AP Precalc 1.7 Rational End BehaviorAP Precalc 1.7 Rational End Behavior

A P P R E C A L C U L U S · UNIT1

Rational Functions and End Behavior

Topics 1.7A and 1.7B Study Guide

In one sentence: a rational function is a polynomial divided by a polynomial, its domain skips every input that makes the denominator 0, and its end behavior is decided by comparing the degrees of the top and bottom.

Why it works, in plain language. A fraction can only be as big as its top divided by its bottom. When x is enormous, the highest power in each polynomial swamps everything else. In (3x² + 5)/(x² − 9), at x = 1,000 the 5 and the 9 are tiny next to 3,000,000 and 1,000,000, so the fraction is almost exactly 3x²/x² = 3. That is why only the leading terms matter for end behavior. If the bottom grows faster (higher degree), the fraction shrinks toward 0. If the top grows faster, the fraction blows up to ±∞. Step by step: domain. (1) Write the denominator. (2) Set it equal to 0 and solve, factoring if needed. (3) Those x-values are the only ones you remove. (4) Write the real line with those points cut out, using ∪ between pieces.

Step by step: horizontal asymptote. (1) Find the degree of the top and the bottom. (2) Compare. (3) Bottom bigger: y = 0. Equal: divide the leading coefficients. Top bigger: none. (4) For ±∞ limits, plug ±∞ into the leading terms and track signs: a negative number to an odd power stays negative, an even power turns positive.

1.7A Domain of a rational function

Rational function: r(x) = f(x) / g(x) where f and g are polynomials and g(x) ≠ 0. Domain: set the denominator ≠ 0, factor, and exclude those x-values. Write the answer in interval notation.

Only the denominator matters for the domain. Zeros of the numerator do not remove anything. Examples. f(x) = (x + 1)/(2x + 5): 2x + 5 ≠ 0 so x ≠ −5/2. Domain (−∞, −5/2) ∪ (−5/2, ∞). g(n) = (n² − 4)/((n + 3)(n − 7)): n ≠ −3, 7. Domain (−∞, −3) ∪ (−3, 7) ∪ (7, ∞). h(x) = (x³ + 7x² + 12x)/(x² + 2x − 15): denominator = (x + 5)(x − 3), so (−∞, −5) ∪ (−5, 3) ∪ (3, ∞). Traps. (1) Do not cancel first and then read the domain. A cancelled factor still stays excluded. (2) If r(x) = f(x)/g(x) with f and g given, the domain of r also needs g(x) ≠ 0. For f(x) = x² + a², g(x) = x² − a²: g ≠ 0 means x ≠ ±a, so the domain is (−∞, −a) ∪ (−a, a) ∪ (a, ∞). (3) On a graph, the domain skips every vertical asymptote and every hole.

AP Precalculus 1 of 4

End behavior and horizontal asymptotes

End behavior describes f(x) as x increases or decreases without bound. Write it with limits: limx→∞ f(x) and limx→−∞ f(x).

Horizontal asymptote: the line y = b is a horizontal asymptote when limx→∞ f(x) = b or limx→−∞ f(x) = b.

Table method (calculator): evaluate at x = ±10, ±100, ±1,000, ±10,000. If the outputs settle toward a number, that is the limit.

Table example. f(x) = x²/(x² − 4x). Outputs at x = −10,000, −1,000, −100, 100, 1,000, 10,000 are 0.9996, 0.99602, 0.96154, 1.0417, 1.004, 1.0004. They approach 1, so both limits are 1 and the horizontal asymptote is y = 1.

Graph example. A graph that levels off at y = 4 on both ends gives lim = 4 on both sides, so y = 4. A graph whose right end climbs without bound and left end rises too has lim = ∞ on both sides and no horizontal asymptote. A graph that hugs the x-axis has y = 0.

1.7B The degree rules

Compare the degree of the numerator (N) and denominator (D). Only the leading terms matter for large |x|.

Degrees End behavior Horizontal asymptote

D bigger than N ("denominator

dominates")

both limits = 0 y = 0

Equal degrees both limits = (leading coefficient of N) ÷

(leading coefficient of D)

y = ratio of leading

coefficients

N bigger than D ("numerator

dominates")

limits are ±∞ (check signs) none

Examples.

(−2x³ + 5)/(4x⁵ − 8x³ + 2x): D bigger, y = 0.

(x² − 9)/(4x² + 2x − 15): equal, y = 1/4.

(3n² − 4)/((n + 3)(n − 7)): denominator is n² − 4n − 21, equal degrees, y = 3. (−3x⁴ + x² + 2)/(2x⁴ + 7x² − 5): y = −3/2.

(4 − t²)/(2t² + t): y = −1/2. And (x² − 1)/(4x(x² + 1)): D is bigger, y = 0.

When the numerator dominates, find the signs

Replace each side with its leading term and substitute ±∞. Example: f(x) = (x² − 3x + 1)/(2x + 3). Left: (−∞)²/(−∞) = (+)/(−) = −∞. Right: (+∞)²/(+∞) = +∞. So limx→−∞ f = −∞ and limx→∞ f = ∞. No horizontal asymptote.

AP Precalculus 2 of 4

Another. f(x) = (−x³ + 3x² + x − 5)/(5x² + 7x − 2): left: −(−∞)³/(+∞) = −(−)/(+) = +∞. Right: −(+)/(+) = −∞. So limx→−∞ f = ∞ and limx→∞ f = −∞.

Limits at infinity, fast

Limit Value Why

lim (3x² + 2)/(x² − 9) 3 equal degrees, 3/1

limx→−∞ 1/(x − 4) 0 denominator dominates

lim (2n⁵ − 3n³ + 2n)/(5n⁵ + 3n − 2) 2/5 equal degrees

lim (x + 3)(2x − 1)/(4x² − 9) 1/2 numerator expands to 2x² + …

limt→−∞ (2t³ − 3t)/(5t² − 4t) −∞ numerator bigger, (−)/(+)

Traps. (1) "No horizontal asymptote" does not mean the limit is 0. (2) A horizontal asymptote can be crossed. (3) A graph can have a horizontal asymptote on one side only. (4) Do the degree comparison after multiplying out factored forms. (5) Calculator tables with huge outputs (like 1.25 × 10⁷) signal ∞. Test-prep patterns

"For x > 0, y = 4 is a horizontal asymptote. What must be true?" Only limx→∞ f(x) = 4. f(4) can be anything, and f(x) can even equal 4 at some points.

"y = a is a horizontal asymptote for x < 0." Only limx→−∞ g(x) = a.

Find a constant: f(x) = (ax³ − 2x² + 5)/(2x³ − 8) has HA y = 3. Equal degrees, so a/2 = 3 and a = 6. Domain plus HA: f(x) = (x − 8)(x + 3)/(x² + 5x − 24) = (x − 8)(x + 3)/((x + 8)(x − 3)). Domain (−∞, −8) ∪ (−8, 3) ∪ (3, ∞), HA y = 1.

Quick check

1. Domain of (x + 2)/(x² − 25)? All reals except x = ±5

2. Horizontal asymptote of (6x² + 1)/(3x² − x)? y = 2

3. limx→∞ (x³ + 1)/(2x² − 5)? ∞

4. If the denominator's degree is larger, what is the horizontal asymptote? y = 0 AP Precalculus 3 of 4

Practice set

Try every question before you look at the key. Show your work.

1. State the domain of f(x) = (x − 3)/(x² − 16) in interval notation.

2. State the domain of g(x) = (2x + 1)/(x² + 4).

3. State the domain of h(x) = x/(x³ − 9x).

4. Find the horizontal asymptote of (5x² − 3)/(2x² + x).