Lesson 2: Polynomial and Rational Functions — End Behavior and Asymptotes

Polynomials and rational functions are where graph-reading becomes a science: factorisation tells you exactly where a graph crosses or touches the axis, degree tells you where the ends go, and for rational functions the factors you *cancel* are as informative as the ones you keep. These are the functions calculus will differentiate first, so their behaviour must become automatic.

End behaviour: the leading term rules

For large |x|, a polynomial is dominated by its leading term. The end behaviour of a·x^n depends only on the sign of a and the parity of n:

Numerical view. For f(x) = 2x^3 − 3x^2 − 11x + 6, the lower terms matter near the origin, but at x = 100 the cubic term contributes 2,000,000 against −30,000 − 1,100 + 6 from the rest — the leading term wins, and both ends behave like 2x^3. End behaviour is a statement about dominance, not about the middle of the graph.

Zeros, factors and multiplicity

By the factor theorem, x = r is a zero exactly when (x − r) is a factor. Worked example: factor p(x) = 2x^3 − 3x^2 − 11x + 6. Testing x = 3: 2(27) − 3(9) − 33 + 6 = 0, so (x − 3) divides p. Polynomial division gives p(x) = (x − 3)(2x^2 + 3x − 2) = (x − 3)(2x − 1)(x + 2), with zeros 3, 1/2 and −2 — each of multiplicity 1, so the graph crosses the axis at all three. Multiplicity controls the local picture: at a zero of odd multiplicity the graph crosses; at a zero of even multiplicity it touches and turns around. For q(x) = (x + 1)^2(x − 4), the graph crosses at x = 4 but bounces off the axis at x = −1 — the same bounce you see at the vertex of a parabola, because locally the factor behaves like a square.

Rational functions: holes and vertical asymptotes

A rational function is a ratio of polynomials. Zeros of the denominator are excluded from the domain, but they come in two kinds. Worked example: f(x) = (2x^2 − 8)/(x^2 − 5x + 6). Factor both parts: numerator 2(x − 2)(x + 2), denominator (x − 2)(x − 3). The shared factor (x − 2) cancels — but x = 2 stays excluded, leaving a hole in the graph. On the simplified function 2(x + 2)/(x − 3), the hole sits at x = 2 with height 2(4)/(−1) = −8: the point (2, −8) is missing. The unshared factor (x − 3) gives a vertical asymptote x = 3, where the function blows up. The rule: cancel first, then classify — shared factors make holes, surviving denominator factors make asymptotes.

Horizontal and slant asymptotes

Compare degrees. If deg(num) < deg(den), the horizontal asymptote is y = 0. If the degrees are equal, y = (leading coefficient ratio) — for the function above, y = 2. If deg(num) = deg(den) + 1, polynomial division reveals a slant asymptote. Worked example: (x^2 + 2x + 5)/(x + 1) = x + 1 + 4/(x + 1). As x → ±∞, the remainder term 4/(x + 1) → 0, so the graph settles onto the line y = x + 1. A graph may cross a horizontal or slant asymptote in the middle (they describe ends, not barriers); it can never cross a vertical asymptote, because the function is undefined there.

Modelling: average cost settles down

A workshop has fixed costs of $500 per day plus $8 per item. The average cost per item when making x items is C(x) = (500 + 8x)/x = 8 + 500/x. Numerically: C(50) = 900/50 = $18, C(500) = 4500/500 = $9, C(5000) = $8.10. The horizontal asymptote y = 8 says the long-run average cost approaches the $8 marginal cost as the fixed $500 is spread thinner — asymptotes are economic predictions, not just graph decorations. Setting a target: C(x) ≤ 10 requires 500/x ≤ 2, i.e. at least 250 items a day.

Common mistakes

• Cancelling terms instead of factors. (x + 2)/(x + 3) does not simplify: there are no factors to cancel. Cancelling is division of numerator and denominator by a common factor only.
• Calling every excluded value an asymptote. Cancelled factors produce holes, as (2, −8) above — misreading a hole as a vertical asymptote breaks limit calculations in calculus.
• Dividing wrong for the slant asymptote. Using (3x^2 − x)/(x + 2): the quotient is 3x − 7 with remainder 14, since (x + 2)(3x − 7) = 3x^2 − x − 14. Check by multiplying back — every time.
• Assuming graphs cannot cross horizontal asymptotes. They describe end behaviour only; the function might cross y = 8 several times for small x before settling.
• Getting the even-degree sign wrong. For −3x^4 both ends fall to −∞; the negative coefficient flips both ends of the x^4 picture, it does not make them opposite.

Exam Experience

1. The end behaviour of f(x) = −2x^5 + 3x^2 − 1 is

• (A) as x → −∞, f → −∞; as x → +∞, f → −∞
• (B) as x → −∞, f → +∞; as x → +∞, f → −∞
• (C) as x → −∞, f → −∞; as x → +∞, f → +∞
• (D) as x → −∞, f → +∞; as x → +∞, f → +∞

Answer: (B). The leading term −2x^5 has odd degree and negative coefficient, so the ends are opposite and both flipped from y = x^5: rising on the left, falling on the right. The +3x^2 term is irrelevant at the ends.

2. The graph of f(x) = (x^2 − 9)/(x − 3) has

• (A) a vertical asymptote at x = 3
• (B) a hole at (3, 6)
• (C) a hole at (3, 0)
• (D) an x-intercept at x = 3

Answer: (B). Factor: (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3. The cancelled factor leaves a hole where x + 3 = 6, i.e. at (3, 6). There is no vertical asymptote since the function approaches a finite value, and x = 3 is not an intercept because f(3) is undefined.

3. The slant asymptote of f(x) = (3x^2 − x)/(x + 2) is

• (A) y = 3x − 7
• (B) y = 3x − 1
• (C) y = 3x + 5
• (D) there is no slant asymptote

Answer: (A). Divide: 3x^2 − x = (x + 2)(3x − 7) + 14, so f(x) = 3x − 7 + 14/(x + 2) and the remainder term vanishes at the ends. (B) comes from stopping the division after one step and ignoring the −6x the second term must absorb.

Constructed response. Let f(x) = (x^2 − 2x − 3)/(x^2 − 4x + 3). (a) State the domain of f. (b) Find all x- and y-intercepts. (c) Find every vertical asymptote and every hole, giving the full coordinates of each hole. (d) Find the horizontal asymptote and describe the end behaviour.

Scoring guide. Factor first: f(x) = (x − 3)(x + 1)/((x − 3)(x − 1)). (a) 1 point: domain excludes x = 3 and x = 1. (b) 2 points: x-intercept where the numerator vanishes (and f defined): x = −1 — 1 point; y-intercept f(0) = −3/3 = −1 — 1 point. (c) 2 points: the shared factor (x − 3) gives a hole at x = 3 with height (3 + 1)/(3 − 1) = 2, the point (3, 2) — 1 point; the surviving denominator factor gives the vertical asymptote x = 1 — 1 point. (d) 1 point: equal degrees with leading ratio 1, so y = 1, and f(x) → 1 at both ends.

Going deeper: sketching a rational function completely

Worked example — full protocol. Sketch f(x) = (2x − 6)/(x + 1). Domain: x ≠ −1. Intercepts: x-intercept where 2x − 6 = 0, x = 3; y-intercept f(0) = −6. Asymptotes: vertical x = −1 (denominator zero, numerator −8 there, nonzero); horizontal y = 2 (equal degrees, ratio 2/1). Behaviour near the vertical asymptote: at x = −0.9, f = (−7.8)/(0.1) = −78 (plunging); at x = −1.1, f = (−8.2)/(−0.1) = 82 (soaring) — so the right branch enters from −∞ and the left branch rises to +∞. Sign chart on the intervals cut by −1 and 3: f is positive on (−∞, −1) (check f(−2) = (−10)/(−1) = 10 > 0 ✓), negative on (−1, 3) (f(0) = −6 ✓), positive again on (3, ∞) (f(4) = 2/5 ✓). The sketch assembles itself: on (−∞, −1) the graph approaches y = 2 from above as x → −∞, passes through (−2, 10), and climbs to +∞ at the asymptote; on (−1, ∞) it rises from −∞ through (0, −6) and (3, 0), settling towards y = 2 from below. Every feature came from the analysed form, not from plotting dozens of points.

Sign charts solve inequalities too. (x − 1)/(x + 3) > 0: the expression changes sign only at x = 1 (zero) and x = −3 (undefined). Testing intervals: positive on (−∞, −3), negative on (−3, 1), positive on (1, ∞). Solution: x < −3 or x > 1. The same chart technology will solve polynomial and rational inequalities throughout calculus optimisation.

Workshop: inverting a rational function

Rational functions with matching degrees often have rational inverses. For f(x) = (2x − 6)/(x + 1) (Lesson X's sketch): set y = (2x − 6)/(x + 1), swap and solve: x = (2y − 6)/(y + 1) → x(y + 1) = 2y − 6 → xy − 2y = −6 − x → y(x − 2) = −(x + 6) → y = (x + 6)/(2 − x). So f^−1(x) = (x + 6)/(2 − x). Verify: f^−1(−6) = 0/8 = 0 = f(0) inverted ✓. And observe the symmetry of the features: f has vertical asymptote x = −1 and horizontal y = 2; f^−1 has vertical asymptote x = 2 and horizontal y = −1 — inversion swaps asymptotes exactly as it swaps domain and range.