 \implies x = 500 \ \text{miles} \] This is the maximum driving range on a full tank.
Problem 9: Changing dimensions over time

Question: A rectangular wire frame has one dimension moving at 0.5 cm/s. Its width is constant at 4 cm. At \(t = 0\), the length is 10 cm.

Rectangle with one edge moving over time and constant width
  1. What is the length at time \(t\)?
  2. Write a formula for the area \(A\) in terms of \(t\).
  3. Write a formula for the perimeter \(P\) in terms of \(t\).
  4. As \(t\) increases, which grows faster: the perimeter or the area?

Solution:

  1. \(L(t) = 10 + 0.5t\) cm.
  2. Area: \(A(t) = 4(10 + 0.5t) = 2t + 40\) cm².
  3. Perimeter: \[ P(t) = 2(4) + 2(10 + 0.5t) = 28 + t \ \text{cm} \]
  4. The slope of \(A(t)\) is \(2\), and the slope of \(P(t)\) is \(1\). Because \(2 > 1\), the area increases faster than the perimeter.

Key Takeaways

  • Two points are enough to determine a linear function.
  • The slope is the rate of change; the \(y\)-intercept is the initial value.
  • In real-world contexts, always state the units of the slope.
  • Intercepts often have practical meaning (starting value, break-even point, capacity limit).
  • Comparing slopes tells you which quantity grows (or shrinks) faster.

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