Using direct and inverse variation. Includes both models and the constant of variation, finding the constant from a single pair of values, comparing the two models numerically over the same inputs, comparing them graphically as a line and a hyperbola, and building an inverse variation model relating a bicycle's banking angle to its turning radius.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Direct and Inverse Variation
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 639-645 — the lesson these objectives are drawn from
Warm-up
Lesson 4.6 introduced direct variation as a line through the origin. This lesson adds a second kind of relationship built on the same idea.
Discussion prompt
If y varies directly with x and y is six when x is three, what is the equation? What stays constant as the two change?
Hint: Divide one by the other.
Answer:
\[ \tfrac{y}{x} = 2 \;\Longrightarrow\; y = 2x \]
The quotient of y by x stays at two whatever the values are. This lesson introduces the other possibility: a pair of quantities whose product stays constant instead, which behaves quite differently.
Concept
Two variables vary directly if their quotient is a constant, and inversely if their product is a constant. In both cases that constant is called the constant of variation.
inverse variation — Two variables vary inversely if their product is a non-zero constant k, so that x times y equals k, or equivalently y equals k over x.
The constant may not be nought in either model.
Figure (svg): The direct and inverse variation models side by side
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 639-639
Section
Section 1
Concept
Two variables vary directly when y over x equals a constant k, which can be written as y equals k x. One pair of values is enough to find k.
\[ \dfrac{y}{x} = k, \quad \text{or} \quad y = kx, \quad k \ne 0 \]
Doubling x doubles y.
Figure (svg): The graph of a direct variation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 639-639 — the Models for Direct and Inverse Variation box and Example 1
Picture it
Slope k, intercept nought.
Figure (svg): The graph of a direct variation
The graph passes through the origin because x equal to nought forces y to be nought as well. That is the feature which distinguishes direct variation from any other line.
Worked example
This is Example 1 from the textbook.
\[ \text{Find an equation with } x \text{ and } y \text{ varying directly, given } y = 4 \text{ when } x = 2. \]
Write the model
Why: The quotient form.
\[ \tfrac{y}{x} = k \]
Substitute the pair
Why: Four over two.
\[ \tfrac{4}{2} = k \]
Simplify
Why: The constant of variation.
\[ k = 2 \]
Write the equation
Why: In either form.
\[ y = 2 x \]
Figure (svg): Finding the constant of variation from one known pair
\[ y = 2x \]
Verify: test the given pair
Why: At x equal to two the equation gives four, which is the pair we were given. Substituting the known pair back is the only check available when a single pair defines the model.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 639-639
Faded example
Divide y by x.
Fill in the blanks
y = 4 \text4 x = 2: \quad k = \dfrac2}___ = ___, \text___ y = 2x
Why: The constant is the quotient of the pair, and it is the slope of the resulting line. One pair is enough because the line is forced through the origin.
Worked example
Guided Practice 1, part a.
\[ \text{Find a direct variation equation given } y = 6 \text{ when } x = 2. \]
Write the model
Why: Quotient equals the constant.
\[ \tfrac{y}{x} = k \]
Substitute
Why: Six over two.
\[ k = 3 \]
Write the equation
Why: Slope three.
\[ y = 3 x \]
Predict another value
Why: At x equal to five.
\[ y = 15 \]
Figure (svg): The graph of a direct variation
\[ y = 3x \]
Verify: check the doubling property
Why: At x equal to four the equation gives twelve, which is twice the value at x equal to two. In a direct variation, doubling the input always doubles the output, and testing that is a quick confirmation of the model's form.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Trap
\[ y = 2x + 1 \]
Fit a line through the given point
Why: A line through two and four could have any slope with the right intercept.
A direct variation must pass through the origin, so its intercept is nought. This line gives y equal to one when x is nought, which means the quotient is not constant.
\[ y = 2x \]
Use the model with no constant term
Why: Direct variation is y equals k x and nothing else.
Checking that the quotient is the same for two different pairs settles it immediately.
Sorting
Check the quotient and the intercept.
Sort into buckets
Sort each equation by whether it is a direct variation.
The negative constant still counts: a direct variation may have a negative k, in which case y falls as x rises but the quotient is still fixed. Only a constant term or a non-linear form disqualifies it.
Elimination
y is 6 when x is 2, varying directly.
Eliminate the wrong options
Which is the direct variation equation?
Survives elimination: A
Why: Option B is worth noticing: the same pair of values fits both a direct and an inverse model, with different constants. The wording of the problem is what decides which one is meant.
Socratic
Two points normally define a line.
Discussion prompt
Explain why a single pair of values is enough to fix a direct variation. Then say what the hidden second point is.
Hint: Where must the graph pass?
Answer:
The model has only one unknown, the constant k, so one equation determines it. Any pair of corresponding values gives that equation, and every other pair then follows from the model.
Geometrically the hidden second point is the origin: a direct variation must pass through it, because x equal to nought forces y equal to nought. So one given pair plus the origin makes two points, which is exactly what a line needs — the constraint is built into the model rather than supplied by the problem.
Section
Section 2
Concept
Two variables vary inversely when x times y equals a constant k, which can be written as y equals k over x. Neither variable may be nought.
\[ xy = k, \quad \text{or} \quad y = \dfrac{k}{x}, \quad k \ne 0 \]
Doubling x halves y.
Figure (svg): The graph of an inverse variation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 639-640 — the inverse variation model and Example 2
Picture it
Approaching both axes, touching neither.
Figure (svg): The graph of an inverse variation
The curve never crosses an axis because neither variable can be nought — nought times anything is nought, not k. That restriction shapes the entire graph.
Worked example
This is Example 2 from the textbook.
\[ \text{Find an equation with } x \text{ and } y \text{ varying inversely, given } y = 4 \text{ when } x = 2. \]
Write the model
Why: The product form.
\[ x y = k \]
Substitute the pair
Why: Two times four.
\[ (2) (4) = k \]
Simplify
Why: The constant of variation.
\[ k = 8 \]
Write the equation
Why: In either form.
\[ y = \tfrac{8}{x} \]
Figure (svg): Finding the constant of variation from one known pair
\[ xy = 8, \quad y = \dfrac{8}{x} \]
Verify: test the halving property
Why: At x equal to four the equation gives two, which is half the value at x equal to two. Doubling the input halves the output, which is what a constant product means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Faded example
Inverse variation uses the product.
Fill in the blanks
y = 4 \text4 x = 2: \quad k = (2)(8) = ___, \text___ y = \tfrac______
Why: The constant is the product of the pair rather than their quotient. That single difference is what separates the two models, and it changes every prediction the model makes.
Worked example
Comparing Examples 1 and 2 directly.
\[ \text{Why does the pair } x = 2, \; y = 4 \text{ give } k = 2 \text{ in one model and } k = 8 \text{ in the other?} \]
Take the direct model
Why: The constant is the quotient.
\[ 4 \div 2 = 2 \]
Take the inverse model
Why: The constant is the product.
\[ 2 \times 4 = 8 \]
Note both fit the pair
Why: Each passes through it.
\[ y = 2x, \; y = \tfrac{8}{x} \]
Note they differ elsewhere
Why: At x equal to four.
\[ 8 \text{ against } 2 \]
Figure (svg): Finding the constant of variation from one known pair
\[ k = \tfrac{y}{x} = 2 \qquad k = xy = 8 \]
Verify: compare at a second value
Why: At x equal to one the direct model gives two and the inverse gives eight. A single pair cannot distinguish the two models, which is why the problem has to say which kind of variation is meant.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Error analysis
The student was told that y varies inversely with x, with y equal to six when x is two.
Annotate
On: \( \begin{aligned} \text{inverse: } \frac{y}{x} &= k \\ \frac{6}{2} &= k \\ k &= 3, \text{ so } y = 3x \end{aligned} \)
The failure is invisible at the one point that was checked, which makes it particularly worth watching for. Testing at a second value settles it: the inverse model gives three when x is four, and the student's answer gives twelve.
Matching
Quotient or product.
Match the pairs
Why: Each pair of values yields two different constants depending on the model. Reading which model is intended is therefore the first step, not an afterthought.
Prediction
In each of the two models.
Predict first
If x doubles, what happens to y?
Correct: It doubles under direct variation and halves under inverse variation.
\[ y = 2x: \; 2 \to 4 \text{ gives } 4 \to 8 \qquad y = \tfrac{8}{x}: \; 2 \to 4 \text{ gives } 4 \to 2 \]
Why: In a direct variation y is k times x, so multiplying x by two multiplies y by two. In an inverse variation y is k over x, so multiplying x by two divides y by two — the product must stay at k, and if one factor grows the other must shrink to compensate. This trade-off is the defining behaviour of inverse variation and the quickest way to identify it in a description.
Socratic
The restriction is stated but not explained.
Discussion prompt
Explain why an inverse variation forbids x and y from being nought. Then say what that means for the graph.
Hint: What would the product be?
Answer:
The product of x and y must equal k, and k is not nought. If either variable were nought the product would be nought, which cannot equal a non-zero constant — so neither is ever permitted. For x there is a second reason: dividing by nought is undefined in the form y equals k over x.
Graphically that means the curve cannot touch either axis, since a point on the vertical axis has x equal to nought and one on the horizontal axis has y equal to nought. The curve approaches both axes without reaching them, and it comes in two separate branches with nothing at all in between.
Section
Section 3
Concept
Tabulating both models over the same values of x shows the difference directly. With a positive constant, direct variation makes y rise as x rises and inverse variation makes it fall.
Both models are functions, since each x gives one y.
Figure (svg): A table comparing the two models over the same values of x
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640 — Example 3, part a, and its Study Tips on functions and domain
Picture it
Steady steps against a falling curve.
Figure (svg): A table comparing the two models over the same values of x
The direct row's differences are constant and the inverse row's are not. Looking at differences is the quickest numerical test of which model a table follows.
Worked example
This is Example 3, part a, from the textbook.
\[ \text{Compare } y = 2x \text{ and } y = \tfrac{8}{x} \text{ at } x = -4, -3, -2, -1, 1, 2, 3, 4. \]
Compute the direct row
Why: Twice each value.
\[ -8, -6, -4, -2, 2, 4, 6, 8 \]
Compute the inverse row
Why: Eight divided by each.
\[ -2, -\tfrac{8}{3}, -4, -8, 8, 4, \tfrac{8}{3}, 2 \]
Describe the direct row
Why: Even steps of two.
Describe the inverse row
Why: Falling, with a jump at nought.
Figure (svg): A table comparing the two models over the same values of x
\[ \text{direct: } +2 \text{ each step}; \quad \text{inverse: falls} \]
Verify: check one entry of each
Why: At x equal to negative three the direct model gives negative six and the inverse gives negative eight thirds, about negative 2.67. Both are defined and both are negative, which is what a positive constant does on the negative side.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Faded example
Eight divided by each value.
Fill in the blanks
x = 2 \to y = 4, \quad x = 4 \to y = 2
Why: Doubling x from two to four halved y from four to two, keeping the product at eight. Checking the product of each column is the fastest way to verify an inverse table.
Worked example
The Study Tip on domain.
\[ \text{Why does the table have no entry for } x = 0 \text{ in the inverse row?} \]
Substitute nought
Why: Into the inverse model.
\[ y = \tfrac{8}{0} \]
Note the problem
Why: Division by nought.
Check the direct model
Why: Two times nought.
\[ y = 0 \]
State the difference
Why: The domains differ.
Figure (svg): A table comparing the two models over the same values of x
\[ \tfrac{8}{0} \text{ is undefined} \]
Verify: look at values near nought
Why: At x equal to 0.1 the inverse model gives eighty, and at 0.01 it gives eight hundred. The values grow without bound as x approaches nought, which is why there is no sensible value to fill the gap with.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Trap
The inverse row falls from left to right, so y always decreases as x increases.
Read the description across the whole table
Why: The two halves both fall, so the whole row must.
Between x equal to negative one and x equal to one the values jump from negative eight to positive eight, which is an increase. The description holds on each branch separately, not across the gap.
On each branch, y decreases as x increases.
Describe each branch separately
Why: The gap at nought separates them.
This is why the graph is drawn as two curves rather than one.
Sorting
Check the quotients and the products.
Sort into buckets
Sort each pair of x and y values by which model it fits.
Every pair here was built from the same numbers arranged two ways, which shows that a single pair never decides. Two pairs are what reveal which quantity is being held constant.
Elimination
Given several pairs of values.
Eliminate the wrong options
What test identifies an inverse variation?
Survives elimination: A
Why: The defining property is the constant product, so computing it for every column is the direct test. Merely decreasing is far too weak a condition, and it is the one people reach for first.
Socratic
The Study Tip says they are.
Discussion prompt
Explain why each model assigns exactly one y to each permitted x. Then say how their domains differ.
Hint: Could one x give two different y values?
Answer:
For a given x, k times x is a single number and k over x is also a single number wherever it is defined, so each input produces exactly one output. That is precisely the requirement for a function from Lesson 4.8.
Their domains differ at one point: the direct model accepts every real number, including nought, while the inverse model excludes nought because division by it is undefined. That single missing point is what splits the hyperbola into two branches, and it is the first time in this course that a function's domain has had a hole in it.
Section
Section 4
Concept
The graph of a direct variation is a line through the origin. The graph of an inverse variation is a hyperbola, a curve with two branches that meets neither axis.
A hyperbola is studied properly in later courses.
Figure (svg): The graph of an inverse variation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640 — Example 3, part b, and its Vocabulary Tip on hyperbolas
Picture it
One line, two branches.
Figure (svg): Two columns of phrases signalling each kind of variation
The graphs are as different as the tables, and either one identifies the model at a glance. A curve avoiding both axes is an inverse variation and almost nothing else in this course.
Worked example
This is Example 3, part b, from the textbook.
\[ \text{Describe the graphs of } y = 2x \text{ and } y = \tfrac{8}{x}. \]
Take the direct model
Why: A constant multiple of x.
Locate it
Why: At x equal to nought, y is nought.
Take the inverse model
Why: Two separate pieces.
Note where it does not go
Why: Neither variable can be nought.
Figure (svg): The graph of an inverse variation
\[ \text{line through } (0,0); \quad \text{hyperbola, two branches} \]
Verify: check a point on each
Why: Both graphs pass through the point at two, four, which is the pair they were built from. That is the only place they meet, and everywhere else they diverge sharply.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Matching
Line or hyperbola.
Match the pairs
Why: Two features belong to each model and none is shared. Any one of the four identifies which model a graph shows, which makes recognition immediate.
Worked example
Reading the graph's behaviour from the equation.
\[ \text{What happens to } y = \tfrac{8}{x} \text{ as } x \text{ grows large, and as } x \text{ approaches nought?} \]
Take x large
Why: Eight over a hundred.
\[ y = 0.08 \]
Take x larger still
Why: Eight over a thousand.
\[ y = 0.008 \]
Take x small
Why: Eight over a tenth.
\[ y = 80 \]
Take x smaller still
Why: Eight over a hundredth.
\[ y = 800 \]
Figure (svg): The graph of an inverse variation
\[ x \to \text{large}: \; y \to 0; \quad x \to 0: \; y \to \text{large} \]
Verify: confirm neither value is ever reached
Why: The output never actually reaches nought, since eight divided by anything is not nought, and it never becomes infinite because every input is a real number. The curve approaches both axes forever without touching either.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 640-640
Trap
Draw the hyperbola as one continuous curve passing through the middle of the plane.
Connect the plotted points in order
Why: The points are joined for every other graph in the course.
There is no value at x equal to nought, so nothing exists between the branches to join. Drawing a connection would claim the function has values it does not have.
Draw two separate branches with a gap at the vertical axis.
Stop each branch as it approaches the axis
Why: The gap is part of the graph's meaning.
This is the first graph in the course with a genuine break in it.
Faded example
Large x and small x.
Fill in the blanks
For y equal to 8 over x, as x grows very large y approaches zero, and as x approaches zero y grows very large.
Why: Neither limit is ever reached, which is why the curve approaches the axes without meeting them. That behaviour is what makes a hyperbola look the way it does.
Hypothesis
For an inverse variation with k negative.
Predict first
Where would the two branches lie?
Correct: In the two quadrants where x and y have opposite signs.
\[ xy = -8: \; (2, -4) \text{ and } (-2, 4) \]
Why: The product of x and y must equal k, so a negative k requires one variable positive and the other negative. That places the branches in the upper-left and lower-right regions rather than the upper-right and lower-left. The shape is the same and only its position changes, in the same way that a negative slope reflects a direct variation's line without altering that it is a line.
Socratic
Every earlier graph was continuous.
Discussion prompt
Explain what the gap in the graph represents. Then say why the two branches cannot be joined by any curve.
Hint: What value would a joining point have?
Answer:
The gap represents a value of x for which the function has no output at all, because dividing by nought is undefined. It is not that the value is very large or very small — there simply is no number there, so the graph has nothing to plot.
Joining the branches would require the curve to pass through some point on the vertical axis, and any such point would have x equal to nought and therefore a product of nought rather than k. Worse, the values grow without bound on one side and towards negative without bound on the other, so there is no height at which a join could even be attempted.
Section
Section 5
Concept
When a situation is known to be an inverse variation, reading one pair of values off a graph or a table determines the constant and therefore the whole model.
The model then describes every other pair.
Figure (svg): Banking angle against turning radius for a bicycle
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 641-641 — Example 4, Write and Use a Model, on bicycle banking angle
Picture it
Tighter turns need steeper leans.
Figure (svg): Banking angle against turning radius for a bicycle
The steepness of the curve at small radii is the model's practical warning: below a certain turning radius the required lean becomes impossible.
Worked example
This is Example 4, parts a and b, from the textbook.
\[ \text{Banking angle } B \text{ varies inversely with turning radius } r, \text{ and } B = 32 \text{ when } r = 3.5. \text{ Find } B \text{ at } r = 5. \]
Write the model
Why: The inverse form.
\[ B = \tfrac{k}{r} \]
Substitute the pair
Why: Thirty-two and three and a half.
\[ 32 = \tfrac{k}{3.5} \]
Solve for the constant
Why: Multiply by three and a half.
\[ k = 112 \]
Use the model
Why: Substitute five for r.
\[ B = \tfrac{112}{5} = 22.4 \]
Figure (svg): Banking angle against turning radius for a bicycle
\[ B = \dfrac{112}{r}, \quad B(5) = 22.4^\circ \]
Verify: check the given pair
Why: At r equal to 3.5 the model gives a hundred and twelve over three and a half, which is thirty-two — the value read off the graph. The model reproduces its own data point, which is the minimum a fitted model must do.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 641-641
Faded example
Multiply the pair.
Fill in the blanks
B = 32 \text3.5 r = 3.5: \quad k = (32)(112) = ___
Why: For an inverse variation the constant is the product of the pair, so one point read off a graph is enough. Every other point on the curve then follows from the model.
Worked example
This is Example 4, part c.
\[ \text{How does the banking angle change as the turning radius gets smaller?} \]
Take a smaller radius
Why: Four feet.
\[ B = 28^\circ \]
Take a smaller one still
Why: Two feet.
\[ B = 56^\circ \]
And smaller again
Why: One foot.
\[ B = 112^\circ \]
Describe the trend
Why: The angle grows without bound.
Figure (svg): Banking angle against turning radius for a bicycle
\[ r \to \text{small} \;\Longrightarrow\; B \to \text{large} \]
Verify: say where the model stops being physical
Why: A banking angle of a hundred and twelve degrees would have the bicycle past horizontal, which is impossible. The model describes the relationship well over a sensible range of radii and predicts absurdities outside it, exactly as the exponential models of Chapter 8 did.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 641-641
Trap
\[ B = kr \;\Longrightarrow\; 32 = 3.5k \;\Longrightarrow\; k \approx 9.14 \]
Use the direct variation model
Why: One pair of values fits either model, so the first one came to mind.
This predicts a banking angle of about forty-six degrees at a five-foot radius, when the true relationship gives 22.4 — and it says the lean increases on wider turns, which is the opposite of what a cyclist experiences.
\[ B = \dfrac{112}{r} \]
Read whether the quantities rise together or trade off
Why: The problem says the relationship is inverse.
A quick sanity check against experience — tighter turns need more lean — confirms which model is meant.
Sorting
Do the quantities rise together or trade off?
Sort into buckets
Sort each pair of quantities by the kind of variation.
In each inverse case there is a fixed total being shared out — a fixed distance, a fixed amount of work — which is the usual signature of inverse variation in a real situation.
Prediction
From five feet to two and a half.
Predict first
What happens to the banking angle?
Correct: It doubles, to about 44.8 degrees.
\[ \tfrac{112}{2.5} = 44.8 \]
Why: The product of the angle and the radius is fixed at a hundred and twelve, so halving one factor must double the other to compensate. That is a large lean — nearly forty-five degrees from vertical — which is why tight turns at speed are difficult and why the model's steepness at small radii matters practically as well as mathematically.
Socratic
It predicts angles above ninety degrees.
Discussion prompt
Say over what range the banking model is sensible and what happens outside it. Then say what a second data point would add.
Hint: What is the largest lean a bicycle can manage?
Answer:
It is sensible over the range of turning radii a bicycle actually uses at that speed — perhaps three to ten feet, giving angles between about eleven and thirty-seven degrees. Below about two feet the model demands a lean past sixty degrees, and below one foot it exceeds ninety, which would put the bicycle beyond horizontal and is physically impossible.
A second data point would test whether the relationship really is inverse. If two widely separated points had the same product, that would support the model; if their products differed, the relationship would be something else and a single-point fit would have been misleading. The problem states the relationship is inverse, which is an assumption supplied rather than one demonstrated.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Direct | Inverse | |
|---|---|---|
| What stays constant | the quotient y over x | the product xy |
| Doubling x | doubles y | halves y |
| The graph | a line through the origin | a hyperbola meeting neither axis |
Every row is decided by the top one. Whether the quotient or the product is held fixed determines the behaviour and the picture completely.
Pattern
To build and use either variation model, these five moves cover it.
Step one is the whole problem in disguise: a single pair of values fits both models with different constants, so the wording is the only thing that decides.
OpenStax Elementary Algebra 2e, §8.9 Use Direct and Inverse Variation §8.9
Check
Direct uses the quotient.
Check your understanding
If x and y vary directly and y = 6 when x = 2, what is the equation?
Answer: A
Why: The constant is the quotient six over two, which is three, so the equation is y equals three x.
Check
Inverse uses the product.
Check your understanding
If x and y vary inversely and y = 4 when x = 2, what is the equation?
Answer: A
Why: The constant is the product two times four, which is eight, so the equation is y equals eight over x.
Check
Read the graph.
Check your understanding
A graph has two branches and never touches either axis. Which model is it?
Answer: A
Why: Avoiding both axes means neither variable can be nought, which is exactly the restriction an inverse variation carries.
Real world
This is the bicycle question from the lesson opener. A cyclist leans into a turn, and the angle from vertical is called the banking angle. At a fixed speed it varies inversely with the turning radius.
Discussion prompt
A graph shows a banking angle of 32 degrees at a turning radius of 3.5 feet. Find the model, use it to find the angle at a radius of 5 feet, and say what happens on very tight turns.
Hint: Multiply the pair to get the constant.
Answer:
\[ B = \dfrac{k}{r}, \quad 32 = \dfrac{k}{3.5} \;\Longrightarrow\; k = 112, \quad B = \dfrac{112}{r} \]
At a radius of five feet the model gives 22.4 degrees, a noticeably gentler lean than on the tighter turn.
As the radius shrinks the required angle grows without bound: two feet needs fifty-six degrees and one foot would need a hundred and twelve, which would put the bicycle past horizontal. So the model is describing a real limit rather than merely misbehaving — there is a turning radius below which the necessary lean cannot be achieved at that speed, and every cyclist has felt it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
y varies inversely with x, and y is 6 when x is 2. What is y when x is 4?
Correct: 3, since y halves when x doubles.
\[ xy = 12: \; (2)(6) = 12, \; (4)(3) = 12 \]
Why: Inverse variation holds the product constant, and here that product is two times six, or twelve — so the model is y equals twelve over x, and at x equal to four the value is three. The first option applies direct variation reasoning, which is what a single pair of values invites, since the same pair fits both models with different constants. The third option uses the direct model outright and does not even pass through the given pair at x equal to four. And a single pair is quite enough here, because the model has only one unknown constant; what a single pair cannot do is tell you which of the two models is meant, and the problem supplies that in words.
Explain it
They used the quotient to find the constant for an inverse variation.
Discussion prompt
In no more than four sentences, explain which operation each model uses and why. Then give them a test that catches the mistake.
Hint: What is being held constant in each?
Answer:
A usable answer: direct variation keeps the quotient constant, so you divide y by x, while inverse variation keeps the product constant, so you multiply them. The words direct and inverse are telling you whether the quantities rise together or trade off against each other.
The test is to check a second value. Both models agree at the pair you were given, so the mistake is invisible there — but at any other value the inverse model gives a smaller y where the direct one gives a larger one, and the difference is obvious.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Choosing the model is fixed by asking whether doubling one quantity doubles or halves the other. The constant is fixed by remembering that direct divides and inverse multiplies. Table behaviour is fixed by checking whether the quotients or the products are constant down the columns. Graph recognition is fixed by looking for the origin: a line through it is direct, and a curve avoiding both axes is inverse. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write both models in both of their forms, and label the constant of variation in each, noting that it may not be nought. Underneath, take a single pair of values and build both models from it, showing the division for one and the multiplication for the other, and write the two different constants side by side so the contrast is visible. In the middle, tabulate both models over the same eight values of x, including negatives, and write beneath each row how it behaves — one in even steps, the other falling — marking the gap where the inverse model has no value. Beneath that, graph both on one set of axes, drawing the line through the origin and the hyperbola as two separate branches with a clear break, and write beside the hyperbola why it never meets an axis. In the lower corner, fit an inverse model to a single data point, use it to predict two further values, and write one sentence saying where the model would stop being physically sensible. Finally, in the margin, write the two-word test for deciding which model a description calls for.
Your two models built from the same pair should agree at exactly one point and nowhere else. If they seem to agree at a second value, one of the two constants has been computed with the wrong operation.
Recap
Five things, and the first decides all the others.
| If the question says | Your first move is |
|---|---|
| Vary directly | Divide the pair to find k |
| Vary inversely | Multiply the pair to find k |
| Doubling one halves the other | Inverse variation |
| The graph passes through the origin | Direct variation |
| The graph avoids both axes | Inverse variation |
Lesson 11.3 turns to the expressions themselves. An inverse variation is the simplest rational expression there is, and simplifying more complicated ones brings back all the factoring of Chapter 10 — along with the excluded values of Lesson 11.1.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.2 Direct and Inverse Variation §11.2, pp. 639-645 — everything on these slides traces back here
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