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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).

Chapter 7 · pp. 169–258

Chapter 3: Polynomial and Rational Functions

This chapter studies two big families of functions. Polynomials are built from power functions, and rational functions are ratios of polynomials. We look at their graphs and long run behavior, find their zeros with factoring, division, and complex numbers, and study quadratics, asymptotes, inverses, and radical functions. This part begins with power functions and the basic vocabulary of polynomials.

Power Functions

A power function has the form $f(x)=x^p$ . The base is the variable $x$ , and the exponent $p$ is a fixed number. (p. 169)

Power functions appear in geometry. A square with side $L$ has area $A(L)=L^2$ . A cube with side $L$ has volume $V(L)=L^3$ . Both are power functions. The exponent can be a whole number, a negative number, or a fraction. (p. 169)

Most toolkit functions are power functions once you write them with an exponent:

constant: $f(x)=x^0$
identity: $f(x)=x^1$
quadratic and cubic: $x^2$ and $x^3$
reciprocal and reciprocal squared: $x^{-1}$ and $x^{-2}$
square root and cube root: $x^{1/2}$ and $x^{1/3}$

The absolute value function is not a power function. (pp. 169–170)

These graphs all pass through $(1,1)$ , because $1$ raised to any power is $1$ . The ones with positive exponents also pass through $(0,0)$ . (p. 170)

Long Run Behavior of Power Functions

Long run behavior describes what the outputs do when the input gets very large and negative ( $x\to-\infty$ ) or very large and positive ( $x\to\infty$ ). The symbol $\infty$ is not a number. It means a quantity grows without bound. So " $x\to\infty$ " says $x$ is getting larger and larger in the positive direction. (pp. 170–171)

Even whole-number exponents. Graphs of $x^2$ , $x^4$ , $x^6$ , and $x^8$ all look like the quadratic. Both ends point up. Squaring or raising to any even power makes a negative input positive. For $f(x)=x^8$ :

$\text{as } x\to\infty,\ f(x)\to\infty \quad\text{and}\quad \text{as } x\to-\infty,\ f(x)\to\infty.$ (pp. 170–171)

Odd whole-number exponents. Graphs of $x^3$ , $x^5$ , and $x^7$ look like the cubic. They fall on the left and rise on the right. An odd power keeps the sign of the input, so a large negative input gives a large negative output:

$\text{as } x\to-\infty,\ f(x)\to-\infty \quad\text{and}\quad \text{as } x\to\infty,\ f(x)\to\infty.$ (p. 170)

Higher exponents. Within each family, a larger exponent makes the graph flatter near the origin and steeper away from it. Compare $x^2$ , $x^4$ , and $x^6$ . When $0<x<1$ , a higher power gives a smaller output, so the graph hugs the horizontal axis. When $x>1$ , a higher power gives a larger output, so the graph climbs faster. (p. 170)

Negative coefficients. A negative coefficient reflects the graph vertically, which reverses the long run behavior.

For $f(x)=-x^9$ , an odd power that is reflected, we get $f(x)\to-\infty$ as $x\to\infty$ and $f(x)\to\infty$ as $x\to-\infty$ .
For $f(x)=-x^4$ , an even power that is reflected, we get $f(x)\to-\infty$ at both ends.

(p. 171)

Polynomials

Polynomials come from combining power functions. Suppose an oil slick is circular. Its radius starts at 24 miles and grows by 8 miles each week, so $r(w)=24+8w$ . The area of a circle is $A(r)=\pi r^2$ . To get the area in terms of weeks, compose the functions by substituting the radius formula into the area formula, then multiply out:

$A(w)=\pi(24+8w)^2=\pi\left(576+384w+64w^2\right).$

The result is a sum of stretched power functions with whole-number exponents. That is what a polynomial is. (pp. 171–172)

A polynomial has the form

$f(x)=a_nx^n+\dots+a_2x^2+a_1x+a_0.$

The numbers $a_i$ are the coefficients. They may be positive, negative, zero, whole numbers, decimals, or fractions. The exponents must be non-negative whole numbers. So $3x^2-4x^3+2$ is a polynomial, but $x^{-1}+2$ is not. (p. 172)

Terminology: Terms, Degree, and Leading Term

A term is one piece of the sum, $a_ix^i$ . Each term is a transformed power function.
The degree is the highest power of the variable.
The leading term is the term with that highest power.
The leading coefficient is the coefficient of the leading term.

(pp. 172–173)

Polynomials are usually rewritten with powers in descending order, so the leading term comes first. For $f(x)=3+2x^2-4x^3$ , rewrite it as $-4x^3+2x^2+3$ . The degree is 3, the leading term is $-4x^3$ , and the leading coefficient is $-4$ . As another example, $g(t)=5t^5-2t^3+7t$ has degree 5, leading term $5t^5$ , and leading coefficient 5. (p. 173)

Long Run Behavior of Polynomials

For very large inputs, the highest power outgrows all the others. So the long run behavior of any polynomial matches the long run behavior of its leading term. A polynomial that begins $0.2x^3+\dots$ acts like $x^3$ at the ends. It goes to $\infty$ as $x\to\infty$ and to $-\infty$ as $x\to-\infty$ . (p. 173)

This also works in reverse. You can read the degree's parity and the sign of the leading coefficient from the ends of a graph:

If the two ends point in opposite directions, the degree is odd. If they point the same way, the degree is even.
If the right end rises, the leading coefficient is positive. If it falls, the leading coefficient is negative.

For example, suppose a graph falls on the left and rises on the right. It has the shape of an unreflected odd power, so its degree is odd and its leading coefficient is positive. In symbols, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to\infty$ as $x\to\infty$ . (p. 173)

Short run behavior and turning points

Long run behavior describes the far ends of a graph. Short run behavior describes its local features: the intercepts and the turning points, where the graph changes direction. (p. 174)

A polynomial is a function, so it has exactly one vertical intercept. It is found by setting the input to zero, which gives $(0,a_0)$ . Horizontal intercepts happen where the output is zero. They are also called zeros or roots, and there can be several. (p. 174)

Example: for $f(x)=(x-2)(x+1)(x-4)$ , we get $f(0)=(-2)(1)(-4)=8$ , so the vertical intercept is $(0,8)$ . Setting each factor to zero gives zeros at $2$ , $-1$ and $4$ . (p. 174)

A polynomial of degree $n$ has at most $n$ horizontal intercepts and at most $n-1$ turning points. If $n$ is odd, there is at least one horizontal intercept. The cubic above has 3 intercepts and 2 turning points. (pp. 174–175)

This works backward too. Suppose a graph rises on both ends, has 2 horizontal intercepts and 3 turning points. Rising on both ends means even degree. Three turning points means degree at least 4. So the degree is probably 4. (p. 175)

Quadratic functions and their forms

A quadratic function is a degree 2 polynomial. Its graph is a parabola, a transformation of $f(x)=x^2$ . Quadratics come up in area and projectile problems. (p. 177)

Example: a garden uses 80 ft of fence on three sides, with the fourth side along a wall. Let $L$ be each side perpendicular to the wall. The side parallel to the wall is $W=80-2L$ . The area is $A(L)=L(80-2L)=80L-2L^2$ . (p. 177)

A quadratic can be written in two forms. (p. 179)

Standard form: $f(x)=ax^2+bx+c$
Transformation (vertex) form: $f(x)=a(x-h)^2+k$ , with vertex $(h,k)$

In vertex form, $h$ shifts the parabola horizontally, $k$ shifts it vertically, and $a$ stretches or compresses it. Expanding the vertex form gives standard form. For example, $\tfrac12(x+2)^2-3=\tfrac12x^2+2x-1$ . (pp. 178–179)

Common mistake: thinking that stretching or compressing moves the vertex. It does not. Only $h$ and $k$ place the vertex. The factor $a$ changes the width and whether the parabola opens up or down. (p. 178)

Writing a quadratic from a graph. Read the vertex to get $h$ and $k$ . Then plug in one other point to solve for $a$ . If the vertex is $(-2,-3)$ , then $g(x)=a(x+2)^2-3$ . The point $(0,-1)$ gives $-1=4a-3$ , so $a=\tfrac12$ . (p. 178)

Finding the vertex

Expand $a(x-h)^2+k$ and match it to $ax^2+bx+c$ . The $x$ -terms give $-2ah=b$ , so

$h=-\frac{b}{2a},\qquad k=f(h).$

It is easier to find $k$ by evaluating the function at $h$ than by using a second formula. (pp. 179–180)

Example: $f(x)=2x^2-6x+7$ . Then $h=\tfrac{6}{4}=\tfrac32$ and $k=f(\tfrac32)=\tfrac92-9+7=\tfrac52$ . So $f(x)=2(x-\tfrac32)^2+\tfrac52$ . (p. 180)

Completing the square is another way to reach vertex form. Take $2x^2-12x+14$ . (pp. 180–181)

Factor the leading coefficient out of the $x^2$ and $x$ terms: $2(x^2-6x)+14$ .
Add the square of half the middle coefficient inside the parentheses: $(-3)^2=9$ .
The 2 outside multiplies that 9, so you actually added $18$ . Subtract $18$ to balance.
Result: $2(x-3)^2-4$ .

Maximum and minimum values

The vertex is the lowest point if $a>0$ (opens up) and the highest point if $a<0$ (opens down). For $g(x)=-x^2+8x+7$ , $a=-1$ , so the vertex is a maximum. (p. 181)

Maximizing area. In $A(L)=80L-2L^2$ , $a=-2$ , $b=80$ , $c=0$ . The vertex is at $L=-\frac{80}{2(-2)}=20$ ft. The third side is $80-2(20)=40$ ft. The maximum area is $A(20)=800$ square feet. (p. 181)

Maximizing revenue. Revenue is price times quantity. A newspaper has 84,000 subscribers at $\$ 30 $. Raising the price to$ $32$ loses 5,000 of them. (p. 182)

Find the linear demand. The slope is $\frac{79{,}000-84{,}000}{32-30}=-2500$ . Then $S=-2500C+159{,}000$ .
Multiply by price: $R=C\cdot S=-2500C^2+159{,}000C$ .
Take the vertex: $C=\frac{159{,}000}{5000}=31.80$ .
The maximum revenue is $\$ 2{,}528{,}100$.

Intercepts of a quadratic

A quadratic has zero, one or two horizontal intercepts, depending on where the parabola sits relative to the axis. The constant $c$ is the vertical intercept $(0,c)$ . (p. 183)

Example: $f(x)=3x^2+5x-2$ has $f(0)=-2$ . Factoring gives $(3x-1)(x+2)=0$ , so the intercepts are $x=\tfrac13$ and $x=-2$ . (p. 183)

If the quadratic does not factor, use vertex form or the quadratic formula. Solving $a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}=0$ once in general gives

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

For $2x^2+4x-4=0$ , this gives $x=\frac{-4\pm\sqrt{48}}{4}=-1\pm\sqrt3$ . (pp. 183–185)

Projectile problems. A ball thrown up from a 40-ft building has height $H(t)=-16t^2+80t+40$ . (p. 185)

The peak time is the vertex: $t=\frac{80}{32}=2.5$ s. The maximum height is $H(2.5)=140$ ft.
To find the landing time, solve $H(t)=0$ with the quadratic formula. This gives $t\approx5.458$ or $t\approx-0.458$ .
A negative time is outside the model's domain, so the ball lands after about 5.458 s.

Zeros of higher-degree polynomials

The vertical intercept of any polynomial is easy: evaluate at $0$ . Horizontal intercepts are harder. Cubics and quartics have formulas that are too long to remember, and higher degrees have no general formula. So we use one of three approaches: factor with known methods (greatest common factor, trinomials), read the zeros from a factored form, or use technology. (p. 191)

When a polynomial has only even powers, factor out the greatest common factor. Then treat what remains as a quadratic in $x^2$ . For $f(x)=x^6-3x^4+2x^2$ :

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

Setting each factor to zero gives $x=0,\pm1,\pm\sqrt2$ , which is 5 horizontal intercepts. (p. 191)

A factored form like $g(t)=(t-2)^2(2t+3)$ gives zeros $t=2$ and $t=-\tfrac32$ directly. Also $g(0)=12$ . (p. 192)

If nothing factors, use a graph. For $h(t)=t^3+4t^2-t-6$ , a graph suggests zeros at $-3$ , $-2$ and $1$ . Check by substituting that each gives $0$ . (p. 192)

Multiplicity and behavior at intercepts

The power on a factor $(x-h)^p$ controls how the graph behaves at $h$ . The number $p$ is the zero's multiplicity. (pp. 192–193)

$p=1$ (single zero): the graph passes straight through, like a line.
$p=2$ (double zero): the graph touches the axis and bounces back, like a parabola.
$p=3$ (triple zero): the graph crosses with a flattened "S" shape, like $x^3$ .

For $f(x)=(x+3)(x-2)^2(x+1)^3$ , the graph crosses at $-3$ , bounces at $2$ , and flattens while crossing at $-1$ . (pp. 192–193)

Higher powers follow the same pattern. Even powers ( $4,6,8,\dots$ ) still bounce, and odd powers ( $5,7,9,\dots$ ) still cross. In both cases the graph gets flatter near the intercept as the power grows. Near $h$ , a factor $(x-h)^4$ looks like $x^4$ , which is flatter than $x^2$ . (p. 193)

Sketching an example. Take $f(x)=-2(x+3)^2(x-5)$ . (p. 184)

The leading term is $-2x^3$ , so the graph rises on the far left and falls on the far right.
At $x=-3$ the factor is squared, so the graph bounces there.
At $x=5$ the factor is single, so the graph crosses there.
$f(0)=-2(9)(-5)=90$ , so the graph crosses the vertical axis at $(0,90)$ .

Put together, the graph rises from the left, touches the axis at $(-3,0)$ , climbs through $(0,90)$ , turns, and falls through $(5,0)$ . (p. 184)

Sketching a factored polynomial

A polynomial in factored form gives you nearly everything you need for a sketch. Work in three steps. First, use the leading term to get the long run behavior. Second, find the intercepts and the multiplicity of each one. Third, connect the pieces.

Take $f(x)=-2(x+3)^2(x-5)$ . If you multiplied it out, the leading term would be $-2x^3$ . That is a reflected cubic, so the graph starts high on the left and ends low on the right. The factor $(x+3)^2$ is squared, so the graph bounces at $x=-3$ . The factor $(x-5)$ appears once, so the graph crosses there. The vertical intercept is $f(0)=-2(9)(-5)=90$ . The path is: come down from the upper left, touch $(-3,0)$ and turn back up, cross the $y$ -axis at $(0,90)$ , turn down, cross at $(5,0)$ , and keep falling. (p. 194)

Polynomial inequalities and square root domains

A polynomial can change sign only at a zero. So to solve an inequality, find the zeros, split the number line into intervals, and test one value in each interval. You can also read the signs from a sketch.

Solve $(x+3)(x+1)^2(x-4)\ge 0$ . The zeros are $-3$ , $-1$ and $4$ , which make four intervals. Test values give:

$f(-4)=72>0$
$f(-2)=-6<0$
$f(0)=-12<0$
$f(5)=288>0$

The function is positive for $x<-3$ and $x>4$ . Because the inequality uses " $\ge$ ", the zeros count too. That includes the bounce point $x=-1$ , which is a single isolated solution. (p. 195)

This idea gives the domain of a square root function. The expression under the root must be at least $0$ . For $v(t)=\sqrt{6-t-t^2}$ , solve $6-t-t^2\ge0$ . It factors as $-(t+3)(t-2)\ge 0$ , with zeros $t=-3$ and $t=2$ . The parabola opens down, so it is nonnegative between the zeros. The domain is $[-3,2]$ . Check with $t=0$ , which gives $\sqrt6$ , a real number. (p. 196)

Writing a polynomial from its intercepts

Each horizontal intercept gives a factor. Follow these steps:

List the zeros.
Choose each multiplicity from the graph: crossing means $1$ , bouncing means $2$ .
Write $f(x)=a(x-x_1)^{p_1}\cdots(x-x_k)^{p_k}$ .
Use one more point to find $a$ .

The degree equals the sum of the multiplicities. Suppose a graph crosses at $-3$ and $5$ , bounces at $2$ , and passes through $(0,-2)$ . Then $f(x)=a(x+3)(x-2)^2(x-5)$ . Substituting gives $-2=a(3)(4)(-5)=-60a$ , so $a=\tfrac1{30}$ . (pp. 196–197)

Estimating maximum and minimum values

For quadratics the vertex gives the extreme value. For higher-degree polynomials, turning points cannot be found by algebra at this level, so we estimate them with technology. (p. 198)

Example: cut squares of side $w$ from the corners of a $14$ by $20$ cm sheet and fold up the sides. The base is $(14-2w)$ by $(20-2w)$ and the height is $w$ , so

$V(w)=(14-2w)(20-2w)w=4w^3-68w^2+280w.$

Only $0<w<7$ makes sense, since the short side must stay positive. Graphing on that interval and zooming in shows a maximum volume of about $339$ cm $^3$ at $w\approx2.7$ cm. (pp. 198–199)

Polynomial long division

Long division of polynomials copies the steps of numeric long division. Divide the leading term of the dividend by the leading term of the divisor. Multiply the result by the whole divisor, subtract, and repeat. Stop when what is left has lower degree than the divisor.

To divide $x^3+4x^2-5x-14$ by $x-2$ , the three quotient terms are $x^2$ , then $6x$ , then $7$ . The remainder is $0$ . So the quotient is $x^2+6x+7$ and

$x^3+4x^2-5x-14=(x-2)(x^2+6x+7).$

The zeros of the quadratic factor come from the quadratic formula: $x=-3\pm\sqrt2$ . With the zero at $x=2$ , this gives all three horizontal intercepts. (pp. 204–205)

In general, dividing $p(x)$ by $d(x)$ gives

$p(x)=d(x)q(x)+r(x),$

where $r$ is zero or has lower degree than $d$ . To check a division, multiply the quotient by the divisor and add the remainder. If the divisor is $x-c$ , the remainder is a constant. (p. 206)

The Remainder and Factor Theorems

Remainder Theorem. Dividing $p(x)$ by $x-c$ leaves remainder $p(c)$ . The reason is that $p(x)=(x-c)q(x)+r$ . Putting in $x=c$ kills the first term, so $p(c)=r$ . (p. 206)

Factor Theorem. The number $c$ is a zero of a nonzero polynomial $p(x)$ exactly when $x-c$ is a factor of $p(x)$ . This follows because $p(c)=0$ means the remainder is $0$ . In the example above, $c=2$ leaves remainder $0$ , so $x-2$ is a factor. (p. 206)

Synthetic division

Synthetic division is a shortcut for dividing by $x-c$ . It uses only coefficients. Write $c$ , then the coefficients of the dividend. Bring down the first coefficient. Then repeat: multiply by $c$ , and add the product to the next coefficient.

For $x^3+4x^2-5x-14$ and $c=2$ , start with $1,4,-5,-14$ . The bottom row is $1,6,7,0$ . The first three numbers are the quotient $x^2+6x+7$ , and the last number is the remainder $0$ . The quotient always has degree one less than the dividend. (pp. 207–209)

Three rules keep the method correct:

It works only for divisors of the form $x-c$ .
A missing power needs a $0$ coefficient.
A divisor $x+k$ uses $c=-k$ .

For $5x^3-2x^2+1$ divided by $x-3$ , use $5,-2,0,1$ . The result is quotient $5x^2+13x+39$ and remainder $118$ , so $x-3$ is not a factor. For $x^3+8$ divided by $x+2$ , use $c=-2$ and coefficients $1,0,0,8$ . The result is $x^2-2x+4$ with remainder $0$ . (pp. 209–210)

Finding the other zeros once one is known

Divide by the known factor, then solve what is left. For a repeated root, divide again. For $4x^4-4x^3-11x^2+12x-3$ with a double zero at $x=\tfrac12$ , two synthetic divisions by $x-\tfrac12$ leave $4x^2-12$ . Setting $4x^2-12=0$ gives $x=\pm\sqrt3$ . (p. 210)

Where to look: Cauchy's Bound and the Rational Roots Theorem

Cauchy's Bound. Let $M$ be the largest absolute value among the coefficients of $f(x)=a_nx^n+\dots+a_0$ . Then every real zero lies in

$\left[-\frac{M}{|a_n|}-1,\ \frac{M}{|a_n|}+1\right].$

For $f(x)=2x^4+4x^3-x^2-6x-3$ , we have $M=6$ and $|a_n|=2$ . So the interval is $[-4,4]$ . Use this as your graphing window and no real zero will be off-screen. (p. 213)

Rational Roots Theorem. If an integer-coefficient polynomial has a rational zero, it has the form $\pm\frac pq$ . Here $p$ divides the constant term and $q$ divides the leading coefficient. For the polynomial above, $p\in\{1,3\}$ and $q\in\{1,2\}$ . The candidates are $\pm1,\pm\frac12,\pm3,\pm\frac32$ . (p. 214)

If every candidate fails, the polynomial either has no real zeros or only irrational ones. A zero like $\sqrt2$ can never appear on the list. (p. 214)

A full procedure for real zeros

Use Cauchy's Bound to set the window.
List the rational candidates.
Use the graph to pick promising ones.
Test them with synthetic division. Each success lowers the degree.
When a quadratic remains, use the quadratic formula.

(pp. 214–216)

Example 1. For $f(x)=2x^4+4x^3-x^2-6x-3$ , the graph suggests $-1$ . Dividing by $x+1$ works twice and leaves $2x^2-3$ . So the other zeros are $\pm\frac{\sqrt6}{2}\approx\pm1.22$ , and $x=-1$ has multiplicity $2$ . (p. 215)

The first calculator picture made the crossing at $-1$ look flat, as if the multiplicity were $3$ . That is a misleading guess, and only the algebra settles multiplicity. A degree 4 polynomial has four zeros counting multiplicity. The double zero at $-1$ plus $\pm\frac{\sqrt6}{2}$ already uses all four. The graph really has a relative maximum at $x=-1$ , which a smaller window hides. (pp. 215–216)

Example 2. For $f(x)=4x^3-10x^2-2x+2$ , Cauchy's Bound gives $\frac{10}{4}+1=3.5$ , so the window is $[-3.5,3.5]$ . The candidates are $\pm1,\pm\frac12,\pm\frac14,\pm2$ . Testing $\frac12$ leaves remainder $-1$ , so it fails. Testing $-\frac12$ leaves remainder $0$ and quotient $4x^2-12x+4=4(x^2-3x+1)$ . The quadratic formula gives $x=\frac{3\pm\sqrt5}{2}\approx2.618,\ 0.382$ . So the only rational zero is $-\frac12$ , and the other two are irrational. (pp. 216–217)

The Imaginary Unit and Complex Numbers

Some equations, such as $x^2=-1$ , have no real solutions. To fix this, we define a new number $i=\sqrt{-1}$ , so $i^2=-1$ . Any real multiple of $i$ , such as $5i$ or $-\tfrac{2}{3}i$ , is called an imaginary number. (p. 220)

The unit $i$ also lets us simplify square roots of negative numbers. Split the root into a positive part and $\sqrt{-1}$ . For example, $\sqrt{-25}=\sqrt{25}\cdot\sqrt{-1}=5i$ , and $\sqrt{-9}=3i$ . (p. 220)

A complex number adds a real number and an imaginary number. It has the form $z=a+bi$ , where $a$ and $b$ are real. We call $a$ the real part and $b$ the imaginary part. Note that the imaginary part is the number $b$ , not $bi$ . For $3-4i$ , the real part is $3$ and the imaginary part is $-4$ . Every real number is also complex, because it is $a+0i$ . (p. 220)

Arithmetic with Complex Numbers

Adding and subtracting. Treat $i$ like a variable and combine like terms. Real parts go with real parts, and imaginary parts go with imaginary parts. (pp. 220–221)

$(3-4i)+(2+5i)=(3+2)+(-4+5)i=5+i$
$(3-4i)-(2+5i)=(3-2)+(-4-5)i=1-9i$

Multiplying. Expand the product as you would with polynomials (FOIL). Then replace every $i^2$ with $-1$ . A real number times a complex number is just the distributive property: $4(2+5i)=8+20i$ . (pp. 221–222)

Example: $(3-4i)(2+3i)=6+9i-8i-12i^2$ . Since $-12i^2=+12$ , this becomes $6+12+i=18+i$ .

Conjugates. The conjugate of $a+bi$ is $a-bi$ . Only the sign of the imaginary part changes. A bar over the number marks the conjugate, so $\overline{4-i}=4+i$ . (p. 222)

The conjugate matters because a number times its conjugate is real:

$(a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2$

The $i$ terms cancel, and $-b^2i^2$ becomes $+b^2$ . For instance, $(4-i)(4+i)=16+1=17$ .

Dividing. A fraction with $i$ in the denominator is not in $a+bi$ form. To fix it, multiply the top and bottom by the conjugate of the denominator. The denominator becomes the real number $a^2+b^2$ . Then split the result into real and imaginary parts. (pp. 221–222)

Example: $\dfrac{2+5i}{4-i}\cdot\dfrac{4+i}{4+i}$ . The numerator is $8+2i+20i+5i^2=3+22i$ . The denominator is $17$ . So the answer is $\dfrac{3}{17}+\dfrac{22}{17}i$ .

Complex Zeros of Quadratics

A quadratic with real coefficients can have no real zeros. The quadratic formula then gives a negative number under the square root. With $i$ , we can finish the calculation. (p. 223)

For $f(x)=x^2-2x+5$ :

$x=\frac{2\pm\sqrt{4-20}}{2}=\frac{2\pm 4i}{2}=1\pm 2i$

The two zeros, $1+2i$ and $1-2i$ , are complex conjugates. This always happens when a quadratic with real coefficients has nonreal zeros. The Factor Theorem still works with these zeros:

$f(x)=\big(x-(1+2i)\big)\big(x-(1-2i)\big)$

Another example is $x^2+4x+13$ . Here $x=\frac{-4\pm\sqrt{-36}}{2}=-2\pm3i$ . (p. 223)

The Fundamental Theorem and Complete Factoring

Fundamental Theorem of Algebra. Every non-constant polynomial with real or complex coefficients has at least one real or complex zero. This is an existence theorem. It promises a zero is there, but it does not tell you how to find it. It also means that once complex numbers are allowed, a polynomial can never have no zeros. (p. 223)

Complex Factorization Theorem. Suppose $f$ has degree $n\ge 1$ . The reasoning goes in steps:

The Fundamental Theorem gives a zero $z_1$ .
The Factor Theorem lets us write $f(x)=(x-z_1)q_1(x)$ , where $q_1$ has degree $n-1$ .
If $q_1$ is not constant, repeat the process on it.
After $n$ rounds, all of $f$ is factored into linear pieces.

So $f$ has exactly $n$ complex zeros when we count multiplicity. If the distinct zeros are $z_1,\dots,z_k$ with multiplicities $m_1,\dots,m_k$ , then

$f(x)=a(x-z_1)^{m_1}(x-z_2)^{m_2}\cdots(x-z_k)^{m_k}$

Here $a$ is the leading coefficient and $m_1+\dots+m_k=n$ . (pp. 223–224)

Conjugate Pairs and Irreducible Quadratics

If a polynomial has real coefficients and $z$ is a nonreal zero, then its conjugate $\bar z$ is also a zero. So nonreal zeros come in pairs. One consequence is that a real-coefficient polynomial of odd degree must have at least one real zero. The nonreal zeros come in pairs, so they use up an even number of the $n$ zeros. (p. 225)

An irreducible quadratic is a quadratic with real coefficients and nonreal zeros. It cannot be factored any further using real numbers. (p. 225)

Worked example: take $f(x)=12x^5-20x^4+19x^3-6x^2-2x+1$ . The Rational Roots Theorem suggests candidates, and synthetic division confirms $x=\tfrac12$ twice and then $x=-\tfrac13$ . What is left is $x^2-x+1$ , which has zeros $\frac{1\pm i\sqrt3}{2}$ . (pp. 224–225)

Over the real numbers: $f(x)=12\left(x-\tfrac12\right)^2\left(x+\tfrac13\right)(x^2-x+1)$ . The process stops at the irreducible quadratic.
Over the complex numbers: the last factor splits into $\left(x-\frac{1+i\sqrt3}{2}\right)\left(x-\frac{1-i\sqrt3}{2}\right)$ . This is called factoring completely.

The multiplicities are $2+1+1+1=5$ , which matches the degree. (p. 225)

The same method works for $f(x)=x^3-4x^2+9x-10$ . Synthetic division shows that $x=2$ is a zero, with quotient $x^2-2x+5$ . That quadratic has zeros $1\pm2i$ , so all three zeros are $2$ , $1+2i$ and $1-2i$ . (p. 226)

Rational functions and inverse proportion

Up to now, the polynomials we built came from power functions with positive whole-number exponents. A rational function comes from power functions with negative exponents and from ratios in general. It is any function that can be written as one polynomial divided by another:

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

Examples are $\frac{3x+7}{x+2}$ and the sugar model $C(t)=\frac{5+t}{100+10t}$ (pp. 228, 231).

A common way to get one is an inverse proportion, where one quantity equals a constant divided by another. Suppose you drive 100 miles at speed $v$ . Since distance equals speed times time, $100=vt$ , so the trip time is $t=\frac{100}{v}$ . Doubling your speed halves your time. The graph of such a relationship is a transformation of the reciprocal toolkit function $\frac1x$ (p. 228).

Some natural quantities depend on the square of the other quantity. The loudness of a sound at distance $d$ from its source is $V=\frac{k}{d^2}$ for a constant $k$ . This is a transformation of the reciprocal squared function $\frac{1}{x^2}$ (p. 228).

Vertical and horizontal asymptotes

The reciprocal function $f(x)=\frac1x$ is undefined at $x=0$ , because we cannot divide by zero. Its short run and long run behavior both produce asymptotes. An asymptote is a line that the graph gets closer and closer to.

Short run (near 0). For small negative inputs, $f(x)$ becomes very large and negative: as $x\to0^-$ , $f(x)\to-\infty$ . For small positive inputs, $f(x)$ becomes very large and positive: as $x\to0^+$ , $f(x)\to\infty$ . The graph hugs the line $x=0$ (p. 229).
Long run (large $|x|$ ). As $x\to\infty$ or $x\to-\infty$ , $f(x)\to0$ . The outputs get close to 0 but never equal it (p. 229).

This gives two definitions:

A vertical asymptote is a vertical line $x=a$ where $f(x)\to\infty$ or $f(x)\to-\infty$ as $x\to a$ .
A horizontal asymptote is a horizontal line $y=b$ where $f(x)\to b$ as $x\to\infty$ or as $x\to-\infty$ .

For $\frac1x$ , the vertical asymptote is $x=0$ and the horizontal asymptote is $y=0$ (p. 229).

Shifting a reciprocal function

When you shift the graph, the asymptotes shift by exactly the same amounts. Move $\frac1x$ left 2 units and up 3 units:

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

Write the terms over a common denominator: $3=\frac{3(x+2)}{x+2}$ , so

$f(x)=\frac{1+3(x+2)}{x+2}=\frac{3x+7}{x+2}.$

This is a rational function. The vertical asymptote moves from $x=0$ to $x=-2$ , and the horizontal asymptote moves from $y=0$ to $y=3$ . Near $x=-2$ , the outputs go to $-\infty$ from the left and to $\infty$ from the right. For large $|x|$ , $f(x)\to3$ (p. 230).

Finding vertical asymptotes, and holes

A rational function is undefined wherever its denominator is zero. So to find vertical asymptotes, set the denominator equal to zero and solve. If the numerator is not zero at that input, there is a vertical asymptote there.

Example. For $k(x)=\frac{5+2x}{x^2-x-2}$ , factor the denominator: $x^2-x-2=(x-2)(x+1)$ . It is zero at $x=2$ and $x=-1$ . The numerator $5+2x$ is $9$ at $x=2$ and $3$ at $x=-1$ , so neither is zero. Both inputs give vertical asymptotes (pp. 231–232).

The exception is when the numerator is also zero at the same input. Then the factor may cancel. A hole is a single missing point in the graph, drawn as an open circle.

Example. Take $k(x)=\frac{x-2}{x^2-4}=\frac{x-2}{(x-2)(x+2)}$ . The denominator is zero at $x=2$ and $x=-2$ . At $x=2$ the numerator is also zero. After cancelling the common factor, we get $\frac{1}{x+2}$ for $x\ne2$ . That formula has no problem at $x=2$ , so the graph has a hole there (at height $\frac14$ ). At $x=-2$ the factor $x+2$ remains in the denominator, so there is a vertical asymptote at $x=-2$ (p. 232).

Common mistake: believing that every zero of the denominator gives a vertical asymptote. Always factor and simplify first. If a factor cancels, that input gives a hole. Only the denominator zeros that remain after cancelling give vertical asymptotes (p. 232).

Horizontal asymptotes: compare the degrees

A polynomial behaves like its leading term in the long run. In the same way, a rational function behaves like the ratio of the leading terms of its numerator and denominator. There are three cases (pp. 233–234).

Denominator degree is bigger. For $f(x)=\frac{3x+2}{x^2+4x-5}$ , the long run behavior is like $\frac{3x}{x^2}=\frac3x$ , which goes to 0. The horizontal asymptote is $y=0$ .
Denominator degree is smaller. For $f(x)=\frac{3x^2+2}{x-5}$ , the long run behavior is like $\frac{3x^2}{x}=3x$ . The outputs keep growing and never level off, so there is no horizontal asymptote.
Degrees are equal. For $f(x)=\frac{3x^2+2}{x^2+4x-5}$ , the long run behavior is like $\frac{3x^2}{x^2}=3$ . The horizontal asymptote is $y=3$ , the ratio of the leading coefficients.

Example with a context. In the mixing tank, the water amount is $100+10t$ gallons and the sugar amount is $5+t$ pounds. The concentration is

$C(t)=\frac{5+t}{100+10t}.$

Both parts have degree 1, so the horizontal asymptote is the ratio of the leading coefficients: $y=\frac{1}{10}$ . In context, the concentration gets closer and closer to 0.1 pound of sugar per gallon as time passes. That matches the tap, which adds 1 pound of sugar for every 10 gallons of water (pp. 231, 234).

Putting it together. Consider $f(x)=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}$ . No input makes both the numerator and the denominator zero, so there are no holes. The denominator is zero at $x=1$ , $x=-2$ and $x=5$ , so those are vertical asymptotes. The numerator has degree 2 and the denominator has degree 3, so the horizontal asymptote is $y=0$ (p. 234).

Intercepts of a rational function

A rational function has a vertical intercept at $x=0$ only if it is defined there. If the denominator is zero at $x=0$ , there is no vertical intercept. Horizontal intercepts come from the numerator. A fraction equals zero only when its numerator is zero, so set the numerator to zero. Skip any input that also makes the denominator zero, because that input is a hole, not an intercept. (p. 235)

Example: For $f(x)=\frac{(x-2)(x+3)}{(x-1)(x+2)(x-5)}$ , the vertical intercept is $f(0)=\frac{-6}{10}=-\frac35$ . The numerator is zero at $x=2$ and $x=-3$ , so those are the horizontal intercepts. (p. 235)

Behavior at intercepts and asymptotes

Factors in a rational function work like factors in a polynomial.

Numerator factors (intercepts): A factor to an odd power, such as $(x-3)$ , makes the graph cross the axis. A squared factor, such as $(x+1)^2$ , makes it bounce off the axis. (p. 236)
Denominator factors (vertical asymptotes): A factor $(x-h)^p$ in the denominator controls the graph near $x=h$ . If $p$ is odd, the graph behaves like $\frac1x$ . It goes to positive infinity on one side and negative infinity on the other. If $p$ is even, it behaves like $\frac1{x^2}$ . It goes to the same infinity on both sides, positive or negative depending on the overall sign. (pp. 236–237)

Example: In $f(x)=\frac{(x+1)^2(x-3)}{(x+3)^2(x-2)}$ , the graph bounces at $x=-1$ and passes through at $x=3$ . It goes to $+\infty$ on both sides of $x=-3$ . It goes to opposite infinities on the two sides of $x=2$ . (pp. 236–237)

Sketching a rational function

Follow these steps:

Find the vertical intercept.
Find the horizontal intercepts from the numerator.
Find the vertical asymptotes from the denominator.
Compare degrees to find the horizontal asymptote.
Use the factor powers to decide the behavior at each feature.
Use the sign of the function between features to fill in the graph.

Example: Let $f(x)=\frac{(x+2)(x-3)}{(x+1)^2(x-2)}$ . Then $f(0)=3$ , the intercepts are $x=-2$ and $x=3$ , and the vertical asymptotes are $x=-1$ and $x=2$ . The denominator has the larger degree, so the horizontal asymptote is $y=0$ . Between the two asymptotes there are no zeros and the vertical intercept is positive, so the graph stays positive there. The squared factor at $x=-1$ sends the graph to $+\infty$ on both sides. The linear factor at $x=2$ makes the two sides go opposite ways. After the intercepts, the graph levels off toward $0$ . (p. 237)

Writing a rational function from a graph

Work backward. Put a numerator factor $(x-x_i)^{p_i}$ for each horizontal intercept. Put a denominator factor $(x-v_j)^{q_j}$ for each vertical asymptote. Choose each power from the graph's behavior there. Then use one more clear point to find the stretch factor $a$ . (p. 238)

Example: Suppose the graph crosses at $-2$ and $3$ . At $x=-1$ it goes to opposite infinities. At $x=2$ it goes to $-\infty$ on both sides. Then $f(x)=a\frac{(x+2)(x-3)}{(x+1)(x-2)^2}$ . The point $(0,-2)$ gives $-2=a\cdot\frac{-6}{4}$ , so $a=\frac43$ . (pp. 238–239)

Oblique asymptotes and other long-run shapes

When the numerator's degree is exactly one more than the denominator's, there is no horizontal asymptote. Divide the numerator by the denominator with polynomial long division. The quotient describes the long-run behavior. The leftover fraction shrinks toward 0 for large $|x|$ . If the quotient is a line, that line is the oblique (slant) asymptote. (pp. 239–240)

Example: $\frac{3x^2+2}{x-5}=3x+15+\frac{77}{x-5}$ . As $x\to\pm\infty$ the fraction part vanishes, so the graph follows $y=3x+15$ . (pp. 239–240)

If the quotient is not linear, it still gives the long-run shape. For $f(x)=\frac{-x^3-x^2+4x+2}{x+1}$ , division gives $-x^2+4-\frac{2}{x+1}$ . For large $|x|$ the graph follows the downward parabola $y=-x^2+4$ . It also has a vertical asymptote at $x=-1$ . (p. 241)

Inverses of quadratics and radical functions

A function has an inverse only if it is one-to-one. A parabola fails this test, because most outputs come from two inputs, one on each side of the vertex. So you must restrict the domain to one side of the vertex first. (pp. 249–250)

Example (trough): A parabolic cross-section with its vertex at the origin is $y=\frac12x^2$ . Restrict to $x\ge0$ and solve for $x$ . This gives $x=\sqrt{2y}$ , the half-width of the water at depth $y$ . For a 36-inch-long trough, the surface area is $36\cdot2x=72\sqrt{2y}$ square inches. (p. 250)

The inverse of a restricted quadratic is a square root function. Functions involving roots are called radical functions. (p. 250)

Steps to invert a quadratic:

Write it in vertex form.
Restrict the domain to one side of the vertex.
Set $y=f(x)$ and solve for $x$ with a square root.
Keep the sign that matches the restricted domain.

For $f(x)=(x-2)^2-3$ with $x\ge2$ , we get $y+3=(x-2)^2$ , so $x=2\pm\sqrt{y+3}$ . Because $x\ge2$ , choose the plus sign, so $f^{-1}(x)=2+\sqrt{x+3}$ . If the quadratic is in standard form, you can instead use the quadratic formula and pick the same sign. (pp. 250–251)

Some toolkit functions need no restriction: identity, cubic, square root and cube root. For $f(x)=5x^3+1$ , solving for $x$ gives $f^{-1}(x)=\sqrt[3]{\frac{x-1}{5}}$ . This is a power function with exponent $\frac13$ . (pp. 251, 254)

Radical functions as models and their domains

The speed of a skidding car is $v=\sqrt{2gfd}$ , with $g=32$ ft/s², friction coefficient $f$ and skid length $d$ . With $f=0.5$ and $d=200$ , $v=\sqrt{6400}=80$ ft/s, about 54.5 miles per hour. (p. 252)

A square root needs a nonnegative input, so the domain of a root of a rational expression is where that expression is $\ge0$ . The sign can change only at horizontal intercepts and vertical asymptotes. Use test values or a sketch to check each interval. For $f(x)=\sqrt{\frac{(x+2)(x-3)}{x-1}}$ , the sign can change at $-2$ , $1$ and $3$ . The expression is nonnegative on $[-2,1)$ and $[3,\infty)$ . That is the domain, and $x=1$ is excluded. (pp. 252–253)

Inverses of rational functions

For a ratio of linear expressions, the inverse comes from algebra:

Multiply both sides by the denominator.
Distribute.
Gather all terms containing the input variable on one side.
Factor out that variable.
Divide.

For $C=\frac{20+0.4n}{100+n}$ , this gives $100C+nC=20+0.4n$ . Then $n(C-0.4)=20-100C$ , so $n=\frac{100C-20}{0.4-C}$ . At $C=0.35$ , $n=\frac{15}{0.05}=300$ mL of 40% solution. (p. 253)

Check yourself

For $f(x)=\frac{(x-1)^2(x+4)}{(x+2)(x-3)^2}$ , state the intercepts, the vertical asymptotes and how the graph behaves at each.
Use long division to find the oblique asymptote of $\frac{2x^2+3x-1}{x+1}$ .
Find the inverse of $f(x)=(x+1)^2+5$ on the domain $x\ge-1$ .
How do the end behavior of $-2x^5$ and of $x^4$ differ, and how does the leading term determine a polynomial's long run behavior?
By the Remainder Theorem, what does dividing $p(x)$ by $x-c$ tell you, and how does this link to the Factor Theorem?