The Central Idea
Two quantities vary inversely when their product remains constant.
The standard equation is
\[ \boxed{y=\frac{k}{x}}, \qquad x\neq0, \]where \(k\) is the constant of variation.
Multiplying both sides by \(x\) gives the equivalent form
\[ \boxed{xy=k}. \]These two equations describe the same relationship.
What Is a Reciprocal?
The reciprocal of a nonzero number \(x\) is
\[ \frac1x. \]A number and its reciprocal multiply to \(1\):
\[ x\cdot\frac1x=1, \qquad x\neq0. \]Examples include
\[ 4\longleftrightarrow\frac14, \] \[ \frac37\longleftrightarrow\frac73, \]and
\[ -5\longleftrightarrow-\frac15. \]The Basic Reciprocal Function
The simplest reciprocal function is
\[ \boxed{f(x)=\frac1x}. \]Its domain excludes zero:
\[ x\neq0. \]Its output can never equal zero because
\[ \frac1x\neq0 \]for every nonzero real \(x\).
Reciprocal Inputs Produce Reciprocal Outputs
For
\[ f(x)=\frac1x, \]a few values are
\[ f(1)=1, \] \[ f(2)=\frac12, \] \[ f(4)=\frac14, \] \[ f\left(\frac12\right)=2. \]Larger positive inputs produce smaller positive outputs, while positive inputs between \(0\) and \(1\) produce outputs greater than \(1\).
Inverse Variation
A variable \(y\) varies inversely with \(x\) when
\[ y=\frac{k}{x} \]for some constant \(k\), usually taken to be nonzero in a genuine inverse variation.
Equivalently,
\[ xy=k. \]The defining feature is therefore not merely that one variable decreases while the other increases. Their product must remain constant.
The Constant of Variation
In
\[ y=\frac{k}{x}, \]the number \(k\) controls the particular inverse relationship.
Because
\[ xy=k, \]the constant can be found from any known pair \((x,y)\):
\[ \boxed{k=xy}. \]Example: Find the Constant of Variation
Suppose \(y\) varies inversely with \(x\), and
\[ y=6 \]when
\[ x=4. \]Use
\[ k=xy. \]Then
\[ k=(4)(6)=24. \]Therefore the inverse-variation equation is
\[ \boxed{y=\frac{24}{x}}. \]Using the Inverse-Variation Equation
If
\[ y=\frac{24}{x}, \]then when \(x=8\),
\[ y=\frac{24}{8}=3. \]The original pair and the new pair have the same product:
\[ 4\cdot6=24, \] \[ 8\cdot3=24. \]This constant product is the signature of inverse variation.
A Table of Inverse Variation
Consider
\[ y=\frac{12}{x}. \]Some corresponding values are
\[ \begin{array}{c|cccccc} x & 1 & 2 & 3 & 4 & 6 & 12\\ \hline y & 12 & 6 & 4 & 3 & 2 & 1 \end{array} \]In every column,
\[ xy=12. \]The values do not change by a constant difference. They change reciprocally.
Recognizing Inverse Variation from a Table
Suppose a table contains the pairs
\[ (2,15),\qquad(3,10),\qquad(5,6). \]Multiply each pair:
\[ 2(15)=30, \] \[ 3(10)=30, \] \[ 5(6)=30. \]Since the products are constant,
\[ \boxed{xy=30} \]and therefore
\[ \boxed{y=\frac{30}{x}}. \]Not Every Decreasing Relationship Is Inverse Variation
Consider
\[ y=10-x. \]As \(x\) increases, \(y\) decreases, but
\[ xy \]is not constant.
For example,
\[ x=2,\ y=8 \]gives
\[ xy=16, \]while
\[ x=4,\ y=6 \]gives
\[ xy=24. \]Therefore the relationship is decreasing but not inverse variation.
Inverse Variation as Constant Product
The equation
\[ y=\frac{k}{x} \]can be understood as a balancing rule:
\[ xy=k. \]If one factor grows, the other must shrink enough to keep the product unchanged.
This constant-product viewpoint is often more intuitive than memorizing the fraction form alone.
What Happens When \(x\) Doubles?
Suppose
\[ y=\frac{k}{x}. \]Replace \(x\) by \(2x\):
\[ y_{\text{new}} = \frac{k}{2x}. \]Therefore,
\[ y_{\text{new}} = \frac12\frac{k}{x} = \frac12y. \]Thus:
\[ \boxed{ x\text{ doubles} \quad\Longrightarrow\quad y\text{ is halved} }. \]What Happens When \(x\) Is Multiplied by a Factor?
If
\[ y=\frac{k}{x} \]and \(x\) is multiplied by a nonzero factor \(a\), then
\[ y_{\text{new}} = \frac{k}{ax} = \frac1a y. \]Therefore,
\[ \boxed{ x\to ax \quad\Longrightarrow\quad y\to\frac{y}{a} }. \]Tripling \(x\) divides \(y\) by \(3\). Multiplying \(x\) by \(1/4\) multiplies \(y\) by \(4\).
Comparing Two Points Without Finding \(k\)
If two pairs \((x_1,y_1)\) and \((x_2,y_2)\) belong to the same inverse variation, then
\[ x_1y_1=k \]and
\[ x_2y_2=k. \]Therefore,
\[ \boxed{x_1y_1=x_2y_2}. \]This relation often solves inverse-variation problems directly.
Example: Use the Constant-Product Relationship
Suppose \(y=10\) when \(x=6\). Find \(y\) when \(x=15\).
Use
\[ x_1y_1=x_2y_2. \]Then
\[ 6(10)=15y. \]So
\[ 60=15y \]and
\[ \boxed{y=4}. \]Direct Variation for Comparison
Direct variation has the form
\[ y=kx. \]In direct variation, the ratio
\[ \frac{y}{x}=k \]is constant.
In inverse variation,
\[ y=\frac{k}{x}, \]the product
\[ xy=k \]is constant.
Direct and Inverse Scaling
Suppose \(x\) doubles.
In direct variation,
\[ y=kx, \]so \(y\) doubles.
In inverse variation,
\[ y=\frac{k}{x}, \]so \(y\) is halved.
This contrast is one of the fastest ways to distinguish the two models.
The Graph of \(y=k/x\)
The graph of
\[ y=\frac{k}{x} \]is a type of curve called a hyperbola.
The graph has two branches because \(x=0\) is excluded from the domain.
The location of the branches depends on the sign of \(k\).
When \(k>0\)
If
\[ y=\frac{k}{x} \]with \(k>0\), then \(x\) and \(y\) have the same sign.
If \(x>0\), then
\[ y>0. \]If \(x<0\), then
\[ y<0. \]Therefore the two graph branches lie in Quadrants I and III.
When \(k<0\)
If
\[ y=\frac{k}{x} \]with \(k<0\), then \(x\) and \(y\) have opposite signs.
If \(x>0\), then
\[ y<0. \]If \(x<0\), then
\[ y>0. \]Therefore the graph branches lie in Quadrants II and IV.
The Axes Are Approached but Never Reached
For
\[ y=\frac{k}{x}, \qquad k\neq0, \]the value \(x=0\) is excluded.
Also,
\[ \frac{k}{x}\neq0 \]for every finite nonzero \(x\).
Thus the graph never crosses either coordinate axis.
Instead, the axes act as asymptotes. The next lesson develops this idea systematically.
Behavior Far from the Origin
For
\[ y=\frac{k}{x}, \]as the magnitude of \(x\) becomes very large, the magnitude of \(y\) becomes very small.
For example, with
\[ y=\frac{12}{x}, \]we have
\[ x=100 \quad\Longrightarrow\quad y=0.12. \]The output approaches zero without becoming zero.
Behavior Near \(x=0\)
For
\[ y=\frac{k}{x}, \]values of \(x\) close to zero produce outputs with large magnitude.
For example,
\[ \frac{1}{0.1}=10, \] \[ \frac{1}{0.01}=100, \] \[ \frac{1}{0.001}=1000. \]This rapid growth near an excluded denominator value is characteristic of reciprocal relationships.
The Domain and Range of a Basic Inverse Variation
For
\[ y=\frac{k}{x}, \qquad k\neq0, \]the real domain is
\[ \boxed{\mathbb{R}\setminus\{0\}}. \]The real range is also
\[ \boxed{\mathbb{R}\setminus\{0\}}. \]Zero is excluded from the domain because division by zero is undefined, and zero is excluded from the range because a nonzero \(k\) divided by a finite nonzero number cannot equal zero.
Inverse Variation Is Symmetric in \(x\) and \(y\)
Starting from
\[ xy=k, \]solve for \(y\):
\[ y=\frac{k}{x}. \]Solve instead for \(x\):
\[ x=\frac{k}{y}. \]Thus the constant-product equation treats the variables symmetrically:
\[ \boxed{ y=\frac{k}{x} \quad\Longleftrightarrow\quad x=\frac{k}{y} }. \]Reciprocal Relationship Is Not the Same as Inverse Function
The words reciprocal and inverse appear in several mathematical contexts, but they do not always mean the same thing.
The reciprocal of \(f(x)\) is
\[ \frac{1}{f(x)}, \]where defined.
The inverse function \(f^{-1}(x)\), when it exists, reverses the input-output mapping of \(f\).
These are different operations.
Example: Reciprocal Versus Inverse Function
Let
\[ f(x)=2x+3. \]Its reciprocal is
\[ \frac{1}{f(x)} = \frac{1}{2x+3}. \]Its inverse function is found by reversing the mapping:
\[ f^{-1}(x)=\frac{x-3}{2}. \]Therefore,
\[ \boxed{ \frac1{f(x)}\neq f^{-1}(x) } \]in general.
Fixed Distance: Speed and Time
If a fixed distance \(d\) is traveled at speed \(v\) for time \(t\), then
\[ d=vt. \]Holding \(d\) constant gives
\[ t=\frac{d}{v}. \]Therefore travel time varies inversely with speed when the distance is fixed.
The constant product is
\[ vt=d. \]Example: Speed and Time
A trip covers \(240\) miles.
At \(60\) miles per hour,
\[ t=\frac{240}{60}=4 \]hours.
At \(80\) miles per hour,
\[ t=\frac{240}{80}=3 \]hours.
In each case,
\[ vt=240. \]Fixed Work: Rate and Time
If a task requires a fixed amount of work \(W\), then
\[ W=rt, \]where \(r\) is the work rate and \(t\) is time.
Holding \(W\) constant gives
\[ t=\frac{W}{r}. \]Thus time varies inversely with work rate for a fixed amount of work.
Workers and Completion Time
In an idealized setting where workers contribute at equal constant rates without interfering with one another, the number of workers \(n\) and the completion time \(t\) satisfy
\[ nt=k. \]Therefore,
\[ t=\frac{k}{n}. \]Doubling the workforce halves the completion time under this model.
Real workplaces may depart from the model because coordination, specialization, congestion, and other effects can change productivity.
Example: Idealized Workforce
Suppose \(8\) equally productive workers can complete a task in \(15\) hours.
The constant product is
\[ k=8(15)=120. \]If \(12\) workers perform under the same idealized assumptions, then
\[ 12t=120. \]Therefore,
\[ \boxed{t=10\text{ hours}}. \]Frequency and Period
The frequency \(f\) of a repeating event measures cycles per unit time.
The period \(T\) measures time per cycle.
They are reciprocal quantities:
\[ \boxed{T=\frac1f} \]and
\[ \boxed{f=\frac1T}. \]Their product is
\[ fT=1. \]Example: Frequency and Period
A signal has frequency
\[ f=50\text{ Hz}. \]Its period is
\[ T=\frac1{50}\text{ s}. \]Therefore,
\[ \boxed{T=0.02\text{ s}}. \]Increasing frequency decreases the time required for each cycle.
Rectangles with Fixed Area
A rectangle with length \(L\), width \(W\), and fixed area \(A\) satisfies
\[ A=LW. \]Holding \(A\) constant gives
\[ W=\frac{A}{L}. \]Thus length and width vary inversely when the area remains fixed.
Example: Fixed-Area Rectangle
Suppose a rectangle has area
\[ 72\text{ square units}. \]Then
\[ LW=72. \]If
\[ L=9, \]then
\[ W=\frac{72}{9}=8. \]If the length increases to \(12\), then
\[ W=\frac{72}{12}=6. \]Pressure and Volume as an Inverse Model
Under idealized conditions with a fixed amount of gas at constant temperature, Boyle's law gives
\[ PV=k. \]Therefore,
\[ P=\frac{k}{V}. \]Pressure and volume vary inversely within the assumptions of the model.
Increasing volume decreases pressure while the product remains constant.
Example: Pressure and Volume
Suppose
\[ P=6 \]when
\[ V=10. \]Then
\[ k=PV=60. \]If the volume becomes
\[ V=15, \]the model predicts
\[ P=\frac{60}{15}=4. \]Thus,
\[ \boxed{P=4}. \]Unit Cost and Quantity Under Fixed Total Cost
If a fixed total cost \(C\) is distributed equally across \(n\) units, then the cost per unit \(p\) satisfies
\[ C=np. \]Therefore,
\[ p=\frac{C}{n}. \]Under this simplified model, cost per unit varies inversely with the number of units sharing the fixed cost.
Dimensional Interpretation
The constant \(k\) inherits units from the product \(xy\).
If
\[ xy=k, \]then the units of \(k\) are
\[ [k]=[x][y]. \]For speed and time,
\[ vt=d, \]so the product of speed units and time units produces distance units.
Unit analysis can therefore help verify an inverse-variation model.
Solving for an Unknown Input
Suppose
\[ y=\frac{48}{x} \]and
\[ y=6. \]Then
\[ 6=\frac{48}{x}. \]Multiply by \(x\):
\[ 6x=48. \]Therefore,
\[ \boxed{x=8}. \]Solving from a Verbal Statement
Suppose: \(y\) varies inversely with \(x\), and \(y=9\) when \(x=4\).
Translate “varies inversely” into
\[ y=\frac{k}{x}. \]Substitute the known values:
\[ 9=\frac{k}{4}. \]Therefore,
\[ k=36. \]The model is
\[ \boxed{y=\frac{36}{x}}. \]From Equation to Verbal Description
Consider
\[ y=\frac{15}{x}. \]A precise verbal description is:
“\(y\) varies inversely with \(x\) with constant of variation \(15\).”
Equivalently:
“The product of \(x\) and \(y\) is always \(15\).”
From Constant Product to Rational Function
Suppose a relationship is given as
\[ xy=20. \]Solve for \(y\):
\[ y=\frac{20}{x}. \]This shows directly why inverse variation belongs to the study of rational relationships: solving the constant-product equation for one variable creates a rational function.
Negative Constants of Variation
Inverse variation does not require \(k\) to be positive.
Suppose
\[ y=-\frac{12}{x}. \]Then
\[ xy=-12. \]If \(x>0\), then \(y<0\).
If \(x<0\), then \(y>0\).
The product remains constant even though the variables have opposite signs.
Example: Negative Inverse Variation
Suppose \(y\) varies inversely with \(x\), and
\[ y=-5 \]when
\[ x=6. \]Then
\[ k=xy=6(-5)=-30. \]Therefore,
\[ \boxed{y=-\frac{30}{x}}. \]At \(x=-10\),
\[ y=-\frac{30}{-10}=3. \]Reciprocal Scaling Is Nonlinear
The graph of
\[ y=\frac{k}{x} \]is not a straight line.
Equal increases in \(x\) do not create equal changes in \(y\).
For
\[ y=\frac{12}{x}, \]increasing \(x\) from \(1\) to \(2\) changes \(y\) from \(12\) to \(6\), a change of \(-6\).
Increasing \(x\) from \(2\) to \(3\) changes \(y\) from \(6\) to \(4\), a change of only \(-2\).
Relative Change Is More Natural Than Additive Change
In inverse variation, multiplicative comparisons are often more useful than differences.
From
\[ x_1y_1=x_2y_2, \]divide by \(x_2y_1\):
\[ \frac{y_2}{y_1} = \frac{x_1}{x_2}. \]Therefore,
\[ \boxed{ \frac{y_2}{y_1} = \frac{x_1}{x_2} }. \]The output changes by the reciprocal of the input's multiplicative change.
Example: Relative Scaling
Suppose \(x\) increases from \(8\) to \(12\).
The multiplicative change in \(x\) is
\[ \frac{12}{8}=\frac32. \]Therefore, under inverse variation, \(y\) is multiplied by
\[ \frac23. \]If the original value was \(y=9\), then
\[ y_{\text{new}} = 9\left(\frac23\right) = 6. \]Inverse Variation with Powers
A broader family of reciprocal relationships has the form
\[ y=\frac{k}{x^n}. \]This is sometimes described as variation inversely as a power of \(x\).
For example,
\[ y=\frac{k}{x^2} \]means
\[ x^2y=k. \]The basic inverse variation \(y=k/x\) is the case \(n=1\).
Example: Inverse-Square Scaling
Suppose
\[ y=\frac{k}{x^2}. \]If \(x\) doubles, then
\[ y_{\text{new}} = \frac{k}{(2x)^2} = \frac{k}{4x^2} = \frac14y. \]Thus doubling the input divides the output by \(4\).
This is stronger decay than ordinary inverse variation.
Joint and Combined Variation
Reciprocal behavior can appear together with direct dependence on other variables.
For example,
\[ y=\frac{kz}{x} \]means \(y\) varies directly with \(z\) and inversely with \(x\).
Equivalently,
\[ xy=kz. \]Such models combine several scaling relationships in one equation.
Example: Combined Variation
Suppose
\[ y=\frac{kz}{x} \]and
\[ y=12,\qquad z=3,\qquad x=5. \]Substitute:
\[ 12=\frac{3k}{5}. \]Multiply by \(5\):
\[ 60=3k. \]Therefore,
\[ \boxed{k=20}. \]Shifted Reciprocal Functions
Not every reciprocal-shaped function is a strict inverse variation.
For example,
\[ y=\frac{k}{x-h}+v \]is a translated reciprocal function.
It has reciprocal structure, but \(xy\) is not generally constant.
Strict inverse variation refers specifically to a constant-product relationship such as
\[ xy=k. \]Why Shifted Reciprocal Functions Matter
The basic function
\[ y=\frac{k}{x} \]has important features centered on the coordinate axes.
A function such as
\[ y=\frac{2}{x-3}+1 \]shifts those features.
The denominator becomes zero at
\[ x=3, \]and the graph approaches
\[ y=1 \]far from the vertical break.
The next lesson formalizes these ideas using holes and asymptotes.
Common Error: Using a Constant Difference
Inverse variation is not identified by a constant difference between successive values.
For
\[ y=\frac{12}{x}, \]the \(y\)-values
\[ 12,\ 6,\ 4,\ 3 \]do not have a constant difference.
Instead, the products
\[ xy \]remain constant.
Common Error: Confusing Direct and Inverse Variation
Direct variation uses
\[ y=kx, \]while inverse variation uses
\[ y=\frac{k}{x}. \]In direct variation,
\[ \frac{y}{x}=k. \]In inverse variation,
\[ xy=k. \]Checking whether the ratio or product is constant resolves the distinction.
Common Error: Saying “Inverse” Means Subtracting
In inverse variation, “inverse” refers to reciprocal scaling, not to subtraction.
If one variable doubles, the other does not decrease by \(2\).
It is multiplied by the reciprocal factor
\[ \frac12. \]The relationship is multiplicative, not additive.
Common Error: Forgetting \(x\neq0\)
In
\[ y=\frac{k}{x}, \]the input \(x=0\) is not permitted.
The constant-product form
\[ xy=k \]does not erase this restriction when it is being used as an equivalent description of a nonzero inverse variation.
Common Error: Confusing Reciprocal with Opposite
The reciprocal of \(x\) is
\[ \frac1x. \]The opposite, or additive inverse, of \(x\) is
\[ -x. \]For example, the reciprocal of \(5\) is
\[ \frac15, \]while its opposite is
\[ -5. \]Common Error: Assuming Every Real Application Is Perfectly Inverse
A mathematical model describes selected relationships under stated or implied assumptions.
For example, doubling workers halves completion time only when total work is fixed and worker productivity scales ideally.
Real systems may include overhead, delays, limits, interactions, or other variables.
Inverse variation should therefore be used when the constant-product model is justified, not merely because two quantities appear to move in opposite directions.
A Reliable Procedure for Inverse-Variation Problems
- Identify the two quantities that vary.
- Translate inverse variation as \(y=k/x\) or \(xy=k\).
- Use a known pair of values to determine \(k\).
- Write the complete variation equation.
- Substitute the new known value.
- Solve for the unknown quantity.
- Check that the constant product is preserved.
- Interpret the answer with appropriate units and model assumptions.
A Reliable Test for Inverse Variation
Given several pairs \((x,y)\):
- verify that the relevant \(x\)-values are nonzero;
- calculate \(xy\) for each pair;
- compare the products;
- if all products equal the same constant \(k\), the data fit \(y=k/x\);
- if the products differ, the data do not represent exact inverse variation.
Example: Full Inverse-Variation Workflow
Suppose \(y\) varies inversely with \(x\), and \(y=14\) when \(x=3\). Find \(y\) when \(x=7\).
Begin with
\[ y=\frac{k}{x}. \]Find \(k\):
\[ 14=\frac{k}{3}. \] \[ k=42. \]Therefore,
\[ y=\frac{42}{x}. \]At \(x=7\),
\[ y=\frac{42}{7}=6. \]Thus,
\[ \boxed{y=6}. \]Example: Solve Using Two-State Form
A fixed-distance trip takes \(5\) hours at \(48\) miles per hour. Assuming constant speed, how long would the same distance take at \(60\) miles per hour?
Since distance is fixed,
\[ v_1t_1=v_2t_2. \]Substitute:
\[ 48(5)=60t. \] \[ 240=60t. \]Therefore,
\[ \boxed{t=4\text{ hours}}. \]Practice 1: Find \(k\)
\(y\) varies inversely with \(x\). If \(y=8\) when \(x=5\), find the constant of variation.
Practice 2: Write the Equation
\(y\) varies inversely with \(x\), and \(y=12\) when \(x=3\). Write the inverse-variation equation.
Practice 3: Find a New Output
If
\[ y=\frac{36}{x}, \]find \(y\) when \(x=9\).
Practice 4: Use Constant Product
In an inverse variation, \(x=4\) corresponds to \(y=15\). Find \(y\) when \(x=10\).
Practice 5: Recognize Inverse Variation
Determine whether the pairs
\[ (2,18),\qquad(3,12),\qquad(6,6) \]represent inverse variation. If so, find \(k\).
Practice 6: Scaling
If \(y\) varies inversely with \(x\), what happens to \(y\) when \(x\) is multiplied by \(5\)?
Practice 7: Fixed Distance
A vehicle travels a fixed route in \(6\) hours at \(50\) miles per hour. How long would the same route take at \(75\) miles per hour, assuming constant speed?
Practice 8: Frequency and Period
A repeating signal has frequency
\[ 200\text{ Hz}. \]Find its period.
Practice 9: Negative Constant
\(y\) varies inversely with \(x\), and \(y=-4\) when \(x=7\). Find \(y\) when \(x=-14\).
Practice 10: Identify the Error
A student says the pairs
\[ (2,10),\qquad(4,8),\qquad(6,6) \]represent inverse variation because \(y\) decreases as \(x\) increases.
Explain the error.
Practice Answers
Practice 1
Use
\[ k=xy. \]Therefore,
\[ k=5(8)=40. \] \[ \boxed{k=40}. \]Practice 2
Find the constant:
\[ k=xy=3(12)=36. \]Therefore,
\[ \boxed{y=\frac{36}{x}}. \]Practice 3
\[ y=\frac{36}{9}=4. \]Thus,
\[ \boxed{y=4}. \]Practice 4
Use
\[ x_1y_1=x_2y_2. \] \[ 4(15)=10y. \] \[ 60=10y. \]Therefore,
\[ \boxed{y=6}. \]Practice 5
Check the products:
\[ 2(18)=36, \] \[ 3(12)=36, \] \[ 6(6)=36. \]The product is constant, so the data represent inverse variation:
\[ \boxed{k=36}. \]Practice 6
If \(x\) is multiplied by \(5\), then \(y\) is multiplied by the reciprocal factor
\[ \frac15. \]Therefore,
\[ \boxed{y\text{ is divided by }5}. \]Practice 7
The distance is fixed, so
\[ v_1t_1=v_2t_2. \] \[ 50(6)=75t. \] \[ 300=75t. \]Therefore,
\[ \boxed{t=4\text{ hours}}. \]Practice 8
Use
\[ T=\frac1f. \] \[ T=\frac1{200}\text{ s}. \]Therefore,
\[ \boxed{T=0.005\text{ s}}. \]Practice 9
Find the constant:
\[ k=xy=7(-4)=-28. \]Thus,
\[ y=-\frac{28}{x}. \]At \(x=-14\),
\[ y=-\frac{28}{-14}=2. \]Therefore,
\[ \boxed{y=2}. \]Practice 10
A decreasing relationship is not automatically an inverse variation.
Check the products:
\[ 2(10)=20, \] \[ 4(8)=32, \] \[ 6(6)=36. \]Since the products are not constant, the data do not represent exact inverse variation.
What to Remember
The defining equations of inverse variation are
\[ \boxed{y=\frac{k}{x}} \]and
\[ \boxed{xy=k}. \]The essential ideas are:
- inverse variation is a constant-product relationship;
- the constant of variation is \(k=xy\);
- multiplying one variable by a factor divides the other by that factor;
- direct variation preserves a ratio, while inverse variation preserves a product;
- the graph of \(y=k/x\) is a two-branch hyperbola;
- the sign of \(k\) determines which quadrants contain the branches;
- \(x=0\) is excluded from the domain;
- for \(k\neq0\), \(y=0\) is excluded from the range;
- reciprocal relationships appear naturally in fixed-product models;
- real applications require attention to the assumptions that make the product approximately constant.
The most useful conceptual test is: if one quantity is multiplied by a factor and the other is divided by the same factor, look for an inverse-variation model.
Where We Go Next
Reciprocal functions already reveal two important features of rational graphs: excluded inputs and values that a curve approaches without reaching. More complicated rational functions can also contain removable gaps created by canceled factors.
The next lesson, Zeros, Holes, and Asymptotes, develops these graphical features systematically and connects them to the factored structure of rational expressions.