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Week 10 · Chapter

Chapter 10 — Polynomial & Rational Functions

College Algebra Free sample edition

The module's primary reading · ~45–55 minutes · Objective 7 (analyze polynomial and rational functions (end behavior, zeros and multiplicity, domain, and asymptotes), and simplify and solve rational and radical expressions and equations, including expressions with rational exponents, checking for extraneous solutions) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

This week you learn to read the shape of a graph from its formula before you plot a single point. A polynomial function is a function built from terms with whole-number powers, like Week 9's parabolas, and you will find where its graph heads and where it hits the x-axis. For a rational function (one polynomial divided by another), you will also find asymptotes, the guide lines a graph creeps toward without arriving. These lines answer real questions, such as how cheap an item can get when a business makes more of them. Every feature you find from the formula, you will confirm on a graph in Desmos or GeoGebra.

Why you already care about this

Picture a small print shop that makes custom tote bags. Setting up the press costs 200 dollars, and each bag then uses 5 dollars of material. If the shop makes 10 bags, the total is 250 dollars, so each bag costs 25 dollars on average. Make 200 bags and the average drops to 6 dollars, less than a quarter of the 25-dollar average.

Does the average keep falling forever? It cannot drop below 5 dollars, because every bag still needs its 5 dollars of material. The graph of the average cost bends down toward a level line at 5 and never touches it. That level line is a horizontal asymptote, and you can read it straight from the formula. A medicine level that climbs and then settles toward a level line is read the same way.

Polynomials tell a related story. A formula like x³ − 5x + 2 has three terms, but far from the origin only one of them matters. Once you know which one, you know where the graph goes at both ends.

By the end of this chapter you will be able to:

  • name a polynomial's leading term, degree, and leading coefficient, and read its end behavior from the leading term alone;
  • sort end behavior into four cases by an even or odd degree and a positive or negative leading coefficient;
  • find the zeros of a polynomial from factored form and state each zero's multiplicity;
  • decide whether a graph crosses or touches the x-axis at each zero, and confirm it with signs;
  • sketch a polynomial from its ends, its zeros, and its y-intercept;
  • find a rational function's domain and tell a vertical asymptote from a hole;
  • find a horizontal asymptote by comparing degrees, and explain what it means in a setting.

Carry this sentence with you: the ends follow the leading term — not the loudest-looking one.

Where we are

  • Already covered: Week 3 introduced domain and rational functions; Week 6 multiplied and factored polynomials; Week 7 solved quadratic equations; Week 9 graphed the parabola.
  • This week: the graphs of polynomial functions of any degree and of rational functions, read from their formulas.
  • How they connect: a zero is a Week 7 solution, a factor is a Week 6 factor, and a domain is a Week 3 domain, now read off a picture.
  • Where we're going next: Week 11, Rational Expressions & Equations, simplifies, adds, and solves the fractions whose graphs you meet here.

Each earlier week hands you a tool. Week 3 said a rational function's domain leaves out any input that makes the denominator 0. Week 6 taught you to factor x² − 4 as (x − 2)(x + 2). Week 7 showed that x² − 6x + 5 = 0 has solutions 1 and 5, and Week 9 drew that parabola with both ends rising. This week you learn why both ends rise, and what changes when the power is 3, 4, or 6.

1. Read the far ends from the leading term

Think about a road trip seen from a plane. Up close, your road bends around every hill. From high above, you only see which way it heads. A graph is the same: far from the origin, only the overall direction shows.

A polynomial function is a function whose rule is a polynomial, such as f(x) = x² − 6x + 5 or x³ − 5x + 2; it accepts every real number as an input.

The leading term is the term of a polynomial with the highest power of the variable, such as −2x⁴ in −2x⁴ + 3x² − 1; far from the origin it outweighs all the other terms.

End behavior is the direction a graph heads at its far left and far right, described as rising or falling on each end; for a polynomial it is decided by the leading term alone.

The leading term carries two pieces of information you met in Week 6. The degree is its exponent. The leading coefficient is the number in front of it.

Worked example — the week's parabola as a polynomial. Read the end behavior of f(x) = x² − 6x + 5.

  1. Find the leading term. The highest power is x², so the leading term is x².
  2. Read the degree. The exponent is 2, so the degree is 2.
  3. Read the leading coefficient. No number is written in front of x², so the leading coefficient is 1.
  4. Decide the ends. The degree 2 is even, so both ends point the same way. The leading coefficient 1 is positive, so both ends rise.
  5. Check with the table. The values at x = 0 through 6 are 5, 0, −3, −4, −3, 0, 5, and their sum is 0. You can see f(0) = 5 and f(6) = 5 sit higher than the middle values. The graph climbs at both edges.

In words: the week's parabola rises on both ends, which is exactly the U shape you drew in Week 9.

Misconception check — the loudest-looking term. Students often think the biggest-looking number decides the ends. In −2x⁴ + 3x² − 1, the 3x² term looks friendly and positive, so it is tempting to say "even degree, positive, both ends rise." That uses the wrong term. The cure: find the highest power first. Here it is −2x⁴, and its negative coefficient sends both ends down. Ignore every lower-degree term when you read the ends.

2. Sort end behavior into four cases

Think of two arms. For an even degree, both arms point the same way, like a person cheering with both hands up, or shrugging with both hands down. For an odd degree, the arms point opposite ways, like someone pointing down to the left and up to the right.

The rule is a property of polynomials, and it has four cases:

End behavior of a polynomial by the parity of its degree and the sign of its leading coefficient
Degree Leading coefficient positive Leading coefficient negative
Even rises on both ends falls on both ends
Odd falls on the left, rises on the right rises on the left, falls on the right

Worked example — even degree, negative. Read the ends of −2x⁴ + 3x² − 1.

  1. Leading term. −2x⁴, so the degree is 4 and the leading coefficient is −2.
  2. Parity. 4 is even, so both ends point the same way.
  3. Sign. −2 is negative, so both ends fall.
  4. Test far out. At x = 10: −2(10)⁴ + 3(10)² − 1 = −20000 + 300 − 1 = −19701. That is far below the axis, as predicted.

Worked example — odd degree, positive. Read the ends of x³ − 5x + 2.

  1. Leading term. x³, so the degree is 3 and the leading coefficient is 1.
  2. Parity and sign. 3 is odd and 1 is positive, so the graph falls on the left and rises on the right.
  3. Why the leading term wins. At x = 10, x³ = 1000 while −5x + 2 = −50 + 2 = −48. Together, x³ − 5x + 2 = 1000 − 50 + 2 = 952. The other terms barely dent the 1000.
  4. Check the left end. At x = −10, (−10)³ = −1000, and x³ − 5x + 2 = −1000 + 50 + 2 = −948. Far left, the graph is far below the axis.

In words: x³ − 5x + 2 comes up from the lower left and leaves toward the upper right.

In practice (your tools). In Desmos, type y = x^3 - 5x + 2 and y = x^3, then zoom out with the minus button. Far from the origin, you see the two curves hug each other.

Misconception check — reading the first term written. Students sometimes read the ends from whichever term you see first. But 3x² − 2x⁴ − 1 has its leading term, −2x⁴, in the middle. The cure: the leading term is the highest power wherever it sits.

3. Find the zeros from factored form

Think of a hiking trail that dips below sea level and climbs back. The spots where it crosses sea level are the interesting ones. On a graph, sea level is the x-axis, and those spots are where the output is 0.

A zero of a function is an input that makes the output of a function equal 0; each real zero shows up on the graph as an x-intercept.

You already know how to find zeros. In Week 7 you solved equations by factoring and setting each factor equal to 0. That is the same move.

Worked example — the zeros of the week's parabola. Find the zeros of f(x) = x² − 6x + 5.

  1. Factor. Two numbers with product 5 and sum −6 are −1 and −5, so f(x) = x² − 6x + 5 = (x − 1)(x − 5).
  2. Set each factor to 0. x − 1 = 0 gives x = 1. x − 5 = 0 gives x = 5.
  3. Check. f(1) = 1 − 6 + 5 = 0 and f(5) = 25 − 30 + 5 = 0.

In words: the zeros are 1 and 5, so the graph meets the x-axis at (1, 0) and (5, 0). These match the two 0s in the table.

Misconception check — the sign inside the factor. Students often read the factor (x − 5) as the zero −5. It is tempting because −5 is the number you see. The cure: set the factor equal to 0 and solve. x − 5 = 0 gives x = 5. A factor (x + 4) gives the zero −4.

4. Use multiplicity: cross or touch

Picture a skateboarder on a ramp that dips to ground level. One path rolls straight through the low spot and keeps going. Another touches the ground and swoops back up. Zeros come in those two kinds.

Multiplicity is the number of times a factor appears in a factored polynomial, which is the exponent on that factor; in (x − 1)²(x + 4) the zero 1 has multiplicity 2.

The rule is a property you can check every time:

  • At a zero of odd multiplicity (1, 3, 5, …), the graph crosses the x-axis. The output changes sign.
  • At a zero of even multiplicity (2, 4, …), the graph touches the x-axis and turns back. The output keeps its sign.

Worked example — the week's parabola crosses. In (x − 1)(x − 5), each factor appears once, so each zero has multiplicity 1. 1 is odd, so the graph crosses at x = 1 and at x = 5. The table shows it: f(0) = 5 is positive and f(2) = −3 is negative, so the graph changes sign at 1. Then f(4) = −3 is negative and f(6) = 5 is positive, so it changes sign again at 5.

Worked example — a double zero. Read the zeros of (x − 1)²(x + 4).

  1. Read the exponents. The factor (x − 1) appears twice, so x = 1 has multiplicity 2. The factor (x + 4) appears once, so x = −4 has multiplicity 1.
  2. Test both sides of x = 1. At x = 0, (0 − 1)²(0 + 4) = 1 · 4 = 4. At x = 2, (2 − 1)²(2 + 4) = 1 · 6 = 6. Both are positive, so the graph touches the x-axis at x = 1 and turns back without crossing.
  3. Test both sides of x = −4. At x = −5, (−5 − 1)²(−5 + 4) = 36 · (−1) = −36. At x = −2, (−2 − 1)²(−2 + 4) = 9 · 2 = 18. The sign changes from negative to positive, so the graph crosses at x = −4.

In words: this graph crosses at −4 and only kisses the axis at 1.

Misconception check — every zero is a crossing. Students often think a graph must cross the axis at every zero. It is tempting because the week's parabola crosses at both of its zeros. The cure: look at the exponent before you decide. In (x − 1)²(x + 4), the exponent on (x − 1) is 2, which is even, so the graph touches at 1 and does not cross. A quick sign test on each side settles any doubt.

5. Sketch a polynomial from its features

Think of giving a friend directions: where to start, where to turn, and where to end up. A polynomial sketch needs the same. The ends tell you where to start and finish, the zeros tell you where you meet the axis, and the y-intercept gives you one more landmark.

Worked example — sketch (x − 1)²(x + 4). You gather four features, then draw.

  1. Leading term. Multiply the leading terms of the factors: x² · x = x³. The degree is 3 and the leading coefficient is 1.
  2. Ends. Odd degree and positive coefficient, so the graph falls on the left and rises on the right.
  3. Zeros. −4 with multiplicity 1 (cross) and 1 with multiplicity 2 (touch).
  4. y-intercept. (0 − 1)²(0 + 4) = 1 · 4 = 4, so the point is (0, 4).
  5. Draw it. Coming up from the lower left, the graph crosses at (−4, 0), passes through (−2, 18), comes down through (0, 4), touches (1, 0), and rises to the upper right.

Worked example — three multiplicities at once. Read x²(x − 3)(x + 2)³.

  1. Zeros. 0 with multiplicity 2 (touches), 3 with multiplicity 1 (crosses), and −2 with multiplicity 3 (crosses, flattening as it passes through).
  2. Degree. Add the exponents: 2 + 1 + 3 = 6. The degree 6 is even and the leading coefficient is 1, so both ends rise.
  3. Check the crossing at −2. At x = −3, (−3)²(−3 − 3)(−3 + 2)³ = 9 · (−6) · (−1) = 54. At x = −1, (−1)²(−1 − 3)(−1 + 2)³ = 1 · (−4) · 1 = −4. The sign changes, so the graph crosses at x = −2.

In practice (your tools). In Desmos or GeoGebra, type y = (x-1)^2(x+4). Click the curve where it meets the x-axis. You will see (−4, 0), where the graph passes through, and (1, 0), where it only touches. Then type y = x^2(x-3)(x+2)^3 and look near (−2, 0): the curve flattens out as it crosses.

Honesty note: a sketch from features shows the shape, not the exact height of every bump. The highest point between −4 and 1 is a little above 18; click the bump in Desmos to see it.

Misconception check — the degree is the largest exponent you see. Students sometimes call x²(x − 3)(x + 2)³ a degree-3 polynomial because 3 is the biggest exponent written. The cure: the degree comes from multiplying everything out, so you add the exponents of the factors. Here that is 2 + 1 + 3 = 6.

6. Find a rational function's domain: vertical asymptotes versus holes

Picture a bridge with one plank missing: you cannot stand on that spot, but the bridge around it is fine. Now picture a cliff edge, where the ground drops away for good. A rational function can have both kinds of trouble, and factoring tells you which.

In Week 3, a rational function was one polynomial divided by another. Its domain leaves out every input that makes the denominator 0. This week you sort those left-out inputs into two kinds.

A vertical asymptote is a vertical line x = c that a graph approaches but never reaches, created by a zero of the denominator that is still there after common factors are cancelled.

A hole is a single missing point on a graph, drawn as an open circle, at an input where a factor in the numerator and denominator cancelled out.

Worked example — the domain and asymptotes of R(x) = (x + 1) ⁄ (x² − 4). You factor before you decide anything.

  1. Factor the denominator. x² − 4 = (x − 2)(x + 2). It is 0 at x = 2 and x = −2.
  2. State the domain. All real numbers except −2 and 2: (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).
  3. Look for a common factor. The numerator x + 1 shares no factor with (x − 2)(x + 2). Nothing cancels.
  4. Classify. Both excluded inputs give vertical asymptotes: x = −2 and x = 2.

In words: the graph of R dives or climbs without end as x gets close to −2 or to 2.

Worked example — a hole, not an asymptote. Classify the excluded input of (x² − x) ⁄ (x − 1).

  1. Factor the numerator. x² − x = x(x − 1).
  2. Cancel the common factor. x(x − 1) ⁄ (x − 1) leaves x, for every x ≠ 1.
  3. Classify. The factor x − 1 cancelled, so x = 1 is a hole, not an asymptote.
  4. Find the hole's height. Put 1 into the simplified form x. The y-value is 1, so the hole is at (1, 1).
  5. Check at a legal input. At x = 2, (2² − 2) ⁄ (2 − 1) = 2 ⁄ 1 = 2, which matches the simplified form x = 2.

In words: the graph is the line y = x with an open circle at (1, 1).

Misconception check — every excluded input is an asymptote. Students often see x − 1 in a denominator and draw a vertical asymptote at x = 1. It is tempting because the input really is excluded from the domain. The cure: factor and cancel first. A cancelled factor leaves a hole, one missing point. Only a denominator zero that remains after you cancel gives an asymptote.

7. Find the horizontal asymptote by comparing degrees

Think about a long season of free throws. Early on, one miss swings your percentage a lot. Late in the season, one more shot barely moves it. A horizontal asymptote is the level such a quantity settles toward.

A horizontal asymptote is a horizontal line y = c that a graph approaches at its far left and far right; for a rational function it is found by comparing the degrees of the numerator and the denominator.

The comparison has three cases, and they are a property of rational functions:

The horizontal asymptote of a rational function by comparing the degree of the numerator with the degree of the denominator, with this week's example of each
Degrees Horizontal asymptote This week's example
Numerator smaller y = 0 R(x) = (x + 1) ⁄ (x² − 4) and 2x ⁄ (x² + 1)
Equal the ratio of the leading coefficients (3x + 1) ⁄ (x − 5) has y = 3
Numerator larger none (x² + 1) ⁄ (x − 1) has none

Worked example — numerator smaller. R(x) has degree 1 over degree 2. The numerator's degree is smaller, so the horizontal asymptote is y = 0. The same holds for 2x ⁄ (x² + 1), degree 1 over degree 2, so y = 0. Its denominator x² + 1 is at least 1 for every x, so it is never 0: the domain is all real numbers, and there is no vertical asymptote. At x = 1 its value is 2 ⁄ 2 = 1.

Worked example — degrees equal. Find the asymptotes of (3x + 1) ⁄ (x − 5).

  1. Compare. Degree 1 over degree 1, so the degrees are equal.
  2. Divide the leading coefficients. y = 3 ⁄ 1 = 3.
  3. Vertical asymptote. The denominator is 0 at x = 5, and nothing cancels, so x = 5.
  4. Check far out. At x = 105, (315 + 1) ⁄ (105 − 5) = 316 ⁄ 100 = 3.16. At x = 205, 616 ⁄ 200 = 3.08. The values move closer to 3.

Worked example — numerator larger. (x² + 1) ⁄ (x − 1) has degree 2 over degree 1, so there is no horizontal asymptote. Far out the values keep growing: at x = 11, (121 + 1) ⁄ 10 = 12.2. The vertical asymptote is x = 1, because the numerator at 1 is 1 + 1 = 2, not 0.

Misconception check — borrowing the ratio rule. Students often offer y = 1 as the horizontal asymptote of (x² + 1) ⁄ (x − 1), since both leading coefficients are 1. Others draw a slanted line and call it the horizontal asymptote. The cure: the ratio rule applies only when the degrees are equal. Here the numerator wins, so there is no horizontal asymptote at all. The graph does follow a slanted path far out, but slanted guide lines are outside this course.

8. Interpret a horizontal asymptote and sketch a rational function

Back to the print shop. The setup cost of 200 dollars is paid once, and each tote bag needs 5 dollars of material. Making x bags costs 5x + 200 dollars, and the average cost is the total cost of making a number of items divided by that number of items, so a fixed setup cost is spread thinner as production grows.

Worked example — the average cost per bag. The shop's average cost is u(x) = (5x + 200) ⁄ x dollars per bag.

  1. Evaluate. u(10) = 250 ⁄ 10 = 25, u(50) = 450 ⁄ 50 = 9, u(100) = 700 ⁄ 100 = 7, and u(200) = 1200 ⁄ 200 = 6 dollars per bag.
  2. Compare degrees. Degree 1 over degree 1, so the horizontal asymptote is y = 5 ⁄ 1 = 5.
  3. Interpret. Your average cost falls toward 5 dollars per bag but never reaches it, because u(x) = 5 + 200 ⁄ x and 200 ⁄ x stays positive.

In words: however many bags the shop makes, each still costs a bit more than 5 dollars on average. Zero bags is outside the setting, so the vertical asymptote x = 0 does not matter to the shop.

Worked example — a level the medicine approaches. Caveat: this is a simplified model built for practice, not real dosing advice. In it, a medicine's concentration in the blood, in milligrams per liter, t hours after a steady drip starts is C(t) = 6t ⁄ (t + 1).

  1. Evaluate. C(1) = 6 ⁄ 2 = 3, C(2) = 12 ⁄ 3 = 4, C(5) = 30 ⁄ 6 = 5, and C(11) = 66 ⁄ 12 = 5.5.
  2. Compare degrees. Equal, so the horizontal asymptote is y = 6 ⁄ 1 = 6.
  3. Interpret. The concentration rises toward 6 milligrams per liter and levels off there.

Worked example — sketch R(x) = (x + 1) ⁄ (x² − 4). Put every feature together.

  1. Asymptotes. Vertical at x = −2 and x = 2; horizontal at y = 0.
  2. x-intercept. The numerator is 0 when x + 1 = 0, so x = −1 and the point is (−1, 0).
  3. y-intercept. R(0) = (0 + 1) ⁄ (0 − 4) = 1 ⁄ (−4) = −1⁄4, so the point is (0, −1⁄4).
  4. Test value. R(3) = (3 + 1) ⁄ (9 − 4) = 4 ⁄ 5, which is positive. To the right of x = 2 the graph sits above the x-axis and sinks toward y = 0.

In practice (your tools). In Desmos, type y = (x+1)/(x^2-4). You will see three separate pieces divided by the vertical lines x = −2 and x = 2. Click the curve near the axes to read (−1, 0) and (0, −0.25), the decimal form of −1⁄4. For the print shop, type y = (5x+200)/x and add the line y = 5; zoom out to the right and watch the curve settle onto the line without touching it.

Misconception check — never crossing. Students often think a graph can never touch its horizontal asymptote. R(x) proves otherwise: R(−1) = 0, so the graph meets y = 0 at (−1, 0). The cure: a horizontal asymptote describes the far ends only.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

  • (a) Think of something in your life with a one-time cost and a per-use cost, such as a bike, a phone, or a pass. Describe how the average cost per use changes as you use it more, and name the level it can never drop below.
  • (b) Pick the graph feature from this chapter you find easiest to read from a formula and the one you find hardest. For each, name one function from this chapter where it shows up, and say what you look at first.

Key terms

  • Polynomial function — A function whose rule is a polynomial, such as f(x) = x² − 6x + 5 or x³ − 5x + 2; it accepts every real number as an input. (first used: 1. Read the far ends from the leading term)
  • Leading term — The term of a polynomial with the highest power of the variable, such as −2x⁴ in −2x⁴ + 3x² − 1; far from the origin it outweighs all the other terms. (first used: 1. Read the far ends from the leading term)
  • End behavior — The direction a graph heads at its far left and far right, described as rising or falling on each end; for a polynomial it is decided by the leading term alone. (first used: 1. Read the far ends from the leading term)
  • Zero of a function — An input that makes the output of a function equal 0; each real zero shows up on the graph as an x-intercept. (first used: 3. Find the zeros from factored form)
  • Multiplicity — The number of times a factor appears in a factored polynomial, which is the exponent on that factor; in (x − 1)²(x + 4) the zero 1 has multiplicity 2. (first used: 4. Use multiplicity: cross or touch)
  • Vertical asymptote — A vertical line x = c that a graph approaches but never reaches, created by a zero of the denominator that is still there after common factors are cancelled. (first used: 6. Find a rational function's domain: vertical asymptotes versus holes)
  • Hole — A single missing point on a graph, drawn as an open circle, at an input where a factor in the numerator and denominator cancelled out. (first used: 6. Find a rational function's domain: vertical asymptotes versus holes)
  • Horizontal asymptote — A horizontal line y = c that a graph approaches at its far left and far right; for a rational function it is found by comparing the degrees of the numerator and the denominator. (first used: 7. Find the horizontal asymptote by comparing degrees)
  • Average cost — The total cost of making a number of items divided by that number of items, so a fixed setup cost is spread thinner as production grows. (first used: 8. Interpret a horizontal asymptote and sketch a rational function)

Summary

  • The leading term alone decides a polynomial's end behavior.
  • Even degree: both ends match. Odd degree: the ends point opposite ways.
  • A positive leading coefficient sends the right end up; a negative one sends it down.
  • Zeros come from setting each factor equal to 0.
  • Odd multiplicity crosses the x-axis; even multiplicity touches and turns.
  • Factor and cancel first: a cancelled factor is a hole, and a remaining denominator zero is a vertical asymptote.
  • Compare degrees for the horizontal asymptote: smaller gives y = 0, equal gives the ratio, larger gives none.
  • In a setting, a horizontal asymptote is the level a quantity approaches but does not reach.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. Describe the end behavior of h(x) = −3x⁵ + 2x² + 7.
  2. For g(x) = (x + 3)²(x − 2), list each zero with its multiplicity, say whether the graph crosses or touches there, and find the y-intercept.
  3. For k(x) = (x + 5) ⁄ (x² + 5x), find the domain, any hole, any vertical asymptote, and the horizontal asymptote.
  4. A basketball player has made 6 of her first 10 free throws, and then she makes every shot she takes. After x more shots, her success rate is p(x) = (x + 6) ⁄ (x + 10). Find p(10) and p(30), and name the level her rate approaches.

Answers (worked reasoning — read after attempting)

  1. The leading term is −3x⁵, so the degree is 5 and the leading coefficient is −3. The degree is odd, so the ends point opposite ways. The coefficient is negative, so the graph rises on the left and falls on the right.
  2. Setting each factor to 0 gives x = −3 (the factor appears twice, multiplicity 2) and x = 2 (multiplicity 1). At −3 the multiplicity is even, so the graph touches: g(−4) = (−1)²(−6) = −6 and g(−2) = (1)²(−4) = −4 are both negative. At 2 the multiplicity is odd, so it crosses: g(1) = 16 · (−1) = −16 and g(3) = 36 · 1 = 36 change sign. The y-intercept is g(0) = (3)²(−2) = 9 · (−2) = −18, so the point is (0, −18).
  3. Factor the denominator: x² + 5x = x(x + 5). It is 0 at x = 0 and x = −5, so the domain is all real numbers except −5 and 0. The factor (x + 5) cancels, leaving 1 ⁄ x for x ≠ −5. So x = −5 is a hole, at height 1 ⁄ (−5) = −1⁄5, the point (−5, −1⁄5). The factor x remains, so x = 0 is a vertical asymptote. The numerator has degree 1 and the denominator degree 2, so the horizontal asymptote is y = 0. Calling x = −5 an asymptote is the hole trap from Section 6.
  4. p(10) = (10 + 6) ⁄ (10 + 10) = 16 ⁄ 20 = 0.8, and p(30) = 36 ⁄ 40 = 0.9. The degrees are equal and both leading coefficients are 1, so the horizontal asymptote is y = 1 ⁄ 1 = 1. Her rate approaches 1, a perfect 100 percent, but never reaches it, because the 4 early misses stay in the count.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion — and next week the factor-and-cancel rule returns to simplify and solve rational expressions.

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