Unit 1: Polynomial and Rational Functions

Functions and How Quantities Change in Tandem

A function is a mathematical relationship that maps a set of input values to a set of output values such that each input value is mapped to exactly one output value. The input values are the domain (often the independent variable, $x$), and the output values are the **range** (often the dependent variable, $y$). The input and output of a function vary according to the function rule, and that relationship can be represented graphically, verbally, analytically, or numerically.

A graph is a visual display of input-output pairs. It helps you see how outputs change as inputs change, including where the function increases/decreases, where it crosses the axes, and how its overall shape behaves.

A function is increasing over an interval of its domain if, as input values increase, output values always increase. Formally, for all $a$ and $b$ in the interval, if $a<b$, then:

$f(a)<f(b)$

A function is decreasing over an interval of its domain if, as input values increase, output values always decrease. Formally, for all $a$ and $b$ in the interval, if $a<b$, then:

$f(a)>f(b)$

x-intercepts are points where the graph meets the x-axis. They correspond to zeros of the function (values of $x$ where the output is 0).

Concavity describes how the rate of change is changing:

• Concave up means the rate of change is increasing.
• Concave down means the rate of change is decreasing.

Exam Focus

• Typical question patterns:
  • Identify domain/range, intercepts, and intervals of increase/decrease from a graph or table.
  • Translate between representations (graph, equation, table, context).
• Common mistakes:
  • Mixing up domain vs range.
  • Thinking “increasing” means “positive outputs” (increasing is about how outputs change as inputs increase).

Rates of Change and Concavity

The average rate of change (AROC) of a function over a closed interval $[a,b]$ is the **slope of the secant line** through $(a,f(a))$ and $(b,f(b))$. The formula is:

$\text{AROC on }[a,b]=\frac{f(b)-f(a)}{b-a}$

The rate of change of a function at a point can be approximated by the average rate of change over small intervals containing that point.

Interpreting sign:

• Positive rate of change: when one quantity increases, the other also increases.
• Negative rate of change: when one quantity increases, the other decreases.

AROC patterns by function type:

• For a linear function, AROC is constant (it changes at a rate of 0).
• For a quadratic function, the AROC over equal-length intervals changes at a constant rate (the secant slopes form a linear pattern). If AROC is increasing over an interval, the function is concave up; if AROC is decreasing, the function is concave down.

Exam Focus

• Typical question patterns:
  • Compute and interpret average rate of change from a table, graph, or formula.
  • Use how AROC changes to make conclusions about concavity.
• Common mistakes:
  • Using $f(a)-f(b)$ instead of $f(b)-f(a)$.
  • Confusing “decreasing function” with “negative rate of change at a point” (a function can be above the x-axis and still be decreasing).

Polynomial Functions: Building Blocks and Key Features

A polynomial function is one of the most important families of functions in precalculus because it’s both structured (so you can predict its behavior) and flexible (so it can model many real situations). You can think of polynomials as being built by adding scaled power functions like $x$, $x^2$, $x^3$, and so on.

What a polynomial is (and what it is not)

A polynomial function has the form:

$f(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0$

where:

• $n$ is a nonnegative integer called the degree
• $a_n,\dots,a_0$ are real coefficients
• $a_n\neq 0$ (otherwise the true degree is smaller)

Key restriction: polynomials allow only whole-number exponents that are not negative. This is what makes polynomial graphs smooth and continuous everywhere.

Not polynomials:

$f(x)=\frac{1}{x}=x^{-1}$

$f(x)=\sqrt{x}=x^{1/2}$

$f(x)=\lvert x\rvert$

Degree, leading term, and end behavior

Two pieces of information control the big-picture shape of a polynomial:

• Degree $n$: affects end behavior and limits how wiggly the graph can be.
• Leading coefficient $a_n$: controls whether the ends go up or down.

The leading term $a_n x^n$ dominates for large $\lvert x\rvert$, so it determines end behavior.

End behavior rules:

• If $n$ is even, both ends go the same way.
  • If $a_n>0$, both ends go up.
  • If $a_n<0$, both ends go down.
• If $n$ is odd, ends go opposite ways.
  • If $a_n>0$, left end down and right end up.
  • If $a_n<0$, left end up and right end down.

A useful global conclusion: a polynomial with even degree must have a global (absolute) maximum or a global (absolute) minimum (because both ends head in the same direction), while an odd-degree polynomial typically does not have an absolute max/min over all real numbers.

Smoothness, continuity, turning points, extrema, and inflection points

Polynomials are continuous for all real $x$. They have no holes and no vertical asymptotes.

A polynomial of degree $n$ has **at most** $n-1$ turning points.

A local (relative) maximum/minimum occurs where the polynomial changes between increasing and decreasing. If the domain is restricted, an endpoint of that restricted domain can also act as a local max/min.

A global (absolute) maximum/minimum is the greatest local maximum or least local minimum over the entire domain in question.

If a polynomial has two distinct real zeros, then between those zeros there is at least one extremum (a peak or a valley). For polynomials, this is guaranteed because they are smooth and continuous.

A point of inflection is where the function’s rate of change switches from increasing to decreasing or from decreasing to increasing, meaning the graph changes concavity.

Zeros, x-intercepts, factors, and multiplicity

A zero of $f$ is a value $x=r$ such that:

$f(r)=0$

On a graph, zeros correspond to x-intercepts. Algebraically, real zeros correspond to linear factors: if $r$ is a real zero, then $x-r$ is a factor.

If a factor repeats, the zero has a multiplicity. If:

$f(x)=(x-r)^k\cdot g(x)$

with $g(r)\neq 0$, then $r$ is a zero with multiplicity $k$.

• If $k$ is **odd**, the graph **crosses** the x-axis at $x=r$ (the function changes sign).
• If $k$ is even, the graph touches and turns around (it does not change sign).

Even multiplicity guarantees a bounce, but the “flatness” depends on how large $k$ is (for example, multiplicity 4 looks flatter than multiplicity 2).

A polynomial function of degree $n$ has exactly $n$ complex zeros counting multiplicity.

Standard form vs factored form

• Standard form (expanded) is best for degree, leading coefficient, and sometimes the y-intercept.
• Factored form is best for zeros and multiplicities.

A typical factored structure is:

$f(x)=a(x-r_1)^{m_1}(x-r_2)^{m_2}\cdots$

Even and odd symmetry (for monomial parents)

For parent monomials $p(x)=a_n x^n$ with $n\ge 1$ and $a_n\neq 0$:

• If $n$ is even, $p$ is an even function, meaning:

$p(-x)=p(x)$

Its graph is symmetric about the y-axis.

• If $n$ is odd, $p$ is an odd function, meaning:

$p(-x)=-p(x)$

Its graph is symmetric about the origin.

Finding degree from a table (finite differences)

If input values are evenly spaced, the degree of a polynomial can be determined from successive differences of output values: the degree is the least value of $n$ for which the successive $n$th differences are constant.

Worked example: build a polynomial from intercept behavior

Problem. Find a polynomial of least degree with x-intercepts at $x=-2$ (crosses), $x=1$ (touches), and end behavior up on both ends.

Step 1: Translate intercept behavior to factors.

• Cross at $x=-2$ means odd multiplicity; choose 1:

$x+2$

• Touch at $x=1$ means even multiplicity; choose 2:

$(x-1)^2$

So far:

$f(x)=a(x+2)(x-1)^2$

Step 2: Use end behavior to choose the sign of $a$.
Degree is $1+2=3$ (odd). For odd degree:

• $a>0$ gives left down, right up.
• $a<0$ gives left up, right down.

But we want both ends up, which is impossible for odd degree. The correct conclusion is that no polynomial exists with exactly those features.

Worked example: determine a possible polynomial from a graph description

Problem. A polynomial has real zeros at $x=-3$ (touches) and $x=2$ (crosses). The y-intercept is $f(0)=-12$. Find one possible polynomial.

Step 1: Write factored form with multiplicities.
Touch at $x=-3$ means even multiplicity (use 2). Cross at $x=2$ means odd multiplicity (use 1):

$f(x)=a(x+3)^2(x-2)$

Step 2: Use the y-intercept to solve for $a$.

$-12=a(3)^2(-2)$

$-12=a\cdot 9\cdot(-2)$

$-12=-18a$

$a=\frac{2}{3}$

So one possible polynomial is:

$f(x)=\frac{2}{3}(x+3)^2(x-2)$

Exam Focus

• Typical question patterns:
  • Given zeros/multiplicities/end behavior, construct or identify a polynomial in factored form.
  • Given a polynomial (often factored), interpret x-intercepts and whether the graph crosses or touches.
  • Use turning-point limits to eliminate impossible sketches.
  • Use finite differences (with equal x-steps) to identify the degree from a table.
• Common mistakes:
  • Forgetting that even degree means same-direction end behavior and odd degree means opposite-direction end behavior.
  • Treating “touches” as “crosses” (mixing up even vs odd multiplicity).
  • Assuming every listed feature is automatically consistent (sometimes the right answer is “impossible”).
  • Trying finite differences when the x-values are not equally spaced.

Transformations and Graphing Strategy for Polynomial Functions

Graphing polynomials efficiently isn’t about plotting lots of points. The goal is to combine transformations, intercepts, multiplicity behavior, sign changes, and end behavior into a correct qualitative sketch.

Parent functions and transformations

A general transformation template is:

$g(x)=a\,f(b(x-h))+k$

Interpretation:

• $h$ shifts the graph right (positive) or left (negative).
• $k$ shifts the graph up (positive) or down (negative).
• $a$ vertically stretches/compresses by $\lvert a\rvert$ and reflects across the x-axis if $a<0$.
• $b$ horizontally stretches/compresses by a factor $\frac{1}{\lvert b\rvert}$ and reflects across the y-axis if $b<0$.

Common special cases written in a quick-recognition way:

• Vertical shift:

$g(x)=f(x)+k$

• Horizontal shift:

$g(x)=f(x+h)$

(this shifts by $-h$)

• Vertical dilation/reflection:

$g(x)=a f(x)$

• Horizontal dilation/reflection:

$g(x)=f(bx)$

The domain and range of a transformed function may differ from the parent function.

A reliable graphing workflow

A consistent workflow for polynomial sketches:

1. Determine degree and leading coefficient for end behavior.
2. Find intercepts: x-intercepts from factors (or solving), y-intercept from $f(0)$.
3. Use multiplicities to decide cross vs touch and the amount of flattening.
4. Use sign analysis between zeros to see where the graph is above/below the x-axis.
5. Check turning-point plausibility (at most $n-1$).

Sign charts (why the graph is above or below)

If a polynomial is factored:

$f(x)=a(x-r_1)^{m_1}(x-r_2)^{m_2}$

then on any interval not containing a zero, each factor has a constant sign, so the entire product’s sign is predictable. Real zeros also act as endpoints when solving polynomial inequalities.

Worked example: sketch using factor/multiplicity and sign

Problem. Sketch key features of:

$f(x)=-(x+1)^2(x-3)$

Step 1: Degree and end behavior.
Degree is $3$. Leading coefficient is negative, so left end up and right end down.

Step 2: x-intercepts and multiplicities.

• $x=-1$ has multiplicity 2 (touch)
• $x=3$ has multiplicity 1 (cross)

Step 3: y-intercept.

$f(0)=-(1)^2(-3)=3$

So it passes through $(0,3)$.

Step 4: Sign between intercepts.
Test points:

• For $x<-1$, use $x=-2$:

$f(-2)=-(-1)^2(-5)=5>0$

• For $-1<x<3$, use $x=0$: $f(0)=3>0$
• For $x>3$, use $x=4$:

$f(4)=-(5)^2(1)=-25<0$

So the graph is above the x-axis before $3$, touches at $-1$ and stays above, then crosses at $3$ and goes below.

Modeling interpretation: why transformations matter

In context, parameters often mean something real:

• Vertical shift $k$ can represent a baseline (starting height, initial profit, initial population offset).
• Horizontal shift $h$ can represent a delayed start.
• Vertical scale $a$ can represent changing units or intensity.

Exam Focus

• Typical question patterns:
  • Describe how the graph of $f$ changes to become $g(x)=a f(b(x-h))+k$.
  • Given a transformed polynomial in factored form, identify intercepts and end behavior.
  • Use sign analysis to determine where the polynomial is positive/negative.
• Common mistakes:
  • Mixing up horizontal vs vertical changes (especially the inside shift $x-h$).
  • Forgetting that a negative inside factor reflects across the y-axis.
  • Drawing a bounce at an odd multiplicity or a cross at an even multiplicity.

Equivalent Representations, Expansion, and Division Tools

A lot of Unit 1 success comes from switching forms strategically. Standard form, factored form, and quotient-remainder forms each reveal different features.

Standard form vs factored form (polynomials and rationals)

• Standard form is especially useful for end behavior because the leading term dominates.
• Factored form is especially useful for identifying:
  • x-intercepts (zeros)
  • multiplicities (cross vs touch)
  • (for rational functions) vertical asymptotes, holes, and domain restrictions

Binomial Theorem and Pascal’s Triangle

The Binomial Theorem expands powers like $(a+b)^n$:

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$

The coefficients $\binom{n}{k}$ can be found using Pascal’s Triangle. The first several rows are:

• Row 0: 1
• Row 1: 1 1
• Row 2: 1 2 1
• Row 3: 1 3 3 1
• Row 4: 1 4 6 4 1
• Row 5: 1 5 10 10 5 1

This is particularly handy for expanding expressions like $(x+c)^n$.

Polynomial division (why it matters)

If you divide a polynomial $P(x)$ by a polynomial $D(x)$ with $D(x)\neq 0$, you can write:

$P(x)=D(x)Q(x)+R(x)$

where the degree of $R(x)$ is less than the degree of $D(x)$. Polynomial long division is also a core tool for finding slant (oblique) asymptotes of rational functions.

The Remainder Theorem

When you divide $P(x)$ by $x-c$, the remainder is $P(c)$.

The Factor Theorem

$x-c$ is a factor of $P(x)$ if and only if:

$P(c)=0$

Synthetic division

Synthetic division is an efficient way to divide by a linear divisor $x-c$. Conceptually, it computes the quotient coefficients and remainder without writing the full long-division layout.

Worked example: use Remainder Theorem

Problem. Find the remainder when:

$P(x)=2x^3-3x^2+4x-7$

is divided by $x-2$.

By the Remainder Theorem, the remainder is $P(2)$:

$P(2)=2(8)-3(4)+4(2)-7$

$P(2)=16-12+8-7$

$P(2)=5$

So the remainder is $5$.

Worked example: factor using Factor Theorem + division

Problem. Factor:

$P(x)=x^3-4x^2-x+4$

given that $x=4$ is a zero.

Step 1: Confirm the factor.

$P(4)=64-64-4+4=0$

So $x-4$ is a factor.

Step 2: Divide to find the remaining factor.
The quotient is:

$x^2-1$

So:

$P(x)=(x-4)(x^2-1)$

Factor further:

$x^2-1=(x-1)(x+1)$

Final factorization:

$P(x)=(x-4)(x-1)(x+1)$

Exam Focus

• Typical question patterns:
  • Use Pascal’s Triangle or the Binomial Theorem to expand $(x+c)^n$.
  • Use the Remainder Theorem to find a remainder quickly.
  • Use the Factor Theorem to test factors and then factor completely using division.
  • Use long division to rewrite a rational function and identify slant (or higher-degree) end behavior asymptotes.
• Common mistakes:
  • Sign mistakes with factors: mixing up $x+c$ vs $x-c$.
  • Forgetting missing terms when dividing (for example, skipping an $x^2$ term should be treated as coefficient 0).
  • Treating expansion coefficients as exponents (coefficients come from combinations, not from distributing powers incorrectly).

Complex Zeros and Conjugate Pairs (Real-Coefficient Polynomials)

Even though AP Precalculus focuses heavily on real graphs and real intercepts, complex zeros matter because a polynomial may not have enough real zeros to factor fully over the reals.

Conjugate pairs

If a polynomial has real coefficients and $a+bi$ (with $b\neq 0$) is a zero, then its conjugate $a-bi$ is also a zero.

Quadratic factor from a conjugate pair

If $a+bi$ and $a-bi$ are zeros, then the polynomial has a real quadratic factor:

$\bigl(x-(a+bi)\bigr)\bigl(x-(a-bi)\bigr)$

Multiplying simplifies to:

$\bigl(x-a\bigr)^2+b^2$

Worked example: build a real polynomial from a complex zero

Problem. Find a quadratic polynomial with real coefficients that has zero $2+3i$.

Because coefficients are real, $2-3i$ is also a zero. So:

$f(x)=(x-(2+3i))(x-(2-3i))$

Rewrite:

$f(x)=(x-2-3i)(x-2+3i)$

Difference of squares idea:

$f(x)=(x-2)^2-(3i)^2$

Since $i^2=-1$:

$(3i)^2=9i^2=-9$

So:

$f(x)=(x-2)^2+9$

Expand if needed:

$f(x)=x^2-4x+13$

Exam Focus

• Typical question patterns:
  • Given one complex zero of a real-coefficient polynomial, state the other zero.
  • Convert conjugate-pair factors into a real quadratic factor.
  • Reason about the number of real zeros vs total complex zeros (degree).
• Common mistakes:
  • Forgetting the conjugate must also be a zero when coefficients are real.
  • Sign errors when multiplying conjugates.
  • Assuming complex zeros show up as x-intercepts (they do not).

Rational Functions: Structure, Domain, Discontinuities, and Intercepts

A rational function is a ratio of two polynomials:

$f(x)=\frac{P(x)}{Q(x)}$

where $Q(x)\neq 0$.

Domain restrictions

The domain is all real numbers except where the denominator is zero. Find restrictions by solving:

$Q(x)=0$

Discontinuities: holes vs vertical asymptotes

Rational functions can have discontinuities at excluded x-values. The type depends on factor cancellation:

1. Removable discontinuity (hole): occurs when the numerator and denominator share a factor that cancels.
2. Non-removable discontinuity (vertical asymptote): occurs when, after simplifying, the denominator is still zero at that x-value.

Multiplicity viewpoint (useful for factoring quickly):

• If a factor appears more times in the denominator than the numerator, there is a vertical asymptote at that zero (after canceling common factors).
• If a factor appears in both, canceling creates a hole at that x-value. If the factor’s multiplicity is larger in the numerator than in the denominator, the simplified function may still have a zero there; the hole’s y-value is still found from the simplified expression, and the removed point may lie on the x-axis.

How to find holes and vertical asymptotes reliably

1. Factor numerator and denominator completely.
2. Cancel common factors.
3. Canceled x-values are holes.
4. Remaining denominator zeros are vertical asymptotes.

Intercepts

• x-intercepts (real zeros) occur where the numerator is zero and the denominator is not zero (so the x-value is in the domain).
• y-intercept occurs at $x=0$ if defined; compute $f(0)$.

Real zeros of the numerator that are in the domain are the real zeros of the rational function. When solving rational inequalities, real zeros of the numerator and real zeros of the denominator serve as critical points (endpoints or asymptotes) for sign-chart intervals.

Worked example: domain, holes, vertical asymptotes, intercepts

Problem. Analyze:

$f(x)=\frac{(x-2)(x+1)}{(x-2)(x-3)}$

Step 1: Domain restriction from denominator.
Denominator is zero at $x=2$ and $x=3$, so exclude both.

Step 2: Cancel common factors.

$f(x)=\frac{x+1}{x-3}$

But the original function is still undefined at $x=2$.

Step 3: Identify the hole and vertical asymptote.

• Hole at $x=2$
• Vertical asymptote at $x=3$

Step 4: Find intercepts.
x-intercept from $x+1=0$ gives $x=-1$, so $(-1,0)$.

y-intercept:

$f(0)=\frac{1}{-3}=-\frac{1}{3}$

So $(0,-1/3)$.

Step 5: Hole’s y-value.

$y=\frac{2+1}{2-3}=-3$

Hole at $(2,-3)$.

Exam Focus

• Typical question patterns:
  • Identify domain restrictions and classify discontinuities (hole vs vertical asymptote).
  • Determine intercepts and whether an apparent intercept is removed by a hole.
  • Match an algebraic expression to a graph with the correct asymptotes/discontinuities.
• Common mistakes:
  • Declaring a vertical asymptote at a canceled factor (it should be a hole).
  • Finding x-intercepts without checking the denominator isn’t zero there.
  • Computing $f(0)$ when $x=0$ is not in the domain.

Asymptotes and End Behavior of Rational Functions

Asymptotes provide the “skeleton” of many rational graphs. End behavior is largely controlled by the highest-degree (leading) terms.

Vertical asymptotes

A vertical asymptote occurs at $x=c$ if, after simplification, the denominator is zero at $c$ and the function grows without bound near that x-value. One-sided behavior can differ:

• As $x\to c^+$, the function may approach $\infty$ or $-\infty$.
• As $x\to c^-$, it may approach a different infinity.

Sign reasoning near $x=c$ comes from checking signs of numerator and denominator just to the left and right of $c$.

Horizontal asymptotes (degree comparison)

For:

$f(x)=\frac{P(x)}{Q(x)}$

let $n$ be degree of $P$ and $m$ be degree of $Q$.

• If $n<m$, the denominator dominates and:

$y=0$

• If $n=m$, neither dominates; the horizontal asymptote is the ratio of leading coefficients:

$y=\frac{a_n}{b_m}$

• If $n>m$, there is no horizontal asymptote.

This can be understood by examining the quotient of the leading terms (the terms of greatest degree): they dominate for large $\lvert x\rvert$.

Slant (oblique) asymptotes and polynomial asymptotes

If the numerator degree is exactly one more than the denominator degree, the quotient of leading terms is linear, and the rational function has a slant asymptote. Use polynomial division:

$\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}$

where $S(x)$ is linear and $\frac{R(x)}{Q(x)}\to 0$ as $x\to\pm\infty$. The slant asymptote is $y=S(x)$.

More generally, if the numerator degree exceeds the denominator degree by more than 1, division produces a higher-degree polynomial asymptote.

Worked example: horizontal vs slant asymptote

Problem. Find the end behavior asymptote of:

$f(x)=\frac{2x^2+3x-1}{x-2}$

Degrees: numerator 2, denominator 1, so expect a slant asymptote. Division gives:

$\frac{2x^2+3x-1}{x-2}=2x+7+\frac{13}{x-2}$

So the slant asymptote is:

$y=2x+7$

because:

$\frac{13}{x-2}\to 0$

as $x\to\pm\infty$.

Worked example: vertical asymptote side behavior

Problem. Describe the behavior near $x=1$ for:

$f(x)=\frac{x+2}{(x-1)(x+3)}$

Near $x=1$, the factor $x-1$ controls the blow-up.

• As $x\to 1^+$: numerator positive, denominator positive, so:

$f(x)\to \infty$

• As $x\to 1^-$: numerator positive, denominator negative, so:

$f(x)\to -\infty$

Exam Focus

• Typical question patterns:
  • Identify horizontal asymptotes quickly by comparing degrees and leading coefficients.
  • Use division to find a slant (or polynomial) asymptote and interpret end behavior.
  • Determine one-sided behavior near a vertical asymptote using sign reasoning.
• Common mistakes:
  • Assuming every rational function has a horizontal asymptote.
  • Using degree rules before simplifying (canceled factors can change degrees and asymptotes).
  • Forgetting that behavior can differ on the left and right sides of a vertical asymptote.

Graphing Rational Functions: A Feature-by-Feature Approach

Unlike polynomials, rational graphs can break into multiple branches, have holes, and approach asymptotes. A correct sketch comes from assembling required features.

A practical graphing algorithm

For:

$f(x)=\frac{P(x)}{Q(x)}$

1. Factor and simplify (to reveal holes).
2. State domain exclusions.
3. Identify holes (canceled factors) and find the missing y-values from the simplified form.
4. Identify vertical asymptotes (remaining denominator zeros).
5. Find horizontal/slant asymptote (degree comparison or division).
6. Find intercepts (x-intercepts from remaining numerator zeros; y-intercept if defined).
7. Use test points/sign analysis to understand branch behavior.

Worked example: full feature analysis

Problem. Sketch key features of:

$f(x)=\frac{x^2-1}{x^2-4x+3}$

Step 1: Factor.

$x^2-1=(x-1)(x+1)$

$x^2-4x+3=(x-1)(x-3)$

So:

$f(x)=\frac{(x-1)(x+1)}{(x-1)(x-3)}$

Cancel $x-1$:

$f(x)=\frac{x+1}{x-3}$

Step 2: Domain exclusions (from original denominator).

$x\neq 1,\;x\neq 3$

Step 3: Hole at $x=1$ and its y-value.

$y=\frac{1+1}{1-3}=-1$

Hole at $(1,-1)$.

Step 4: Vertical asymptote at $x=3$.

Step 5: Horizontal asymptote.
Degrees are equal after simplification (or cancellation changes the effective degrees), leading coefficients match, so:

$y=1$

Step 6: Intercepts.
x-intercept: $x+1=0$ gives $(-1,0)$.

y-intercept:

$f(0)=\frac{1}{-3}=-\frac{1}{3}$

Step 7: Branch behavior.
Near $x=3$, it behaves like $\frac{x+1}{x-3}$, so it goes to $\infty$ as $x\to 3^+$ and $-\infty$ as $x\to 3^-$. For large $\lvert x\rvert$ it approaches $y=1$.

Can a rational function cross a horizontal asymptote?

Yes. A horizontal asymptote describes end behavior, not a barrier. You can test crossing by solving $f(x)=L$.

Example:

$f(x)=\frac{x}{x+1}$

has horizontal asymptote $y=1$, but it never equals 1 for real $x$ because $x=x+1$ has no solution. Other rational functions do cross their horizontal asymptotes.

Exam Focus

• Typical question patterns:
  • Given a formula, identify all asymptotes, holes, and intercepts, then choose the matching graph.
  • Given a graph, infer a possible rational function structure (factors in numerator/denominator).
  • Decide whether an apparent point is included or removed by a discontinuity.
• Common mistakes:
  • Using the simplified function to claim the hole point is included (it is still excluded).
  • Forgetting to recompute asymptotes after simplifying.
  • Treating horizontal asymptotes as “uncrossable.”

Operations with Rational Expressions and Solving Rational Equations/Inequalities

Rational expressions are algebraic fractions involving polynomials. Strong algebra skills here prevent hidden mistakes in graphing and modeling.

Simplifying rational expressions (factor first)

To simplify:

$\frac{P(x)}{Q(x)}$

1. Factor numerator and denominator.
2. Cancel common factors.
3. State domain restrictions from the original denominator.

Cancellation changes the expression but not the original exclusions.

Adding and subtracting rational expressions

To add:

$\frac{A(x)}{B(x)}+\frac{C(x)}{D(x)}$

use a common denominator, ideally the least common denominator (LCD) formed by taking each distinct denominator factor to the highest power needed.

Multiplying and dividing

• Multiply: factor and cancel before multiplying.
• Divide: multiply by the reciprocal.

Solving rational equations (avoid extraneous solutions)

Steps:

1. Note domain restrictions.
2. Multiply both sides by the LCD.
3. Solve the resulting equation.
4. Check against restrictions to eliminate extraneous solutions.

Worked example: solve a rational equation

Problem. Solve:

$\frac{x}{x-1}+\frac{2}{x+1}=3$

Step 1: Domain restrictions.

$x\neq 1,\;x\neq -1$

Step 2: Multiply by the LCD $(x-1)(x+1)$.

$x(x+1)+2(x-1)=3(x-1)(x+1)$

Step 3: Expand.

$x^2+x+2x-2=3(x^2-1)$

$x^2+3x-2=3x^2-3$

Step 4: Solve.

$0=2x^2-3x-1$

Factor:

$2x^2-3x-1=(2x-1)(x-1)$

So:

$x=\frac{1}{2}$

or

$x=1$

Step 5: Check restrictions.
$x=1$ is not allowed, so the solution is:

$x=\frac{1}{2}$

Rational inequalities (sign analysis)

To solve:

$\frac{P(x)}{Q(x)}>0$

1. Factor numerator and denominator.
2. Identify critical points: zeros of numerator and zeros of denominator.
3. Make a sign chart on the intervals.
4. Apply strict vs non-strict inequality rules.

Denominator zeros are never included in solutions (even for $\geq$), because the expression is undefined there.

Exam Focus

• Typical question patterns:
  • Simplify rational expressions while stating domain restrictions.
  • Solve rational equations and justify rejecting extraneous solutions.
  • Solve rational inequalities using sign charts (zeros and asymptotes as interval boundaries).
• Common mistakes:
  • Canceling terms that are not factors.
  • Forgetting to state/apply domain restrictions after simplifying.
  • Keeping an extraneous solution that makes a denominator zero.

Modeling with Polynomial and Rational Functions: Selecting, Building, and Interpreting Models

Modeling means choosing a function type that matches a situation’s behavior, fitting parameters from conditions or data, and interpreting features (intercepts, asymptotes, extrema) in context.

Choosing an appropriate model type

• Linear functions: contexts with roughly constant rates of change.
• Quadratic functions: contexts with roughly linear rates of change or roughly symmetrical data with a unique maximum/minimum.
• Geometric (area/volume) relationships: two-dimensional growth often relates to quadratics; three-dimensional growth often relates to cubics.
• Polynomial functions: scenarios with multiple zeros or multiple extrema.
• Rational functions: scenarios with division, inverse variation, limiting values, or undefined boundaries. Data sets that are inversely proportional are naturally modeled by rational functions.
• Piecewise functions: used when a scenario has different characteristics over different intervals, with different rules on non-overlapping domain intervals.

Assumptions and restrictions in models

A model may require you to articulate:

• what is being assumed to stay consistent,
• how quantities change together,
• domain restrictions (from math, context clues, or physical limits),
• range restrictions (for example, rounding, or outputs that cannot be negative).

Why polynomials are useful models

Polynomials are good when the relationship is smooth and continuous, with no blow-ups or undefined points. Zeros can represent break-even times or when height is 0; local maxima/minima can represent optimal profit or peak height.

Caution: high-degree polynomials can overfit (match points but behave unrealistically between or outside them). Often the “best” exam model is the simplest degree that matches the given information.

Why rational functions are useful models

Rational models fit contexts involving ratios (average cost, density, rate) and naturally represent:

• vertical asymptotes as boundaries or impossibilities,
• horizontal asymptotes as long-term steady levels.

Building a polynomial model from conditions

Example. A projectile hits the ground at $t=0$ and $t=6$ (height 0), and its height at $t=2$ is 16 meters. Model height with a quadratic.

Use factored form:

$h(t)=a t(t-6)$

Use $(2,16)$:

$16=a\cdot 2\cdot(-4)$

$16=-8a$

$a=-2$

So:

$h(t)=-2t(t-6)$

Building a rational model from asymptotic behavior

Example. A model has vertical asymptote at $x=4$ and horizontal asymptote at $y=2$.

A simple model is:

$f(x)=2+\frac{1}{x-4}$

Equivalent form:

$f(x)=\frac{2(x-4)+1}{x-4}$

How models can be constructed and applied

Models can be constructed:

• from restrictions in a scenario,
• using transformations of parent functions,
• using technology and regressions (linear, quadratic, cubic, quartic),
• by combining methods into a piecewise model.

Once a model is built, use it to draw conclusions, and include appropriate units when interpreting results.

Exam Focus

• Typical question patterns:
  • Select an appropriate model type (linear/quadratic/polynomial/rational/piecewise) and justify the choice.
  • Construct a polynomial model from intercepts and one additional point.
  • Construct a rational model from asymptotes (and possibly one point).
  • Interpret zeros, extrema, and asymptotes in context with correct units.
• Common mistakes:
  • Using a polynomial when the context suggests a limiting value or undefined boundary.
  • Ignoring domain restrictions that represent meaningful constraints.
  • Overfitting by choosing a degree higher than necessary.
  • Giving only algebraic transformation language when the question asks for contextual meaning.