Chapter 3

Polynomial and Rational Functions

Functions defined by polynomial expressions are called polynomial functions. Their graphs can have many peaks and valleys, which makes them excellent models for real-world situations — from a factory's productivity curve (degree 2) to the multi-stage growth of animal populations (degree 3). This chapter studies how to graph, divide, factor, and find the zeros of polynomial and rational functions, and how to solve inequalities built from them.

Instructor Note

This chapter is the backbone for a lot of engineering math. Quadratics show up everywhere in optimization (max/min), and rational functions describe real physical systems like parallel resistances and the Doppler effect. Master Section 3.4 (finding real zeros) early — you'll lean on it constantly when factoring and solving inequalities later.

3.1

Quadratic Functions and Models

A polynomial function of degree $n$ has the form $P(x)=a_n x^n + a_{n-1}x^{n-1}+\cdots+a_1 x + a_0$ with $a_n \neq 0$. Degree 0 and 1 polynomials are lines; this section studies degree 2, the quadratic functions, whose graphs are parabolas.

Quadratic Function

A quadratic function is a polynomial function of degree 2 — a function of the form $f(x)=ax^2+bx+c$ with $a\neq 0$. Every quadratic graphs as a parabola, obtainable from $y=x^2$ by transformations.

Standard Form of a Quadratic Function

\[ f(x)=a(x-h)^2+k \]

Obtained by completing the square. The graph is a parabola with vertex $(h,k)$; it opens upward if $a>0$ and downward if $a<0$.

EXAMPLE 1 Standard Form of a Quadratic Function

Let $f(x)=2x^2-12x+13$. Express $f$ in standard form, then find its vertex and intercepts.

Solution

Factor 2 from the $x$-terms: $f(x)=2(x^2-6x)+13$.

Complete the square (add 9 inside, subtract $2\cdot 9$ outside): $f(x)=2(x^2-6x+9)+13-18$.

Simplify to standard form: $f(x)=2(x-3)^2-5$.

Vertex is $(3,-5)$. The $y$-intercept is $f(0)=13$.

Set $f(x)=0$ and use the Quadratic Formula: $x=\dfrac{6\pm\sqrt{10}}{2}\approx 1.42 \text{ and } 4.58$.

Domain: $(-\infty,\infty)$. Range: $[-5,\infty)$.

Maximum and Minimum Values

If a quadratic has vertex $(h,k)$, the vertex gives a minimum value when the parabola opens upward and a maximum value when it opens downward.

Maximum or Minimum Value of a Quadratic

\[ x=-\frac{b}{2a} \]

The extreme value occurs at $x=-\dfrac{b}{2a}$. If $a>0$, $f\!\left(-\dfrac{b}{2a}\right)$ is the minimum; if $a<0$, it is the maximum.

EXAMPLE 2 Minimum Value of a Quadratic Function

Consider $f(x)=5x^2-30x+49$. Express $f$ in standard form and find its minimum value.

Solution

Factor 5 from the $x$-terms: $f(x)=5(x^2-6x)+49$.

Complete the square: $f(x)=5(x^2-6x+9)+49-45=5(x-3)^2+4$.

Vertex $(3,4)$, opening upward. Since $a>0$, the minimum value is $f(3)=4$.

EXAMPLE 4 Finding Maximum and Minimum Values

Find the extreme value of (a) $f(x)=x^2+4x$ and (b) $g(x)=-2x^2+4x-5$.

Solution

(a) Here $a=1,\ b=4$, so $x=-\dfrac{4}{2(1)}=-2$. Since $a>0$, minimum value $f(-2)=(-2)^2+4(-2)=-4$.

(b) Here $a=-2,\ b=4$, so $x=-\dfrac{4}{2(-2)}=1$. Since $a<0$, maximum value $g(1)=-2(1)^2+4(1)-5=-3$.

Modeling with Quadratic Functions

EXAMPLE 5 Maximum Gas Mileage for a Car

Gas mileage is modeled by $M(s)=-\frac{1}{28}s^2+3s-31$ for $15\le s\le 70$, where $s$ is speed in mi/h. Find the best mileage and the speed at which it occurs.

Solution

With $a=-\frac{1}{28},\ b=3$: $s=-\dfrac{3}{2(-\frac{1}{28})}=42$ mi/h.

Maximum mileage: $M(42)=-\frac{1}{28}(42)^2+3(42)-31=32$ mi/gal.

Best gas mileage is 32 mi/gal, attained at 42 mi/h.

EXAMPLE 6 Maximizing Revenue from Ticket Sales

An arena seats 15,000. At a $14 ticket, attendance is 9500; each $1 price drop adds 1000 attendees. Find the revenue model and the price that maximizes revenue.

Solution

Let $x$ = ticket price. Attendance $=9500+1000(14-x)$.

Revenue $R(x)=x[\,23{,}500-1000x\,]=23{,}500x-1000x^2$.

With $a=-1000,\ b=23{,}500$: $x=-\dfrac{23{,}500}{2(-1000)}=\$11.75$ maximizes revenue.

Setting $R(x)=0$ gives $x=0$ or $x=23.5$; a $23.50 ticket is so high no one attends.

Instructor Note

The revenue example is the classic "express attendance in words, then in algebra" trick. In engineering economics you'll set up profit and cost models the exact same way: write the quantity in plain words first, translate to algebra, then optimize with $x=-b/2a$. Don't skip the word step — it prevents sign errors.

3.2

Polynomial Functions and Their Graphs

Polynomial Function (Terminology)

A polynomial function of degree $n$ is $P(x)=a_n x^n + a_{n-1}x^{n-1}+\cdots+a_1 x + a_0$, where $n$ is a nonnegative integer and $a_n\neq 0$. The $a_i$ are the coefficients; $a_0$ is the constant term; $a_n$ is the leading coefficient, and $a_n x^n$ is the leading term. A single-term polynomial is a monomial.

Smooth and continuous. The graph of every polynomial function is a smooth, continuous curve — no breaks, holes, corners, or cusps.

End Behavior

The end behavior of a polynomial — what happens as $x\to\infty$ or $x\to-\infty$ — is determined entirely by its leading term, because for large $x$ the other terms are insignificant.

End Behavior of Polynomials

For $P(x)=a_n x^n + \cdots + a_0$, behavior depends on degree parity and the sign of $a_n$:

Odd degree, $a_n>0$: $y\to-\infty$ as $x\to-\infty$, $y\to\infty$ as $x\to\infty$.
Odd degree, $a_n<0$: $y\to\infty$ as $x\to-\infty$, $y\to-\infty$ as $x\to\infty$.
Even degree, $a_n>0$: $y\to\infty$ at both ends.
Even degree, $a_n<0$: $y\to-\infty$ at both ends.

EXAMPLE 2 End Behavior of a Polynomial

Determine the end behavior of $P(x)=-2x^4+5x^3+4x+7$.

Solution

Degree 4 (even) with leading coefficient $-2$ (negative).

Therefore $y\to-\infty$ as $x\to-\infty$ and $y\to-\infty$ as $x\to\infty$ — both ends point downward.

Using Zeros to Graph Polynomials

Real Zeros of Polynomials

If $P$ is a polynomial and $c$ a real number, the following are equivalent: (1) $c$ is a zero of $P$; (2) $x=c$ is a solution of $P(x)=0$; (3) $x-c$ is a factor of $P(x)$; (4) $c$ is an $x$-intercept of the graph.

Intermediate Value Theorem for Polynomials

If $P$ is a polynomial and $P(a)$ and $P(b)$ have opposite signs, then there is at least one value $c$ between $a$ and $b$ with $P(c)=0$. Consequently, between successive zeros the graph lies entirely above or entirely below the $x$-axis.

EXAMPLE 5 Finding Zeros and Graphing

Let $P(x)=x^3-2x^2-3x$. Find the zeros and describe the graph.

Solution

Factor out $x$: $P(x)=x(x^2-2x-3)$.

Factor the quadratic: $P(x)=x(x-3)(x+1)$.

Zeros: $x=0,\ 3,\ -1$. The $y$-intercept is $P(0)=0$.

Odd degree, positive leading coefficient: $y\to-\infty$ as $x\to-\infty$, $y\to\infty$ as $x\to\infty$.

EXAMPLE 7 Factoring by Grouping to Find Zeros

Let $P(x)=x^3-2x^2-4x+8$. Find the zeros.

Solution

Group and factor: $P(x)=x^2(x-2)-4(x-2)$.

Factor out $(x-2)$: $P(x)=(x^2-4)(x-2)$.

Difference of squares: $P(x)=(x+2)(x-2)(x-2)=(x+2)(x-2)^2$.

Zeros: $x=-2$ and $x=2$. The graph touches but does not cross at $x=2$ (even-power factor).

Shape of the Graph Near a Zero

Multiplicity of a Zero

If the factor $x-c$ appears exactly $m$ times in the factorization of $P(x)$, then $c$ is a zero of multiplicity $m$. The graph crosses the $x$-axis at $c$ if $m$ is odd, and touches (does not cross) if $m$ is even. Near $c$, the graph resembles $y=A(x-c)^m$.

EXAMPLE 8 Graphing Using Zeros and Multiplicities

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

Solution

Zeros: $0$ (multiplicity 4), $2$ (multiplicity 3), $-1$ (multiplicity 2).

At $x=2$ (odd multiplicity) the graph crosses; at $x=0$ and $x=-1$ (even) it touches and turns back.

Degree $4+3+2=9$ (odd) with positive leading coefficient: $y\to-\infty$ as $x\to-\infty$, $y\to\infty$ as $x\to\infty$.

Local Maxima and Minima

Local Extrema of Polynomials

A polynomial of degree $n$ has at most $n-1$ local extrema (local maxima and minima combined). It may have fewer.

Instructor Note

Multiplicity is one of those ideas students underestimate. "Cross vs. touch" tells you the shape instantly without plugging in dozens of points. When you sketch a transfer function or a load curve by hand, knowing whether each root crosses saves real time. Quick check: count up all multiplicities — they must equal the degree.

3.3

Dividing Polynomials

To factor polynomials algebraically, we first need to divide them. Polynomial division mirrors the familiar long division of numbers.

Division Algorithm

\[ \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)} \quad\text{or}\quad P(x)=D(x)\cdot Q(x)+R(x) \]

For polynomials $P$ (dividend) and $D\neq 0$ (divisor), there exist unique $Q$ (quotient) and $R$ (remainder), where $R=0$ or $\deg R<\deg D$.

Long Division

EXAMPLE 1 Long Division of Polynomials

Divide $6x^2-26x+12$ by $x-4$.

Solution

Divide leading terms: $6x^2/x=6x$. Multiply: $6x(x-4)=6x^2-24x$; subtract to get $-2x+12$.

Divide again: $-2x/x=-2$. Multiply: $-2(x-4)=-2x+8$; subtract to get remainder 4.

Result: $\dfrac{6x^2-26x+12}{x-4}=6x-2+\dfrac{4}{x-4}$, or $6x^2-26x+12=(x-4)(6x-2)+4$.

EXAMPLE 2 Long Division with Missing Terms

For $P(x)=8x^4-6x^2-3x+1$ and $D(x)=2x^2-x-2$, find $Q$ and $R$.

Solution

Insert $0x^3$ so columns align: $8x^4+0x^3-6x^2-3x+1$.

Carry out the division to obtain $Q(x)=4x^2+2x$ and $R(x)=7x+1$.

So $8x^4-6x^2-3x+1=(2x^2-x-2)(4x^2+2x)+(7x+1)$.

Synthetic Division

A shortcut for dividing by a linear divisor of the form $x-c$. Write only the coefficients of the dividend and the value $c$; bring down, multiply by $c$, and add column by column. The last entry is the remainder; the rest form the quotient's coefficients.

EXAMPLE 3 Synthetic Division

Use synthetic division to divide $2x^3-7x^2+5$ by $x-3$.

Solution

Coefficients (insert 0 for missing $x$): $2,\ -7,\ 0,\ 5$; use $c=3$.

Bring down 2; $3\cdot2=6$, add to $-7\Rightarrow -1$; $3\cdot(-1)=-3$, add to $0\Rightarrow -3$; $3\cdot(-3)=-9$, add to $5\Rightarrow -4$.

Quotient $2x^2-x-3$, remainder $-4$: $2x^3-7x^2+5=(x-3)(2x^2-x-3)-4$.

Remainder and Factor Theorems

Remainder Theorem

If $P(x)$ is divided by $x-c$, the remainder equals $P(c)$. So synthetic division by $x-c$ is a fast way to evaluate $P(c)$.

Factor Theorem

\[ c \text{ is a zero of } P \iff (x-c) \text{ is a factor of } P(x) \]

EXAMPLE 5 Factoring Using the Factor Theorem

Let $P(x)=x^3-7x+6$. Show $P(1)=0$ and factor $P$ completely.

Solution

Check: $P(1)=1-7+6=0$, so $x-1$ is a factor.

Divide: $P(x)=(x-1)(x^2+x-6)$.

Factor the quadratic: $P(x)=(x-1)(x-2)(x+3)$.

EXAMPLE 6 Finding a Polynomial with Specified Zeros

Find a degree-4 polynomial with zeros $-3, 0, 1, 5$ and coefficient of $x^3$ equal to $-6$.

Solution

By the Factor Theorem: $P(x)=(x+3)(x)(x-1)(x-5)=x^4-3x^3-13x^2+15x$.

The $x^3$ coefficient is $-3$, but we need $-6$; multiply by 2 (degree unchanged).

$Q(x)=2x^4-6x^3-26x^2+30x$ has all required properties.

3.4

Real Zeros of Polynomials

Finding zeros is the same as factoring into linear factors. This section gives algebraic tools for locating the real zeros of a polynomial.

Rational Zeros Theorem

\[ \text{rational zero}=\frac{p}{q} \]

If $P(x)=a_n x^n+\cdots+a_0$ has integer coefficients ($a_n\neq0,\ a_0\neq0$), then every rational zero $p/q$ (in lowest terms) has $p$ a factor of the constant term $a_0$ and $q$ a factor of the leading coefficient $a_n$.

EXAMPLE 2 Finding Rational Zeros

Write $P(x)=2x^3+x^2-13x+6$ in factored form and find all zeros.

Solution

Possible zeros: factors of 6 over factors of 2 $\Rightarrow \pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}$.

Test by synthetic division: $x=2$ gives remainder 0, so it is a zero.

$P(x)=(x-2)(2x^2+5x-3)=(x-2)(2x-1)(x+3)$.

Zeros: $2,\ \tfrac{1}{2},\ -3$.

EXAMPLE 3 Rational Zeros Theorem with the Quadratic Formula

Find the zeros of $P(x)=x^4-5x^3-5x^2+23x+10$.

Solution

Leading coefficient 1, so candidate zeros are divisors of 10: $\pm1,\pm2,\pm5,\pm10$.

$x=5$ is a zero: $P(x)=(x-5)(x^3-5x-2)$.

For the cubic, $x=-2$ is a zero: $P(x)=(x-5)(x+2)(x^2-2x-1)$.

Quadratic Formula on the last factor: $x=1\pm\sqrt{2}$.

Zeros: $5,\ -2,\ 1+\sqrt2,\ 1-\sqrt2$.

Descartes' Rule of Signs

A variation in sign occurs where adjacent coefficients (in descending order, omitting zeros) have opposite signs. The number of positive real zeros equals the number of sign variations in $P(x)$ or is less by an even whole number. The number of negative real zeros equals the sign variations in $P(-x)$ or less by an even whole number. A zero of multiplicity $m$ is counted $m$ times.

EXAMPLE 4 Using Descartes' Rule

Determine the possible number of positive and negative real zeros of $P(x)=3x^6+4x^5+3x^3-x+3$.

Solution

$P(x)$ has 1 sign variation $\Rightarrow$ exactly 1 positive real zero.

$P(-x)=3x^6-4x^5-3x^3+x+3$ has 3 sign variations $\Rightarrow$ 3 or 1 negative real zeros.

Upper and Lower Bounds

Upper and Lower Bounds Theorem

Divide $P(x)$ by $x-b$ ($b>0$) using synthetic division. If the quotient-and-remainder row has no negative entry, then $b$ is an upper bound for the real zeros. Dividing by $x-a$ ($a<0$): if the row entries alternate nonpositive and nonnegative, then $a$ is a lower bound.

EXAMPLE 5 Bounds for the Zeros of a Polynomial

Show that all real zeros of $P(x)=x^4-3x^2-2x-5$ lie between $-3$ and $2$.

Solution

Synthetic division by $x-2$: all entries nonnegative $\Rightarrow 2$ is an upper bound.

Synthetic division by $x+3$: entries alternate in sign $\Rightarrow -3$ is a lower bound.

Neither is a zero (remainders $\neq0$), so all real zeros lie strictly between $-3$ and $2$.

EXAMPLE 9 Determining the Size of a Fuel Tank

A tank is a 4-ft cylinder capped by two hemispheres (a full sphere). For volume $100\text{ ft}^3$, find the radius $r$.

Solution

Volume equation: $\frac{4}{3}\pi r^3+4\pi r^2=100$.

Only positive $r$ is meaningful; $r=3$ gives over $226\text{ ft}^3$, too large, so the root lies in $(0,3)$.

Graphing $y=\frac{4}{3}\pi x^3+4\pi x^2$ against $y=100$, the intersection is at $r\approx 2.15$ ft.

Instructor Note

The fuel-tank problem is exactly the kind of geometry-to-polynomial setup you'll see in fluid storage and pressure-vessel design. Build the volume from its parts (cylinder + sphere), set equal to the target, and solve. The Upper/Lower Bounds Theorem is your friend for picking a sensible graphing window before you hunt for a root numerically.

3.5

Complex Zeros and the Fundamental Theorem of Algebra

In the complex number system, an $n$th-degree polynomial has exactly $n$ zeros (counting multiplicity) and factors into exactly $n$ linear factors.

Fundamental Theorem of Algebra

Every polynomial $P(x)=a_n x^n+\cdots+a_0$ ($n\ge1,\ a_n\neq0$) with complex coefficients has at least one complex zero. (Proved by Gauss in 1799.)

Complete Factorization Theorem

\[ P(x)=a(x-c_1)(x-c_2)\cdots(x-c_n) \]

Every degree-$n$ polynomial ($n\ge1$) factors into $n$ linear factors over the complex numbers, where $a$ is the leading coefficient and $c_1,\dots,c_n$ are the zeros.

EXAMPLE 2 Factoring a Polynomial Completely

Let $P(x)=x^3-2x+4$. Find all zeros and the complete factorization.

Solution

Candidate rational zeros $\pm1,\pm2,\pm4$; synthetic division shows $x=-2$ is a zero.

$P(x)=(x+2)(x^2-2x+2)$.

Quadratic Formula on $x^2-2x+2$: $x=\dfrac{2\pm\sqrt{-4}}{2}=1\pm i$.

Zeros: $-2,\ 1+i,\ 1-i$. Factorization: $P(x)=(x+2)(x-1-i)(x-1+i)$.

Zeros Theorem

Every polynomial of degree $n\ge1$ has exactly $n$ zeros, provided a zero of multiplicity $k$ is counted $k$ times.

EXAMPLE 3 Factoring with Complex Zeros of Multiplicity

Find the complete factorization and all five zeros of $P(x)=3x^5+24x^3+48x$.

Solution

Factor out $3x$: $P(x)=3x(x^4+8x^2+16)=3x(x^2+4)^2$.

Since $x^2+4=(x-2i)(x+2i)$: $P(x)=3x(x-2i)^2(x+2i)^2$.

Zeros: $0$ (multiplicity 1), $2i$ (multiplicity 2), $-2i$ (multiplicity 2) — five in all.

Conjugate Zeros

Conjugate Zeros Theorem

If $P$ has real coefficients and the complex number $z$ is a zero of $P$, then its conjugate $\bar z$ is also a zero. Complex zeros come in conjugate pairs.

EXAMPLE 6 A Polynomial with a Specified Complex Zero

Find a degree-3 polynomial with integer coefficients and zeros $\tfrac{1}{2}$ and $3-i$.

Solution

Since $3-i$ is a zero, so is $3+i$. The pair multiplies to $(x-3)^2-i^2=x^2-6x+10$.

$P(x)=a\!\left(x-\tfrac12\right)(x^2-6x+10)=a\!\left(x^3-\tfrac{13}{2}x^2+13x-5\right)$.

Set $a=2$ for integer coefficients: $P(x)=2x^3-13x^2+26x-10$.

Linear and Quadratic Factors

Linear and Quadratic Factors Theorem

Every polynomial with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients. (An irreducible quadratic has no real zeros.)

EXAMPLE 7 Linear and Quadratic Factors

For $P(x)=x^4-2x^2-8$: (a) factor into real linear and irreducible quadratic factors; (b) factor completely over the complex numbers.

Solution

(a) $P(x)=(x^2-4)(x^2+2)=(x-2)(x+2)(x^2+2)$; the factor $x^2+2$ is irreducible over the reals.

(b) Factor the irreducible quadratic: $x^2+2=(x-i\sqrt2)(x+i\sqrt2)$.

Complete factorization: $P(x)=(x-2)(x+2)(x-i\sqrt2)(x+i\sqrt2)$.

Why it matters: A polynomial with real coefficients and odd degree must have at least one real zero — because its non-real zeros come in conjugate pairs, leaving an odd one over.

3.6

Rational Functions

Rational Function

A rational function has the form $r(x)=\dfrac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials with no common factor. Its domain is all real numbers except where $Q(x)=0$.

Vertical and Horizontal Asymptotes

The line $x=a$ is a vertical asymptote if $y\to\pm\infty$ as $x\to a^+$ or $x\to a^-$. The line $y=b$ is a horizontal asymptote if $y\to b$ as $x\to\pm\infty$. An asymptote is a line the graph approaches ever more closely.

EXAMPLE 1 A Simple Rational Function

Graph $f(x)=\dfrac{1}{x}$ and state domain and range.

Solution

As $x\to0^-$, $f(x)\to-\infty$; as $x\to0^+$, $f(x)\to+\infty$. So $x=0$ is a vertical asymptote.

As $x\to\pm\infty$, $f(x)\to0$. So $y=0$ is a horizontal asymptote.

Domain $\{x\mid x\neq0\}$; range $\{y\mid y\neq0\}$.

Finding Asymptotes of Rational Functions

For $r(x)=\dfrac{a_n x^n+\cdots}{b_m x^m+\cdots}$:

Vertical asymptotes occur at the zeros of the denominator.
If \(n
If $n=m$: horizontal asymptote $y=\dfrac{a_n}{b_m}$.
If $n>m$: no horizontal asymptote.

EXAMPLE 4 Asymptotes of a Rational Function

Find the vertical and horizontal asymptotes of $r(x)=\dfrac{3x^2-2x-1}{2x^2+3x-2}$.

Solution

Factor the denominator: $(2x-1)(x+2)$. Vertical asymptotes: $x=\tfrac12$ and $x=-2$.

Degrees equal (both 2), so horizontal asymptote $y=\dfrac{3}{2}$.

Sketching Rational Functions

Guidelines for Graphing Rational Functions

Factor numerator and denominator.
Find intercepts: zeros of numerator ($x$-intercepts) and $r(0)$ ($y$-intercept).
Find vertical asymptotes and test the sign on each side.
Find the horizontal asymptote (if any).
Plot points and sketch a smooth curve respecting the asymptotes.

EXAMPLE 5 Graphing a Rational Function

Graph $r(x)=\dfrac{2x^2+7x-4}{x^2+x-2}$.

Solution

Factor: $r(x)=\dfrac{(2x-1)(x+4)}{(x-1)(x+2)}$.

$x$-intercepts $\tfrac12$ and $-4$; $y$-intercept $r(0)=2$.

Vertical asymptotes $x=1$ and $x=-2$; horizontal asymptote $y=2$ (equal degrees).

Common Factors (Holes)

Holes. If numerator and denominator share a factor, cancel it — but only where that factor isn't zero. The graph then matches the simpler function but has a hole at the cancelled value. Example: $s(x)=\dfrac{x-3}{x^2-3x}=\dfrac{1}{x}$ for $x\neq3$, so it looks like $1/x$ with a hole at $x=3$.

Slant Asymptotes and End Behavior

Slant (Oblique) Asymptote

If the numerator's degree is exactly one more than the denominator's, division gives $r(x)=ax+b+\dfrac{R(x)}{Q(x)}$. Since $R(x)/Q(x)\to0$ for large $|x|$, the line $y=ax+b$ is a slant asymptote.

EXAMPLE 9 A Rational Function with a Slant Asymptote

Graph $r(x)=\dfrac{x^2-4x-5}{x-3}$.

Solution

Factor: $r(x)=\dfrac{(x+1)(x-5)}{x-3}$; $x$-intercepts $-1,5$; $y$-intercept $\tfrac{5}{3}$.

Vertical asymptote $x=3$. Numerator degree exceeds denominator by 1, so a slant asymptote exists.

Long division: $r(x)=x-1+\dfrac{-8}{x-3}$, so the slant asymptote is $y=x-1$.

EXAMPLE 11 Electrical Resistance (Application)

A fixed 8-ohm resistor is in parallel with a variable resistor $x$. The combined resistance is $R(x)=\dfrac{8x}{8+x}$. Interpret the graph for $x>0$.

Solution

For $x>0$ there is no vertical asymptote; $R$ increases as $x$ increases.

Horizontal asymptote $R=8$: no matter how large $x$ gets, combined resistance never exceeds 8 ohms.

Instructor Note

Example 11 is pure engineering payoff. The horizontal asymptote $R=8$ tells you a design limit physically: adding parallel resistance only pushes you toward, never past, the fixed value. Whenever you see a rational model in a system (resistance, flow, concentration), ask "what does the horizontal asymptote mean in the real world?" — that's usually the steady-state or saturation value.

3.7

Polynomial and Rational Inequalities

Core idea: A polynomial does not change sign between successive zeros. So between consecutive zeros, its graph lies entirely above or entirely below the $x$-axis — which lets us solve inequalities with test points.

Solving Polynomial Inequalities

Move all terms to one side.
Factor the polynomial into irreducible factors; find the real zeros.
List the intervals determined by the zeros.
Make a sign table using test values in each interval.
Read off the solution; check endpoints if the inequality is $\le$ or $\ge$.

EXAMPLE 1 Solving a Polynomial Inequality

Solve $2x^3+x^2+6\ge 13x$.

Solution

Move terms: $2x^3+x^2-13x+6\ge0$.

Factor: $(x-2)(2x-1)(x+3)\ge0$; zeros $-3,\tfrac12,2$.

Sign table over $(-\infty,-3),(-3,\tfrac12),(\tfrac12,2),(2,\infty)$: expression $\ge0$ on $[-3,\tfrac12]$ and $[2,\infty)$.

Solution: $\left[-3,\tfrac12\right]\cup[2,\infty)$.

EXAMPLE 2 Inequality with an Irreducible Factor

Solve $3x^4+x^2+4\le 2x^3+12x$.

Solution

Move terms: $3x^4-2x^3+x^2-12x+4\le0$.

Factor: $(x-2)(3x-1)(x^2+x+2)\le0$; the quadratic factor has no real zeros (always positive).

Real zeros from the first two factors: $\tfrac13$ and $2$.

Sign analysis gives solution $\left(\tfrac13,2\right)$ (endpoints excluded since the value is 0 there, but the strict portion satisfies $\le$; checking shows the open interval).

Rational Inequalities

Cut Points

For $r(x)=\dfrac{P(x)}{Q(x)}$, the cut points are the zeros of the numerator and the zeros of the denominator. A rational function can change sign only at its cut points, so we test intervals between successive cut points.

Solving Rational Inequalities

Move all terms to one side; combine over a common denominator.
Factor numerator and denominator; find the cut points.
List intervals determined by the cut points.
Build a sign table with test values.
Read off the solution; never include values that make the denominator 0.

EXAMPLE 3 Solving a Rational Inequality

Solve $\dfrac{1-2x}{x^2-2x-3}\le 1$.

Solution

Move and combine: $\dfrac{(1-2x)-(x^2-2x-3)}{x^2-2x-3}\le0 \Rightarrow \dfrac{4-x^2}{x^2-2x-3}\le0$.

Factor: $\dfrac{(2-x)(2+x)}{(x-3)(x+1)}\le0$; cut points $-2,-1,2,3$.

Sign table over the five intervals shows the expression $\le0$ on $[-2,-1)$ and $[2,3)$.

Solution: $[-2,-1)\cup[2,3)$ — excluding $-1$ and $3$ (denominator zero).

EXAMPLE 5 Rational Inequality (Application: Light Intensity)

Two light sources are 10 m apart, one three times as intense. Intensity at distance $x$ from the weaker source is $L(x)=\dfrac{10}{x^2}+\dfrac{30}{(10-x)^2}$. Find where intensity is $\le 4$ lux.

Solution

Solve $\dfrac{10}{x^2}+\dfrac{30}{(10-x)^2}\le4$ for \(0

Graph $y_1=L(x)$ and $y_2=4$; they intersect at $x\approx1.67$ and $x\approx7.19$.

Between these points $y_1\le y_2$, so the intensity is $\le4$ lux on $(1.67,7.19)$ meters.

Instructor Note

For rational inequalities, the single biggest error I see is treating denominator zeros like numerator zeros. They're both cut points where the sign can flip, but a denominator zero is never part of the solution set — the function isn't even defined there. Always use brackets vs. parentheses deliberately. The light-intensity example also shows the graphical method, which is often faster for messy expressions you can't factor cleanly.