Rational models appear in problems about concentration, cost, and population. Two readings matter most. The initial value is the output at t = 0, so evaluate f(0). The long-run behavior is the limit as t grows without bound, which is the horizontal asymptote. In context, that asymptote is often a maximum that the quantity approaches but never reaches, or a floor such as a lowest average cost.
A rational function is a quotient of two polynomials, f(x)/g(x). When the numerator’s degree is at least the denominator’s, polynomial long division rewrites it as quotient + remainder/divisor, where the remainder has a smaller degree than the divisor. Divide the leading term of what remains by the leading term of the divisor, multiply back, subtract, and repeat until the leftover has a smaller degree. This form shows the end behavior at a glance, because the remainder fraction fades toward 0 and only the quotient is left.
Worked Example: Rewrite (x² + 5x + 7)/(x + 2).
x² ÷ x = x. Then x(x + 2) = x² + 2x, and subtracting leaves 3x + 7.
3x ÷ x = 3. Then 3(x + 2) = 3x + 6, and subtracting leaves a remainder of 1.
(x² + 5x + 7)/(x + 2) = x + 3 + 1x + 2
1. Rewrite (x² + 3x − 1)/(x + 1) using long division.
Explanation: x² ÷ x = x, and x(x + 1) = x² + x leaves 2x − 1. Then 2x ÷ x = 2, and 2(x + 1) = 2x + 2 leaves −1 − 2 = −3. So the result is x + 2 − 3/(x + 1). Check: (x + 1)(x + 2) = x² + 3x + 2, and 2 − 3 = −1.
2. What are the quotient and remainder when 2x³ − x² + 4 is divided by x² + 1?
Explanation: Write the dividend with a 0x term: 2x³ − x² + 0x + 4. First, 2x³ ÷ x² = 2x, and 2x(x² + 1) = 2x³ + 2x leaves −x² − 2x + 4. Next, −x² ÷ x² = −1, and −1(x² + 1) leaves −2x + 5. The remainder has degree 1, which is less than 2, so we stop.
3. Which expression is equivalent to (3x² − 2x + 8)/(x − 2)?
Explanation: 3x² ÷ x = 3x, and 3x(x − 2) = 3x² − 6x leaves 4x + 8. Then 4x ÷ x = 4, and 4(x − 2) = 4x − 8 leaves a remainder of 16. So the result is 3x + 4 + 16/(x − 2).
The real zeros of a rational function come from its numerator, but only at inputs that are allowed, meaning the denominator is not 0 there. So to find x-intercepts, set the numerator equal to 0, solve, and throw out any solution that also makes the denominator 0. For the y-intercept, evaluate f(0), if it is defined.
Worked Example: Find the intercepts of f(x) = (x − 2)(x + 3)/(x² + 1).
Numerator: (x − 2)(x + 3) = 0 gives x = 2 and x = −3. The denominator x² + 1 is never 0, so both count.
y-intercept: f(0) = (−2)(3)/1 = −6.
x-intercepts (2, 0) and (−3, 0); y-intercept (0, −6)
1. What are the intercepts of g(x) = (x² − 25)/(x + 1)?
Explanation: Numerator: x² − 25 = 0 gives x = ±5, and the denominator is not 0 at either. For the y-intercept, g(0) = −25/1 = −25.
2. What is the x-intercept of h(x) = (x − 4)(x + 1)/(x − 4)?
Explanation: The numerator is 0 at x = 4 and x = −1. But x = 4 also makes the denominator 0, so h(4) is undefined and gives no intercept (it is a hole). Only x = −1 works, so the x-intercept is (−1, 0).
3. What are the intercepts of f(x) = (3x − 12)/(x² + 4)?
Explanation: Numerator: 3x − 12 = 0 gives x = 4 (the denominator is 8 there, so it is fine). For the y-intercept, f(0) = −12/4 = −3.
A rational function has a vertical asymptote at x = a when the denominator is 0 at a but the numerator is not. Near it, the outputs grow without bound, written limx→a f(x) = ∞ or −∞. Approaching from the left is written x→a−, and from the right x→a+. To find which way the graph goes, check the sign of the numerator and each denominator factor at a test input just to the left and just to the right of a. The graph never crosses a vertical asymptote.
Worked Example: Find the vertical asymptotes of f(x) = (x + 1)/((x − 1)(x + 3)) and describe the behavior at x = 1.
The denominator is 0 at x = 1 and x = −3, and the numerator (x + 1) is not 0 at either.
Just right of 1 (x = 1.01): the signs are (+)/((+)(+)), so f → +∞. Just left of 1 (x = 0.99): (+)/((−)(+)), so f → −∞.
Vertical asymptotes: x = 1 and x = −3; limx→1− f(x) = −∞ and limx→1+ f(x) = ∞
1. What are the vertical asymptotes of g(x) = (x − 5)/(x² − x − 12)?
Explanation: The denominator factors as (x − 4)(x + 3), which is 0 at x = 4 and x = −3. The numerator x − 5 is not 0 at either, so both are vertical asymptotes. The zero at x = 5 is an x-intercept, not an asymptote.
2. Based on the graph of g(x) = −2/(x − 3)², what is limx→3 g(x)?
Explanation: The graph falls without bound on both sides of x = 3. Algebraically, (x − 3)² is a small positive number, and dividing −2 by it gives a large negative number. So the limit is −∞.
3. For f(x) = x/(x − 2), what happens as x approaches 2 from the left and from the right?
Explanation: Near x = 2 the numerator is about 2, which is positive. From the left (x = 1.99) the denominator is a small negative number, so f is large and negative. From the right (x = 2.01) the denominator is a small positive number, so f is large and positive.
If a factor appears in both the numerator and the denominator, it cancels, and the graph has a hole (a removable discontinuity) where that factor is 0. Always factor first. The hole’s x-coordinate comes from the canceled factor, and its y-coordinate comes from plugging that x into what is left after canceling. Any factor left in the denominator still creates a vertical asymptote.
Worked Example: Find the discontinuity of h(x) = (x² − x − 6)/(x − 3).
Factor: (x − 3)(x + 2)/(x − 3). The factor (x − 3) cancels, so there is a hole at x = 3.
What is left is x + 2, and at x = 3 that equals 5.
Hole at (3, 5), and no vertical asymptote
1. Where is the hole in f(x) = (x² − 16)/(x − 4)?
Explanation: Factor: (x − 4)(x + 4)/(x − 4). After canceling, f(x) = x + 4. At x = 4 that gives 8, so the hole is at (4, 8).
2. Which statement is true about g(x) = (x² − 2x − 3)/((x + 1)(x − 5))?
Explanation: The numerator factors as (x − 3)(x + 1), so (x + 1) cancels and x = −1 is a hole. What is left is (x − 3)/(x − 5), which equals (−4)/(−6) = 2/3 at x = −1. The factor (x − 5) remains in the denominator, so x = 5 is a vertical asymptote.
3. Which function has a hole at x = 3 and a vertical asymptote at x = −2?
Explanation: A hole at 3 needs the factor (x − 3) in both the numerator and denominator, and an asymptote at −2 needs (x + 2) only in the denominator. Choice A does both. B has no cancellation, C has the hole and asymptote swapped, and D has no hole.
Compare the degree n of the numerator with the degree m of the denominator. If n < m, there is a horizontal asymptote at y = 0. If n = m, the horizontal asymptote is y = (leading coefficient of the numerator)/(leading coefficient of the denominator). If n > m, there is no horizontal asymptote. When n = m + 1, long division gives a quotient that is linear, and its line is the slant asymptote. These are the limits at infinity: for example, if the horizontal asymptote is y = b, then limx→∞ f(x) = b and limx→−∞ f(x) = b. A graph can cross a horizontal or slant asymptote, but never a vertical one.
Worked Example 1: Find the horizontal asymptote of f(x) = 2x²/(x² − 4).
Both degrees are 2, so the asymptote is the ratio of the leading coefficients, 2/1.
y = 2, so limx→∞ f(x) = 2 and limx→−∞ f(x) = 2
Worked Example 2: Find the end behavior asymptote of j(x) = (x² + 1)/(x − 1).
The numerator has degree 2 and the denominator degree 1, so there is no horizontal asymptote. Long division gives x + 1 with remainder 2.
j(x) = x + 1 + 2/(x − 1), so the slant asymptote is y = x + 1
1. What is limx→∞ (4x³ − x)/(2x³ + 5)?
Explanation: The degrees are equal (3 and 3), so the limit is the ratio of the leading coefficients: 4/2 = 2.
2. What is limx→∞ (3x + 1)/(x² − 7)?
Explanation: The numerator has degree 1 and the denominator degree 2. With n < m, the denominator grows faster, so the outputs approach 0.
3. Which function has the horizontal asymptote y = 3/2?
Explanation: A horizontal asymptote equal to 3/2 needs equal degrees with leading coefficients 3 over 2. Only choice A has that. B gives y = 0, C has no horizontal asymptote, and D gives y = 2/3.
4. What is the slant asymptote of f(x) = (2x² − x + 3)/(x − 1)?
Explanation: 2x² ÷ x = 2x, and 2x(x − 1) = 2x² − 2x leaves x + 3. Then x ÷ x = 1, and 1(x − 1) leaves 4. So f(x) = 2x + 1 + 4/(x − 1), and the slant asymptote is y = 2x + 1.
Use a seven-step plan: (1) factor and remove any common factors to find holes, (2) find the x- and y-intercepts, (3) find the horizontal or slant asymptote from the degrees, (4) find the vertical asymptotes from the remaining denominator factors, (5) check symmetry by computing f(−x), (6) use test points in each region the vertical asymptotes create, and (7) sketch the curve through the intercepts, approaching each asymptote.
Worked Example: Sketch f(x) = (x² − 4)/(x² − 1).
1. Factor: (x − 2)(x + 2)/((x − 1)(x + 1)). Nothing cancels, so no holes.
2. x-intercepts: x = ±2. y-intercept: f(0) = (−4)/(−1) = 4.
3. Equal degrees, so the horizontal asymptote is y = 1/1 = 1.
4. Vertical asymptotes: x = 1 and x = −1.
5. f(−x) = f(x), so f is even and the graph is symmetric about the y-axis.
6. Test x = 1.5: f = (−1.75)/(1.25) = −1.4, so the graph is below the axis between 1 and 2. Test x = 3: f = 5/8, so it is above the axis (and below y = 1) for x > 2.
The sketch is shown below.
1. Which features describe g(x) = (x − 1)(x + 3)/((x − 1)(x − 2))?
Explanation: (x − 1) cancels, so there is a hole at x = 1. What is left is (x + 3)/(x − 2), which equals 4/(−1) = −4 at x = 1, so the hole is (1, −4). The remaining denominator factor gives the vertical asymptote x = 2, the numerator gives the x-intercept (−3, 0), and the equal degrees give y = 1.
2. What symmetry does h(x) = x/(x² + 1) have?
Explanation: h(−x) = (−x)/((−x)² + 1) = −x/(x² + 1) = −h(x). So h is odd, and its graph is symmetric about the origin.
3. For r(x) = (x − 2)/((x + 1)(x − 4)), on which intervals is the graph above the x-axis?
Explanation: The critical values are −1, 2, and 4. Test x = 0: (−2)/((1)(−4)) = 1/2, which is positive, so (−1, 2) is above. Test x = 3: (1)/((4)(−1)) is negative, so (2, 4) is below. Test x = 5: (3)/((6)(1)) is positive, so (4, ∞) is above. Test x = −2: (−4)/((−1)(−6)) is negative, so (−∞, −1) is below.
A rational inequality asks where the graph of f is above or below the x-axis. (1) Move everything to one side so the other side is 0, and combine into a single fraction. (2) Find the critical values: the zeros of the numerator and the zeros of the denominator. (3) Place them on a number line and test one point in each interval. (4) Keep the intervals that match. Numerator zeros are included for ≤ or ≥ (closed circles). Denominator zeros are never included (open circles), because the function is undefined there.
Worked Example: Solve (x + 1)/(x − 2) ≤ 0.
Critical values: x = −1 (numerator) and x = 2 (denominator). Test x = −2: (−1)/(−4) is positive. Test x = 0: (1)/(−2) is negative. Test x = 3: (4)/(1) is positive.
We want negative or zero values. x = −1 gives 0, so include it. x = 2 is undefined, so exclude it.
Solution: [−1, 2)
1. Solve (x − 3)/(x + 2) > 0.
Explanation: Critical values: −2 and 3. Test x = 0: (−3)/(2) is negative. Test x = −3: (−6)/(−1) is positive. Test x = 4: (1)/(6) is positive. The inequality is strict, so both endpoints are open: (−∞, −2) ∪ (3, ∞).
2. Solve 2x/(x − 1) ≥ 1.
Explanation: Subtract 1 and combine: (2x − (x − 1))/(x − 1) = (x + 1)/(x − 1) ≥ 0. Critical values: −1 and 1. The expression is positive for x < −1 and x > 1, and negative between. x = −1 gives 0, so it is included, but x = 1 makes the denominator 0, so it is excluded.
3. Solve 3/(x + 1) ≤ 1.
Explanation: Subtract 1: (3 − (x + 1))/(x + 1) = (2 − x)/(x + 1) ≤ 0. Critical values: 2 and −1. Test x = 0: 2/1 is positive (not wanted). Test x = 3: (−1)/(4) is negative (wanted). Test x = −2: (4)/(−1) is negative (wanted). x = 2 gives 0, so include it, and x = −1 is undefined, so exclude it.
Rational models appear in problems about concentration, cost, and population. Two readings matter most. The initial value is the output at t = 0, so evaluate f(0). The long-run behavior is the limit as t grows without bound, which is the horizontal asymptote. In context, that asymptote is often a maximum that the quantity approaches but never reaches, or a floor such as a lowest average cost.
Worked Example: A population measured in hundreds is modeled by P(t) = (50t + 20)/(t + 4), where t is in months. Find the initial value and the long-run value.
Initial value: P(0) = 20/4 = 5.
Long run: the degrees are equal, so the horizontal asymptote is 50/1 = 50.
The population starts at 5 hundred and approaches 50 hundred.
1. A drug concentration is C(t) = (9t + 12)/(t + 3), where t is hours. What is the initial concentration?
Explanation: The initial value is C(0) = (9(0) + 12)/(0 + 3) = 12/3 = 4.
2. The average cost per item to make n items is A(n) = (1200 + 5n)/n dollars. What happens to A(n) as n grows very large?
Explanation: The degrees are equal (1 and 1), so the limit is the ratio of leading coefficients: 5/1 = 5. The average cost per item approaches $5, because the fixed cost of $1,200 gets spread over more and more items.
3. A model for the size of a group is P(t) = 300t/(t + 6). Which statement is correct?
Explanation: P(0) = 0/6 = 0. The degrees are equal, so the horizontal asymptote is 300/1 = 300. The size gets close to 300 but never exceeds it, so 300 is the sustainable maximum.
Both happen where the denominator is 0. If the factor also cancels with the numerator, it is a hole, and the graph just has a missing point. If the factor stays in the denominator after canceling, it is a vertical asymptote, and the outputs grow without bound.
When the numerator’s degree is exactly one more than the denominator’s. Divide with long division, and the linear quotient is the slant asymptote.
No. The end behavior is controlled by the degrees, which give exactly one of the cases: horizontal asymptote y = 0, horizontal asymptote y = ratio, or a slant asymptote (or a higher-degree curve).
The Rational Functions quiz in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.
