Proportional Relationships - Comprehensive Page-by-Page Notes

Page 1: Identifying Proportional Relationships

• A proportional relationship is represented by a straight line that goes through the origin (0,0) on a graph.
• If the line is curved, or if the line does not touch the point (0,0), the relationship is not proportional.
• Quick diagnostic question: among several candidate lines, which ones are proportional? The ones that are straight and pass through the origin.
• Important takeaway: for a proportional relationship, the ratio y/x is constant for every point on the graph.
• The constant of proportionality is denoted by k and is defined by the formula
  $k = \frac{y}{x}$
  for any point (x, y) on the line.
• If the line passes through (0,0), you can use any other point on the line to compute k. The resulting k should be the same for all points on the line.
• Example intuition: when the line represents miles vs hours (x = hours, y = miles), a proportional relationship means distance grows linearly with time at a constant rate (the unit rate).
• Common error: remember to compute y/x, not x/y.

Page 2: The Constant of Proportionality (k) and the Unit Rate

• The constant of proportionality k is found by taking any point on the line and computing $k = \dfrac{y}{x}$.
• Worked example 1:
  • Points: (x, y) = (4, 10) and (8, 20).
  • $\dfrac{y}{x} = \dfrac{10}{4} = \dfrac{5}{2} = 2.5$
  • $\dfrac{y}{x} = \dfrac{20}{8} = \dfrac{5}{2} = 2.5$
  • Therefore, $k = \dfrac{5}{2}$ for all points on this line.
• Another explicit example uses k = 5: if a line going through the origin has points (10, 50) and (7, 35):
  • $\dfrac{50}{10} = 5,\quad \dfrac{35}{7} = 5$
  • Hence, for every point on the line, $k = 5$ and the unit rate is 5 units of y per 1 unit of x.
• When units are involved, the unit rate is the same numerical value as k but interpreted with the denominator as the unit of x. For example, if the relation is miles per hour and k = 40, then:
  • If x = 1 hour, y = 40 miles, so the unit rate is 40 miles per hour.
  • In general, the unit rate means y per 1 unit of x.
• Important mental model: the constant of proportionality is the rate that never changes along the line; it is the same across all points.
• Quick rule: if y/x is not constant across points (i.e., you get different values for different points), the relation is not proportional.
• Note: sometimes the same constant is introduced with the variable k or simply called the unit rate when units are in play.
• Clear statement: for a proportional relationship, the formula of the line is
  $y = kx$
  where k is the constant of proportionality.

Page 3: Using the Equation y = kx

• Once k is known, the equation of the proportional relation is simply
  $y = kx.$
• Example with k = 5:
  • The equation is $y = 5x.$
  • Verification with point (3, 15): $15 = 5 \times 3.$
• Applications of the equation:
  • If you know the hours x and want to know miles y, compute $y = kx.$ For x = 20: $y = 5 \times 20 = 100.$
  • If you know miles y and want to know hours x, solve $kx = y \Rightarrow x = \dfrac{y}{k}.$ For y = 200 and k = 5: $x = \dfrac{200}{5} = 40.$
• This equation also serves as the “unit rate” interpretation when x represents time (or other units) and y represents the measured quantity.

Page 4: Proportional Relationships in a Real-World Example (Footballs)

• Problem setup: number of footballs (x) and total cost in dollars (y).
• Since the relationship is proportional (goes through (0,0) and is a straight line), find k = cost per football.
• Quick calculation: pick a convenient point, e.g., (x, y) = (3, 27).
  • $k = \dfrac{y}{x} = \dfrac{27}{3} = 9.$
  • The unit rate is $9 per football, so the equation is
    $y = 9x.$
• Validation with multiple points:
  • (1, 9) → y/x = 9/1 = 9
  • (7, 63) → y/x = 63/7 = 9
  • (10, 90) → y/x = 90/10 = 9
• Practical use of the equation:
  • Cost for 7 footballs: $y = 9 \times 7 = 63.$
  • Cost for 11 footballs: $y = 9 \times 11 = 99.$
  • If you spent $99, number of footballs: $99 = 9x \Rightarrow x = \dfrac{99}{9} = 11.$
• The reason this is useful: the relationship is proportional, so the same k applies to any quantity along the line.
• Additionally, the graph’s intercept is at the origin: the y-intercept is (0,0) for a proportional relation.

Page 5: Proportional Relationships in Tables

• Tables behave like graphs: first column is x, second column is y; for proportional relations, the ratio y/x is constant across all rows.
• Example with a constant k = 6 from a complete row: if a row shows (x, y) = (3, 18) and (6, 36) etc., you can see 18/3 = 6 and 36/6 = 6.
• From a single complete pair, you can deduce the equation: if k = 6, then
  $y = 6x.$
• Fill missing values using the equation:
  • If a row has x = 5, then y = 6(5) = 30.
  • If a row has y = 99 and k = 9, then x = \dfrac{y}{k} = \dfrac{99}{9} = 11.
• A common example from the transcript: if you know (x, y) = (12, 36) → k = 3; then (x, y) with x = 5 yields y = 15; if y is missing with x = 35, then y = kx = 3 × 35 = 105.
• If a row has x = 0, then y must be 0 (zero dollars for zero footballs, zero miles for zero hours, etc.). This reinforces the idea that for a proportional relationship, zero input yields zero output.
• Reminder: if a table includes a zero in x or y, the proportional rule still applies: y = kx with k constant.

Page 6: Finding k from a Graph or Table (Smallest x, clear point)

• When asked to find k from a graph, use a point with a small x value to minimize confusion. For example, pick a point with x = 1 if available.
• Example: Given a point where y = 60 and x = 1 on a graph, compute
  $k = \dfrac{y}{x} = \dfrac{60}{1} = 60.$
  The interpretation: 60 miles per hour, so the equation is $y = 60x.$
• Alternatively, from another point (x, y) = (10, 600) would also yield $k = \dfrac{600}{10} = 60.$
• Important nuance: k is always computed as $k = \dfrac{y}{x}$; it is not $\dfrac{x}{y}$, and it does not have to be an integer (e.g., k = \tfrac{1}{3} is perfectly valid).
• If k = \tfrac{1}{3}, the equation is $y = \tfrac{1}{3}x.$
• The y-intercept for a proportional relation remains (0,0). If the graph starts at a different base, that would indicate a non-proportional relationship (or a different kind of model) and will be discussed later.

Page 7: Zero Intercept and the Y-Intercept in Proportional Relationships

• The y-intercept is the point where the graph crosses the y-axis (x = 0).
• For a proportional relationship, the line passes through the origin, so the y-intercept is 0, i.e., the line hits the axis at (0,0).
• Real-world intuition: if you have zero input, you have zero output (e.g., zero hours → zero miles; zero footballs → zero dollars).
• When filling tables or solving problems, keep in mind that if x = 0 then y must be 0 for proportional relationships.

Page 8: Solving Missing Values in Tables Using k

• If you have a table with missing entries but you know it is proportional, you can deduce missing values using the rule y = kx.
• Example from the transcript: given a partial table where y/x = 3 for some complete rows, you can fill missing values as follows:
  • If x = 5 and k = 3, then y = 3(5) = 15.
  • If x is missing but y = 105 and k = 3, then x = y/k = 105/3 = 35.
• If you encounter a zero in the table, use the fact that y = kx and x = 0 implies y = 0.
• A second example: a table with partial entries can be completed by first determining k from a complete pair (e.g., 36/12 = 3 gives k = 3), then filling in other cells using y = kx or x = y/k as appropriate.

Page 9: Using the Graph and Table Together; Unit Rate vs. k

• The unit rate is the value of y when x is 1. When x = 1, y = k, so the unit rate equals k.
• The constant of proportionality k and the unit rate are the same concept, just described with or without units.
• If you are given a graph or a table and asked for the equation, the process is:
  • Determine k using any point (preferably with small x): k = y/x.
  • Write the equation as $y = kx.$ Then use it to compute any missing y or x via simple algebra:
  • If you know x and want y: $y = kx.$
  • If you know y and want x: $x = \dfrac{y}{k}.$
• Key caveat: ensure you are using y/x, not x/y, when computing k.

Page 10: Quick Practice: Filling a Table Given k

• Example: If a graph shows cost per football as $9 and there is a partial table of footballs (x) and cost (y), then the table should satisfy y = 9x.
• Fill-ins:
  • If x = 7, y = 63 (since 9×7 = 63).
  • If y = 99, x = 11 (since 99/9 = 11).
  • If x = 0, y = 0 (zero footballs cost zero dollars).
• Strategy for quizzes: identify the constant of proportionality from the graph (or from a complete row in the table), then fill in the rest of the table using y = kx. If a blank row exists, you can infer zeros as needed for proportional relationships.

Page 11: Quick Review: What is the Y-Intercept and How to Find It

• The y-intercept is the point where the line crosses the y-axis.
• For a proportional relationship (a line through the origin), the y-intercept is 0 (the origin).
• If you see a line that does not pass through (0,0), the relationship is not proportional (though it could be modeled differently).
• When you have a table and a proportional relationship, you can use the fact that x = 0 yields y = 0 to anchor the table.

Page 12: Quick Practical Strategy for Quizzes and Homework

• Step-by-step approach when you see a proportional relationship: 1) Check if the graph shows a straight line through the origin. If not, it is not proportional (for this unit week we focus on the origin case). 2) If proportional, compute k = y/x for a point on the graph (prefer the point with the smallest nonzero x). 3) Write the equation as $y = kx.$ If units are involved, this is also the unit rate interpretation. 4) Use the equation to find any missing y or x:
  • If x is known: $y = kx.$
  • If y is known: $x = \dfrac{y}{k}.$
    5) In a table, verify that y/x is constant for all provided rows, and use that constant to fill missing values.
    6) Remember the special case: x = 0 implies y = 0 for proportional relationships, reinforcing the idea that the line passes through the origin.

Page 13: Summary of Key Concepts and Practical Tips

• Proportional relationships are linear and pass through (0,0). The graph is a straight line with slope equal to k.
• The constant of proportionality k is computed as
  $k = \dfrac{y}{x}$
  for any point (x, y) on the line, and the equation is $y = kx.$
• The unit rate is the same as k and corresponds to the amount of output per one unit of input (when there are units involved).
• The y-intercept of a proportional relationship is always 0 (the origin).
• In tables, the key check is that $\dfrac{y}{x}$ is the same for all filled-in rows; use this to fill in missing values.
• Common pitfalls to avoid:
  • Using y/x incorrectly as x/y.
  • Assuming a non-origin intercept means non-proportional without checking the graph and the ratio across points.
• Real-world application pattern: distance = rate × time (e.g., miles = speed × hours) where speed is the constant of proportionality; e.g., 60 miles per hour gives $y = 60x.$ If x = 2 hours, y = 120 miles.

Page 14: Quick Practice Reminders and Office Hours

• When you are given a graph and asked for the constant of proportionality, select a point with a small x-value, compute $k = \dfrac{y}{x}$, then use $y = kx$ to fill in the rest of the problem.
• If a table has missing entries, use the following workflow:
  • Determine k from a known pair (x, y): k = y/x.
  • Fill in missing y for a given x with $y = kx.$
  • Fill in missing x for a given y with $x = y/k.$
  • If x = 0 or y = 0, enforce the proportional rule: y = 0 when x = 0.
• Real-world nuance: the teacher notes that some graphs may start at nonzero bases in more advanced contexts; for now, focus on lines through the origin for proportional relationships.
• If you need extra help, the instructor offered office hours for additional questions.