The Central Idea
Consider a rational function
\[ f(x)=\frac{P(x)}{Q(x)}. \]Its graph is strongly influenced by three kinds of algebraic information:
- zeros of the numerator;
- zeros of the denominator;
- factors shared by numerator and denominator.
Factoring reveals which of these zeros create x-intercepts, holes, or vertical asymptotes.
Begin with the Domain
Before interpreting any graph, determine where the rational function is defined.
For
\[ f(x)=\frac{P(x)}{Q(x)}, \]every real zero of \(Q(x)\) must be excluded.
These excluded values may later appear graphically as holes or vertical asymptotes.
Zeros of a Rational Function
A zero of \(f(x)\) is an input value for which
\[ f(x)=0. \]A fraction equals zero when its numerator equals zero and its denominator does not.
Therefore, for
\[ f(x)=\frac{P(x)}{Q(x)}, \]zeros occur where
\[ P(x)=0 \]while
\[ Q(x)\neq0. \]Zeros Become x-Intercepts
If
\[ f(a)=0, \]then the graph contains the point
\[ (a,0). \]Therefore zeros of a rational function correspond to x-intercepts, provided the function is actually defined there.
Example: Find the Zero
Consider
\[ f(x)=\frac{x-4}{x+2}. \]The numerator is zero when
\[ x-4=0. \]Therefore,
\[ x=4. \]Since the denominator at \(x=4\) is nonzero,
\[ \boxed{x=4} \]is a zero, and the graph crosses or touches the x-axis at
\[ \boxed{(4,0)}. \]Several Zeros
Consider
\[ f(x)=\frac{(x-2)(x+5)}{x+1}. \]The numerator is zero at
\[ x=2 \]and
\[ x=-5. \]Neither value makes the denominator zero.
Therefore the x-intercepts are
\[ \boxed{(2,0),\qquad(-5,0)}. \]Multiplicity Still Matters
Rational functions inherit the same basic multiplicity behavior seen in polynomial graphs.
For example,
\[ f(x)=\frac{(x-3)^2}{x+1} \]has a zero of multiplicity \(2\) at
\[ x=3. \]Because the multiplicity is even, the graph typically touches the x-axis and turns rather than crossing there.
A Denominator Zero Is Not a Function Zero
Consider
\[ f(x)=\frac{x+2}{x-3}. \]The denominator is zero at
\[ x=3. \]This does not make \(f(3)=0\).
Instead,
\[ f(3) \]is undefined.
Common Factors Need Special Attention
Suppose numerator and denominator share a factor:
\[ f(x)=\frac{(x-2)(x+3)}{(x-2)(x+5)}. \]The factor \(x-2\) appears in both numerator and denominator.
Algebraically, it can be canceled for \(x\neq2\), but the original function remains undefined at \(x=2\).
This produces a hole.
What Is a Hole?
A hole is a missing point in the graph of a function.
In a rational function, a hole commonly occurs when a factor cancels from numerator and denominator.
The canceled factor identifies the excluded x-value.
The simplified expression identifies the y-value that the graph would have had there.
Example: Find a Hole
Consider
\[ f(x)=\frac{(x-2)(x+3)}{(x-2)(x+5)}. \]The original restriction is
\[ x\neq2,\qquad x\neq-5. \]Cancel \(x-2\):
\[ f(x)=\frac{x+3}{x+5}, \qquad x\neq2,\ -5. \]The canceled factor tells us there is a hole at \(x=2\).
Finding the y-Coordinate of a Hole
Continue with the simplified expression
\[ \frac{x+3}{x+5}. \]Evaluate it at the excluded value \(x=2\):
\[ \frac{2+3}{2+5} = \frac57. \]Therefore the graph has a hole at
\[ \boxed{\left(2,\frac57\right)}. \]Why the Hole Is Not Filled
The simplified formula
\[ \frac{x+3}{x+5} \]is defined at \(x=2\), but the original function
\[ \frac{(x-2)(x+3)}{(x-2)(x+5)} \]is not.
The simplified expression tells us the value approached near \(x=2\), not the actual value of the original function there.
Removable Discontinuity
A hole is also called a removable discontinuity.
The discontinuity is called removable because assigning the missing limiting value at the hole would make the simplified graph continuous at that point.
The original rational function, however, still has the hole unless the function is explicitly redefined.
Vertical Asymptotes
A vertical asymptote occurs where a denominator factor remains after common factors are canceled.
Near such an x-value, the denominator approaches zero while the numerator remains nonzero.
The function's magnitude can therefore grow without bound.
Example: Vertical Asymptote
Consider
\[ f(x)=\frac{x+1}{x-4}. \]The denominator is zero at
\[ x=4. \]No factor cancels.
Therefore,
\[ \boxed{x=4} \]is a vertical asymptote.
Holes and Vertical Asymptotes Both Come from Denominator Zeros
The difference is what happens after factoring.
If a denominator factor cancels, its zero becomes a hole.
If a denominator factor remains, its zero becomes a vertical asymptote.
Thus:
\[ \boxed{ \text{canceled denominator factor}\to\text{hole} } \] \[ \boxed{ \text{uncanceled denominator factor}\to\text{vertical asymptote} }. \]Example: Hole and Vertical Asymptote Together
Consider
\[ f(x)= \frac{(x-1)(x+2)} {(x-1)(x-3)}. \]Restrictions:
\[ x\neq1,\qquad x\neq3. \]Cancel \(x-1\):
\[ f(x)=\frac{x+2}{x-3}, \qquad x\neq1,\ 3. \]Therefore:
- \(x=1\) gives a hole;
- \(x=3\) gives a vertical asymptote.
Find the Hole in the Combined Example
Use the simplified formula
\[ \frac{x+2}{x-3}. \]At \(x=1\),
\[ \frac{1+2}{1-3} = -\frac32. \]Therefore the hole is
\[ \boxed{\left(1,-\frac32\right)}. \]The vertical asymptote remains
\[ \boxed{x=3}. \]Repeated Denominator Factors
Consider
\[ f(x)=\frac{x+1}{(x-2)^2}. \]The denominator factor \(x-2\) remains and has multiplicity \(2\).
Therefore,
\[ x=2 \]is still a vertical asymptote.
Multiplicity affects the behavior on the two sides of the asymptote.
Odd and Even Multiplicity at Vertical Asymptotes
A denominator factor with odd multiplicity changes sign as \(x\) passes through its zero.
A denominator factor with even multiplicity keeps the same sign on both sides.
This often determines whether the graph approaches opposite infinities or the same infinity on the two sides of a vertical asymptote.
Example: Odd-Multiplicity Vertical Asymptote
Consider
\[ f(x)=\frac1{x-2}. \]As \(x\to2^-\), the denominator is a small negative number, so
\[ f(x)\to-\infty. \]As \(x\to2^+\), the denominator is a small positive number, so
\[ f(x)\to+\infty. \]The graph approaches opposite infinities on the two sides.
Example: Even-Multiplicity Vertical Asymptote
Consider
\[ f(x)=\frac1{(x-2)^2}. \]The denominator is positive on both sides of \(x=2\).
Therefore,
\[ f(x)\to+\infty \]from both sides.
Numerator Sign Also Matters Near an Asymptote
For a function such as
\[ f(x)=\frac{x+1}{x-3}, \]the sign near \(x=3\) depends on both numerator and denominator.
Since \(x+1\) is positive near \(3\), the sign behavior is controlled by \(x-3\).
In more complicated examples, factor-by-factor sign analysis is useful.
Horizontal Asymptotes
A horizontal asymptote describes the value a function approaches as \(x\) becomes very large in magnitude.
Rational functions often have horizontal asymptotes determined by the degrees and leading coefficients of the numerator and denominator.
Degree of the Numerator Less Than Degree of the Denominator
Suppose
\[ f(x)=\frac{P(x)}{Q(x)} \]and
\[ \deg P<\deg Q. \]Then the denominator grows faster in magnitude than the numerator.
Therefore,
\[ f(x)\to0 \]as \(|x|\to\infty\).
The horizontal asymptote is
\[ \boxed{y=0}. \]Example: Horizontal Asymptote at Zero
Consider
\[ f(x)=\frac{3x+1}{x^2+4}. \]The numerator has degree \(1\), and the denominator has degree \(2\).
Therefore,
\[ \boxed{y=0} \]is the horizontal asymptote.
Equal Degrees
If
\[ \deg P=\deg Q, \]then the highest-degree terms dominate the behavior for large \(|x|\).
Suppose
\[ P(x)\sim ax^n \]and
\[ Q(x)\sim bx^n. \]Then
\[ \frac{P(x)}{Q(x)} \to \frac{a}{b}. \]The horizontal asymptote is
\[ \boxed{y=\frac{a}{b}}. \]Example: Equal Degrees
Consider
\[ f(x)=\frac{2x^2+3x-1}{5x^2+4}. \]Both numerator and denominator have degree \(2\).
The ratio of leading coefficients is
\[ \frac25. \]Therefore,
\[ \boxed{y=\frac25} \]is the horizontal asymptote.
Why Leading Terms Control End Behavior
Divide numerator and denominator by the highest common power of \(x\).
For
\[ \frac{2x^2+3x-1}{5x^2+4}, \]divide by \(x^2\):
\[ \frac{2+\frac3x-\frac1{x^2}} {5+\frac4{x^2}}. \]As \(|x|\) becomes large,
\[ \frac3x,\quad\frac1{x^2},\quad\frac4{x^2} \]all approach zero.
The remaining ratio approaches
\[ \frac25. \]When the Numerator Degree Is Greater
If
\[ \deg P>\deg Q, \]there is generally no horizontal asymptote.
Instead, polynomial division reveals a polynomial that describes the function's end behavior.
Oblique or Slant Asymptotes
If the numerator degree is exactly one greater than the denominator degree, polynomial division produces a linear quotient.
That line is called an oblique asymptote or slant asymptote.
Example: Find a Slant Asymptote
Consider
\[ f(x)=\frac{x^2+1}{x-1}. \]Divide:
\[ x^2+1=(x-1)(x+1)+2. \]Therefore,
\[ f(x)=x+1+\frac{2}{x-1}. \]As \(|x|\to\infty\),
\[ \frac{2}{x-1}\to0. \]Thus the slant asymptote is
\[ \boxed{y=x+1}. \]Polynomial Asymptotes
If the numerator degree exceeds the denominator degree by more than one, polynomial division produces a higher-degree polynomial quotient.
The rational function approaches that polynomial as \(|x|\) becomes large.
Slant asymptotes are simply the degree-one case of this broader idea.
A Horizontal Asymptote Is Not a Barrier
A rational function may cross a horizontal asymptote.
Horizontal asymptotes describe end behavior, not forbidden y-values.
For example,
\[ f(x)=\frac{x}{x+1} \]has horizontal asymptote
\[ y=1. \]Other rational functions can cross their horizontal asymptotes at finite x-values.
A Vertical Asymptote Is Different
A vertical asymptote occurs at an excluded x-value where a denominator factor remains.
The graph cannot contain a point with that x-coordinate because the function is undefined there.
Thus vertical and horizontal asymptotes describe different types of behavior.
Finding y-Intercepts
Although this lesson focuses on zeros, holes, and asymptotes, the y-intercept is another useful graph feature.
Set
\[ x=0 \]if \(0\) belongs to the domain.
For
\[ f(x)=\frac{x-2}{x+3}, \]we get
\[ f(0)=\frac{-2}{3}. \]Therefore the y-intercept is
\[ \boxed{\left(0,-\frac23\right)}. \]A Full Factored Analysis
Consider
\[ f(x)= \frac{(x-4)(x+1)} {(x-2)(x+1)}. \]From the original denominator,
\[ x\neq2,\qquad x\neq-1. \]Cancel \(x+1\):
\[ f(x)=\frac{x-4}{x-2}, \qquad x\neq2,\ -1. \]We can now classify the major features.
Classify the Features
For
\[ f(x)= \frac{(x-4)(x+1)} {(x-2)(x+1)}, \]- the remaining numerator zero \(x=4\) gives an x-intercept;
- the canceled denominator zero \(x=-1\) gives a hole;
- the remaining denominator zero \(x=2\) gives a vertical asymptote.
Therefore,
\[ \boxed{\text{x-intercept: }(4,0)} \] \[ \boxed{\text{vertical asymptote: }x=2}. \]Find the Hole in the Full Analysis
Use the simplified formula
\[ \frac{x-4}{x-2}. \]At \(x=-1\),
\[ \frac{-1-4}{-1-2} = \frac{-5}{-3} = \frac53. \]Therefore the hole is
\[ \boxed{\left(-1,\frac53\right)}. \]Find the Horizontal Asymptote in the Full Analysis
The simplified function
\[ \frac{x-4}{x-2} \]has equal numerator and denominator degrees.
The leading coefficients are both \(1\).
Therefore,
\[ \boxed{y=1} \]is the horizontal asymptote.
Why Simplification Comes Before Classification
If we looked only at denominator zeros before canceling, we might wrongly call every excluded value a vertical asymptote.
Factoring and cancellation distinguish:
- removable discontinuities from canceled factors;
- vertical asymptotes from remaining denominator factors.
This is why graph analysis begins with algebraic simplification while preserving the original domain.
Zeros Can Disappear Through Cancellation
Consider
\[ f(x)=\frac{(x-3)(x+2)}{x-3}. \]The numerator appears to be zero at \(x=3\), but the denominator is also zero there.
After cancellation,
\[ f(x)=x+2, \qquad x\neq3. \]Therefore \(x=3\) is not a zero of the original function.
It is a hole.
A Zero Must Survive Simplification
For a factored rational function, a numerator factor creates an actual zero only if it remains after all common factors are canceled.
This leads to a useful rule:
\[ \boxed{ \text{remaining numerator factor}\to\text{zero} }. \]A numerator factor that cancels does not create an x-intercept at its zero.
Vertical Asymptotes Can Also Be Removed by Cancellation
Consider
\[ f(x)=\frac{x-5}{(x-5)(x+1)}. \]The original denominator is zero at
\[ x=5 \]and
\[ x=-1. \]Cancel \(x-5\):
\[ f(x)=\frac1{x+1}, \qquad x\neq5,\ -1. \]Therefore \(x=5\) is a hole, not a vertical asymptote.
Graphical Meaning of Sign Changes at Zeros
If a remaining numerator factor has odd multiplicity, the sign of the function often changes as the graph passes through the zero.
If the multiplicity is even, the sign generally stays the same and the graph touches the axis.
This is the rational-function version of polynomial zero behavior.
Graphical Meaning of Sign Changes at Vertical Asymptotes
If a remaining denominator factor has odd multiplicity, its sign changes across the excluded x-value.
If its multiplicity is even, its sign does not.
Combined with the signs of the other factors, this predicts whether the graph approaches \(+\infty\) or \(-\infty\) from each side.
Sign Charts Help Organize Rational Graphs
A sign chart uses zeros, holes, and vertical asymptotes to divide the number line into intervals.
One test point from each interval reveals whether the rational function is positive or negative there.
This information helps sketch the graph between its major structural features.
Example: Sign Analysis
Consider
\[ f(x)=\frac{x-1}{x+2}. \]Critical x-values are
\[ x=-2 \]and
\[ x=1. \]These divide the real line into
\[ (-\infty,-2),\qquad(-2,1),\qquad(1,\infty). \]Testing gives signs
\[ +,\quad-,\quad+. \]Can a Graph Cross a Vertical Asymptote?
No point on the graph can have the x-coordinate of a vertical asymptote, because the function is undefined there.
The graph may have branches on both sides, but the vertical line itself remains excluded.
Can a Graph Cross a Horizontal Asymptote?
Yes.
A horizontal asymptote describes only the function's behavior as \(|x|\) becomes very large.
It does not generally prevent the graph from crossing that horizontal line at finite x-values.
Can a Graph Cross a Slant Asymptote?
Yes.
Like a horizontal asymptote, a slant asymptote describes end behavior.
A rational graph may cross its slant asymptote one or more times at finite x-values.
Common Error: Calling Every Denominator Zero a Vertical Asymptote
Consider
\[ \frac{(x-2)(x+1)}{(x-2)(x-5)}. \]Both \(x=2\) and \(x=5\) make the original denominator zero.
But \(x-2\) cancels.
Therefore:
\[ \boxed{x=2\text{ gives a hole}} \]and
\[ \boxed{x=5\text{ gives a vertical asymptote}}. \]Common Error: Treating a Hole as an x-Intercept
Suppose
\[ f(x)=\frac{(x-3)(x+4)}{x-3}. \]The numerator has a factor \(x-3\), but that factor also appears in the denominator.
Since it cancels, \(x=3\) is excluded.
Therefore the graph does not contain the point \((3,0)\).
Instead, \(x=3\) gives a hole.
Common Error: Forgetting the Original Domain After Simplifying
Consider
\[ \frac{x^2-1}{x-1}. \]Simplifying gives
\[ x+1. \]But the original restriction
\[ x\neq1 \]remains.
The graph is the line \(y=x+1\) with a hole at
\[ \boxed{(1,2)}. \]Common Error: Using Denominator Degree Alone for End Behavior
Horizontal and slant asymptotes depend on comparing numerator and denominator degrees.
Looking only at the denominator degree is insufficient.
Always compare
\[ \deg P \]with
\[ \deg Q. \]Common Error: Assuming Horizontal Asymptotes Cannot Be Crossed
Horizontal asymptotes describe limiting behavior as
\[ x\to\pm\infty. \]They are not graph barriers.
A rational function may cross a horizontal asymptote at finite x-values.
Common Error: Confusing Undefined with Infinite
At a vertical asymptote \(x=a\), the function is not equal to
\[ \infty \]or
\[ -\infty. \]The function is undefined at \(x=a\).
The symbols \(+\infty\) and \(-\infty\) describe unbounded behavior as \(x\) approaches \(a\), not function values at \(a\).
A Reliable Procedure for Graph Structure
- Factor the numerator and denominator completely.
- Record all restrictions from the original denominator.
- Cancel common factors while preserving their excluded x-values.
- Use canceled denominator factors to locate holes.
- Use remaining numerator factors to locate zeros and x-intercepts.
- Use remaining denominator factors to locate vertical asymptotes.
- Find the y-coordinate of each hole using the simplified formula.
- Find the y-intercept when \(x=0\) is allowed.
- Compare degrees or divide polynomials to determine end behavior.
- Use sign analysis when a more complete sketch is needed.
Example: Full Structural Analysis
Analyze
\[ f(x)= \frac{(x-3)(x+2)} {(x-3)(x-1)}. \]Original restrictions:
\[ x\neq3,\qquad x\neq1. \]Cancel \(x-3\):
\[ f(x)=\frac{x+2}{x-1}, \qquad x\neq3,\ 1. \]Complete the Structural Analysis
The remaining numerator zero is
\[ x=-2, \]so the x-intercept is
\[ (-2,0). \]The canceled factor gives a hole at \(x=3\).
The remaining denominator gives a vertical asymptote at
\[ x=1. \]Equal degrees give the horizontal asymptote
\[ y=1. \]Find the Hole and y-Intercept
For
\[ \frac{x+2}{x-1}, \]evaluate at \(x=3\):
\[ \frac{3+2}{3-1} = \frac52. \]Therefore the hole is
\[ \boxed{\left(3,\frac52\right)}. \]At \(x=0\),
\[ f(0)=\frac2{-1}=-2. \]Therefore the y-intercept is
\[ \boxed{(0,-2)}. \]Practice 1: Zero and Vertical Asymptote
Analyze
\[ f(x)=\frac{x-5}{x+2}. \]Find the zero and vertical asymptote.
Practice 2: Hole
Analyze
\[ f(x)=\frac{(x-4)(x+1)}{(x-4)(x+3)}. \]Find the hole and vertical asymptote.
Practice 3: Several Zeros
Find the zeros of
\[ f(x)=\frac{(x-2)(x+6)}{x-1}. \]Practice 4: Horizontal Asymptote
Find the horizontal asymptote of
\[ f(x)=\frac{3x^2+1}{7x^2-5x+2}. \]Practice 5: Horizontal Asymptote at Zero
Find the horizontal asymptote of
\[ f(x)=\frac{2x+1}{x^2+4}. \]Practice 6: Slant Asymptote
Find the slant asymptote of
\[ f(x)=\frac{x^2+3}{x-2}. \]Practice 7: Hole Versus Zero
Analyze
\[ f(x)=\frac{(x-1)(x+5)}{x-1}. \]Does \(x=1\) give a zero or a hole?
Practice 8: Repeated Vertical Asymptote
Analyze the behavior near
\[ x=3 \]for
\[ f(x)=\frac1{(x-3)^2}. \]Practice 9: Full Feature Analysis
Analyze
\[ f(x)= \frac{(x+2)(x-4)} {(x+2)(x-1)}. \]Find the zero, hole, vertical asymptote, and horizontal asymptote.
Practice 10: Identify the Error
A student says that
\[ f(x)=\frac{(x-2)(x+3)}{(x-2)(x+5)} \]has vertical asymptotes at both \(x=2\) and \(x=-5\).
Explain the error.
Practice Answers
Practice 1
The numerator is zero at
\[ x=5, \]so the zero is
\[ \boxed{x=5}. \]The denominator is zero at
\[ x=-2, \]and no factor cancels.
\[ \boxed{x=-2\text{ is a vertical asymptote}}. \]Practice 2
Cancel \(x-4\):
\[ f(x)=\frac{x+1}{x+3}, \qquad x\neq4,\ -3. \]Therefore \(x=4\) gives a hole.
Its y-coordinate is
\[ \frac{4+1}{4+3}=\frac57. \]So the hole is
\[ \boxed{\left(4,\frac57\right)}. \]The vertical asymptote is
\[ \boxed{x=-3}. \]Practice 3
The numerator factors already show zeros at
\[ \boxed{x=2,\qquad x=-6}. \]Practice 4
The numerator and denominator have equal degree.
The ratio of leading coefficients is
\[ \frac37. \]Therefore,
\[ \boxed{y=\frac37}. \]Practice 5
The numerator degree is less than the denominator degree.
Therefore,
\[ \boxed{y=0}. \]Practice 6
Divide:
\[ x^2+3=(x-2)(x+2)+7. \]Therefore,
\[ f(x)=x+2+\frac7{x-2}. \]The slant asymptote is
\[ \boxed{y=x+2}. \]Practice 7
The factor \(x-1\) cancels:
\[ f(x)=x+5, \qquad x\neq1. \]Therefore \(x=1\) gives a hole, not a zero.
The hole is at
\[ \boxed{(1,6)}. \]Practice 8
The denominator is positive on both sides of \(x=3\).
Therefore,
\[ \boxed{ f(x)\to+\infty \text{ as }x\to3^- \text{ and as }x\to3^+ }. \]Practice 9
Cancel \(x+2\):
\[ f(x)=\frac{x-4}{x-1}, \qquad x\neq-2,\ 1. \]The zero is
\[ \boxed{x=4}. \]The hole occurs at \(x=-2\). Its y-coordinate is
\[ \frac{-2-4}{-2-1}=2. \]Therefore the hole is
\[ \boxed{(-2,2)}. \]The vertical asymptote is
\[ \boxed{x=1}. \]Equal degrees with leading coefficient ratio \(1\) give
\[ \boxed{y=1} \]as the horizontal asymptote.
Practice 10
The factor \(x-2\) cancels.
Therefore \(x=2\) produces a hole, not a vertical asymptote.
Only the remaining denominator factor \(x+5\) produces a vertical asymptote:
\[ \boxed{x=-5}. \]What to Remember
Factored form reveals the structural features of a rational graph.
The most important correspondences are:
- remaining numerator zeros give zeros and x-intercepts;
- canceled denominator factors give holes;
- remaining denominator factors give vertical asymptotes;
- hole coordinates come from evaluating the simplified formula at the excluded x-value;
- horizontal asymptotes come from degree comparison when the numerator degree is not greater than the denominator degree;
- slant or polynomial asymptotes come from polynomial division when the numerator degree is greater.
A compact structural summary is:
\[ \boxed{ \begin{aligned} \text{remaining numerator factor} &\to \text{zero},\\ \text{canceled denominator factor} &\to \text{hole},\\ \text{remaining denominator factor} &\to \text{vertical asymptote}. \end{aligned} } \]The central habit is: factor first, preserve the original domain, and classify features only after cancellation is understood.
Where We Go Next
Rational expressions have now provided a complete setting for division, reciprocal relationships, restricted domains, and discontinuities. The next group of lessons turns to another operation that reverses a familiar power relationship: radicalization.
The next lesson, Roots, Radicals, and Radicalization, introduces roots as inverse operations of powers, develops radical notation, and establishes the domain conditions needed for real-valued radical expressions.