Rational Functions & Oblique Asymptotes

Figure 2.2: Rational Function with Vertical and Oblique Asymptotes

1. The Global Anatomy of Rational Functions

A rational function $f(x) = \frac{P(x)}{Q(x)}$ is the quotient of two polynomials $P(x)$ and $Q(x)$ with no common factors:

Vertical Asymptotes (VA): Occur at real zeros of the denominator $Q(c) = 0$. The line $x = c$ is a vertical asymptote because $\lim_{x \to c^\pm} f(x) = \pm\infty$.
Horizontal Asymptotes (HA):
• If $\deg(P) < \deg(Q)$: the horizontal asymptote is $y = 0$ ($x$-axis).
• If $\deg(P) = \deg(Q)$: the horizontal line is $y = \frac{a_n}{b_n}$ (ratio of leading coefficients).
Oblique (Slant) Asymptotes (OA): Occur if and only if the degree of the numerator is exactly 1 greater than the degree of the denominator: $$\deg(P) = \deg(Q) + 1$$

2. Derivation of Oblique Asymptotes via Polynomial Long Division

When $\deg(P) = \deg(Q) + 1$, dividing $P(x)$ by $Q(x)$ yields a linear quotient $(mx + c)$ plus a strictly proper remainder fraction $\frac{R(x)}{Q(x)}$ where $\deg(R) < \deg(Q)$:

$$f(x) = \frac{P(x)}{Q(x)} = mx + c + \frac{R(x)}{Q(x)}$$

Taking the limit as $x \to \pm\infty$:

$$\lim_{x \to \pm\infty} [f(x) - (mx + c)] = \lim_{x \to \pm\infty} \frac{R(x)}{Q(x)} = 0$$

Because the vertical difference between the curve and the line approaches zero, $y = mx + c$ is an oblique asymptote.

3. Direction of Asymptotic Approach & Curve Intersections

Approach from Above vs. Below: The sign of the remainder term $\frac{R(x)}{Q(x)}$ dictates whether the curve lies above or below the asymptote:
- If $\frac{R(x)}{Q(x)} > 0$ as $x \to +\infty$, the curve approaches the oblique asymptote from above ($f(x) > mx + c$).
- If $\frac{R(x)}{Q(x)} < 0$ as $x \to +\infty$, the curve approaches the oblique asymptote from below ($f(x) < mx + c$).
Can a Curve Cross an Asymptote?
• Vertical Asymptote: A curve can NEVER cross a vertical asymptote, because the function is undefined there.
• Oblique / Horizontal Asymptote: A curve CAN cross an oblique asymptote at finite values of $x$. This occurs wherever the remainder term is zero: $$f(x) = mx + c \iff \frac{R(x)}{Q(x)} = 0 \iff R(x) = 0$$

4. Range of Rational Curves via Quadratic Discriminant Method

For rational functions with quadratic numerator and linear denominator $y = \frac{ax^2 + bx + c}{dx + e}$, we can find the range without calculus by rewriting the equation as a quadratic in $x$:

$$y(dx + e) = ax^2 + bx + c \implies ax^2 + (b - dy)x + (c - ey) = 0$$

Because $x$ must be real, the discriminant of this quadratic in $x$ must be non-negative: $\Delta_x \ge 0$:

$$(b - dy)^2 - 4a(c - ey) \ge 0$$

Solving this quadratic inequality in $y$ directly yields the exact range of the function, confirming the locations of local maximum and minimum values!

Worked Examples

Worked Example 1: Long Division & Asymptotic Approach Direction

Question: Consider the rational function $f(x) = \frac{2x^2 + 5x - 1}{x + 2}$.
(a) Find the equations of all asymptotes of $y = f(x)$. [4 marks]
(b) Determine whether the curve approaches the oblique asymptote from above or below as $x \to +\infty$ and as $x \to -\infty$. [2 marks]

Solution:

(a) Finding Asymptotes:
• Vertical Asymptote: Denominator is zero at $x + 2 = 0 \implies x = -2$. [A1]
Numerator at $x = -2$: $2(-2)^2 + 5(-2) - 1 = 8 - 10 - 1 = -3 \neq 0$. Thus, $x = -2$ is a vertical asymptote.
• Oblique Asymptote: Perform polynomial long division: [M1]

$$\begin{array}{r} 2x + 1\phantom{)} \\ x + 2{\overline{\smash{\big)}\,2x^2 + 5x - 1\phantom{)}}} \\ \underline{-(2x^2 + 4x)\phantom{-1)}} \\ x - 1\phantom{)} \\ \underline{-(x + 2)} \\ -3\phantom{)} \end{array}$$ $$f(x) = 2x + 1 - \frac{3}{x + 2} \quad [\text{A1}]$$

As $x \to \pm\infty$, $-\frac{3}{x + 2} \to 0$. Hence the oblique asymptote is the line $y = 2x + 1$. [A1]

(b) Approach Direction: The vertical difference is $d(x) = f(x) - (2x + 1) = -\frac{3}{x + 2}$. [M1]

As $x \to +\infty$: $x + 2 > 0 \implies -\frac{3}{x + 2} < 0$. The curve lies below the oblique asymptote. [A1]
As $x \to -\infty$: $x + 2 < 0 \implies -\frac{3}{x + 2} > 0$. The curve lies above the oblique asymptote. [A1]

Worked Example 2: Curve Crossing its Oblique Asymptote

Question: The function $g(x) = \frac{x^3 + 2x^2 - 5x + 3}{x^2 + 1}$ has an oblique asymptote.
(a) Find the equation of the oblique asymptote. [3 marks]
(b) Determine the exact coordinates of the point where the curve crosses this asymptote. [2 marks]

Solution:

(a) Division: Divide $(x^3 + 2x^2 - 5x + 3)$ by $(x^2 + 1)$: [M1]

$$x^3 + 2x^2 - 5x + 3 = x(x^2 + 1) + 2(x^2 + 1) - 6x + 1 = (x + 2)(x^2 + 1) + (-6x + 1)$$ $$g(x) = x + 2 + \frac{-6x + 1}{x^2 + 1} \quad [\text{A1}]$$

As $x \to \pm\infty$, $\frac{-6x + 1}{x^2 + 1} \to 0$. Oblique asymptote: $y = x + 2$. [A1]

(b) Point of Intersection: Set curve equal to asymptote: $g(x) = x + 2$: [M1]

$$\frac{-6x + 1}{x^2 + 1} = 0 \implies -6x + 1 = 0 \implies x = \frac{1}{6} \quad [\text{A1}]$$

Evaluate $y$ on the asymptote: $y = \frac{1}{6} + 2 = \frac{13}{6}$.
Intersection point: $\left(\frac{1}{6}, \frac{13}{6}\right)$. [A1]

Worked Example 3: Stationary Points & Second Derivative Test

Question: For the function $y = \frac{x^2 - x + 4}{x - 1}$ for $x \neq 1$:
(a) Find the coordinates of all stationary points. [4 marks]
(b) Use the second derivative test to determine the nature of each stationary point. [2 marks]

Solution:

(a) Decompose & Differentiate: Divide first: [M1]

$$y = \frac{x(x - 1) + 4}{x - 1} = x + \frac{4}{x - 1} = x + 4(x - 1)^{-1}$$ $$\frac{dy}{dx} = 1 - 4(x - 1)^{-2} = 1 - \frac{4}{(x - 1)^2} \quad [\text{A1}]$$

Set $\frac{dy}{dx} = 0$: [M1]

$$1 = \frac{4}{(x - 1)^2} \implies (x - 1)^2 = 4 \implies x - 1 = \pm 2$$

$x = 1 + 2 = 3 \implies y = 3 + \frac{4}{3 - 1} = 3 + 2 = 5$. Stationary point: $(3, 5)$. [A1]
$x = 1 - 2 = -1 \implies y = -1 + \frac{4}{-1 - 1} = -1 - 2 = -3$. Stationary point: $(-1, -3)$. [A1]

(b) Nature via Second Derivative: [M1]

$$\frac{d^2 y}{dx^2} = \frac{d}{dx}\left[1 - 4(x - 1)^{-2}\right] = 8(x - 1)^{-3} = \frac{8}{(x - 1)^3}$$

At $x = 3$: $\frac{d^2 y}{dx^2} = \frac{8}{(2)^3} = 1 > 0 \implies (3, 5)$ is a local minimum. [A1]
At $x = -1$: $\frac{d^2 y}{dx^2} = \frac{8}{(-2)^3} = -1 < 0 \implies (-1, -3)$ is a local maximum. [A1]

Examiner Advice: Notice that the local maximum $y = -3$ has a lower $y$-value than the local minimum $y = 5$! This is typical of rational functions split by a vertical asymptote ($x = 1$).

Worked Example 4: Finding Range via the Quadratic Discriminant

Question: Find the exact range of $f(x) = \frac{x^2 - 2x + 5}{x - 1}$ for $x \neq 1$ using the quadratic discriminant method. [5 marks]

Solution:

Let $y = \frac{x^2 - 2x + 5}{x - 1}$. Multiply by $(x - 1)$: [M1]

$$y(x - 1) = x^2 - 2x + 5 \implies yx - y = x^2 - 2x + 5$$

Rearrange into a quadratic equation in $x$: [M1]

$$x^2 - (2 + y)x + (5 + y) = 0 \quad [\text{A1}]$$

Since $x \in \mathbb{R}$, this quadratic must have real roots $\implies \Delta_x \ge 0$: [M1]

$$\Delta_x = [-(2 + y)]^2 - 4(1)(5 + y) \ge 0$$ $$(4 + 4y + y^2) - (20 + 4y) \ge 0$$ $$y^2 - 16 \ge 0 \iff (y - 4)(y + 4) \ge 0 \quad [\text{A1}]$$

The inequality holds outside the roots: [A1]

$$y \le -4 \quad \text{or} \quad y \ge 4$$

Therefore, the exact range of $f$ is $(-\infty, -4] \cup [4, \infty)$.

Worked Example 5: Parameterized Rational Function

Question: A rational curve has equation $y = \frac{2x^2 + kx + 8}{x - 2}$ where $k \in \mathbb{R}$.
(a) Find the oblique asymptote in terms of $k$. [2 marks]
(b) Find the range of values of $k$ for which the curve has two distinct turning points. [3 marks]

Solution:

(a) Oblique Asymptote: Divide $(2x^2 + kx + 8)$ by $(x - 2)$: [M1]

$$2x^2 + kx + 8 = 2x(x - 2) + (k + 4)x + 8 = (2x + k + 4)(x - 2) + (2k + 16)$$ $$y = 2x + k + 4 + \frac{2k + 16}{x - 2} \quad [\text{A1}]$$

Oblique asymptote: $y = 2x + k + 4$. [A1]

(b) Condition for Two Distinct Turning Points: Differentiate $y$: [M1]

$$\frac{dy}{dx} = 2 - \frac{2k + 16}{(x - 2)^2}$$

Set $\frac{dy}{dx} = 0$: [M1]

$$2 = \frac{2k + 16}{(x - 2)^2} \implies (x - 2)^2 = \frac{2k + 16}{2} = k + 8$$

For two distinct real solutions for $x$, the right-hand side must be strictly positive: [R1]

$$k + 8 > 0 \implies k > -8 \quad [\text{A1}]$$

If $k = -8$, the remainder is zero, the fraction simplifies to a straight line with a hole at $x = 2$, and there are no turning points.

Practice Exercises (10 Tiered Questions)

Try these 10 exam-style questions, from core skills to harder problems. Open each question to view its solution.

Exercise 1 · Finding Oblique Equation

Find the equation of the oblique asymptote of $f(x) = \frac{3x^2 - 4x + 5}{x - 1}$.

Show Full Solution

Divide $(3x^2 - 4x + 5)$ by $(x - 1)$: [M1]
$3x^2 - 4x + 5 = 3x(x - 1) - x + 5 = 3x(x - 1) - 1(x - 1) + 4 = (3x - 1)(x - 1) + 4$.
$f(x) = 3x - 1 + \frac{4}{x - 1}$. [A1]
Oblique asymptote: $y = 3x - 1$. [A1]

Exercise 2 · Vertical Asymptote Behavior

For $g(x) = \frac{x^2 + 1}{x - 3}$, evaluate $\lim_{x \to 3^+} g(x)$ and $\lim_{x \to 3^-} g(x)$.

Show Full Solution

At $x \to 3$, numerator $\to 3^2 + 1 = 10 > 0$. [M1]
As $x \to 3^+$, denominator $(x - 3) \to 0^+$, so $\lim_{x \to 3^+} g(x) = +\infty$. [A1]
As $x \to 3^-$, denominator $(x - 3) \to 0^-$, so $\lim_{x \to 3^-} g(x) = -\infty$. [A1]

Exercise 3 · Identifying Asymptote Types

State whether each function has a horizontal or oblique asymptote, and write down its equation:
(a) $y = \frac{4x^2 + 1}{2x^2 - 3}$, (b) $y = \frac{x^3 + 2}{x^2 + 4}$, (c) $y = \frac{2x - 1}{x^2 + 5}$.

Show Full Solution

(a) Degrees equal (2 = 2): Horizontal asymptote $y = \frac{4}{2} = 2$. [A1]
(b) Degree num = 3, den = 2 ($3 = 2 + 1$): Oblique asymptote $y = x$ (since $x^3 + 2 = x(x^2 + 4) - 4x + 2$). [A1]
(c) Degree num < den (1 < 2): Horizontal asymptote $y = 0$. [A1]

Exercise 4 · Slant Asymptote with Negative Leading Term

Find the oblique asymptote of $h(x) = \frac{7 - 4x - 2x^2}{x + 3}$.

Show Full Solution

Rewrite numerator in standard order: $-2x^2 - 4x + 7$. [M1]
Divide by $(x + 3)$: $-2x^2 - 4x + 7 = -2x(x + 3) + 2x + 7 = -2x(x + 3) + 2(x + 3) + 1 = (-2x + 2)(x + 3) + 1$. [M1]
$h(x) = -2x + 2 + \frac{1}{x + 3}$.
Oblique asymptote is $y = -2x + 2$. [A1]

Exercise 5 · Stationary Points & Nature

Find the stationary points of $f(x) = \frac{x^2 + 9}{x}$ and determine their nature using the second derivative.

Show Full Solution

$f(x) = x + 9x^{-1}$. [M1]
$f'(x) = 1 - 9x^{-2} = 1 - \frac{9}{x^2}$.
$f'(x) = 0 \implies x^2 = 9 \implies x = \pm 3$. [A1]
$f''(x) = 18x^{-3} = \frac{18}{x^3}$. [M1]
• At $x = 3$: $f(3) = 6$, $f''(3) = \frac{18}{27} > 0 \implies (3, 6)$ is a local minimum. [A1]
• At $x = -3$: $f(-3) = -6$, $f''(-3) = -\frac{18}{27} < 0 \implies (-3, -6)$ is a local maximum. [A1]

Exercise 6 · Determining Approach Direction

Show that the curve $y = \frac{x^2 - 2x + 5}{x - 1}$ never intersects its oblique asymptote, and determine its position relative to the asymptote for $x > 1$ and $x < 1$.

Show Full Solution

$\frac{x^2 - 2x + 5}{x - 1} = \frac{(x - 1)^2 + 4}{x - 1} = (x - 1) + \frac{4}{x - 1}$. [M1]
Oblique asymptote: $y = x - 1$.
Difference $y - (x - 1) = \frac{4}{x - 1} \neq 0$ for all real $x$. Hence the curve never crosses its asymptote. [A1]
• For $x > 1$: $\frac{4}{x - 1} > 0 \implies$ curve lies strictly above the asymptote. [A1]
• For $x < 1$: $\frac{4}{x - 1} < 0 \implies$ curve lies strictly below the asymptote. [A1]

Exercise 7 · Parameterized Asymptotes

The rational curve $y = \frac{a x^2 + b x - 8}{x - 2}$ has an oblique asymptote $y = 3x + 10$. Find the values of constants $a$ and $b$.

Show Full Solution

Division of $(a x^2 + b x - 8)$ by $(x - 2)$: [M1]
$a x^2 + b x - 8 = (3x + 10)(x - 2) + R$. [M1]
Expand: $(3x + 10)(x - 2) = 3x^2 - 6x + 10x - 20 = 3x^2 + 4x - 20$.
Comparing coefficients of $x^2$ and $x$:
$a = 3$. [A1]
$b = 4$. [A1]
(Remainder $R = -8 - (-20) = 12$).

Exercise 8 · No Real Stationary Points

Show that the curve $f(x) = \frac{x^2 + 2x - 1}{x + 1}$ has no local extrema, and explain this geometrically.

Show Full Solution

$f(x) = \frac{(x+1)^2 - 2}{x + 1} = (x + 1) - \frac{2}{x + 1} = x + 1 - 2(x + 1)^{-1}$. [M1]
$f'(x) = 1 + 2(x + 1)^{-2} = 1 + \frac{2}{(x + 1)^2}$. [M1]
Since $(x + 1)^2 > 0$ for all $x \neq -1$, $\frac{2}{(x + 1)^2} > 0$, which means $f'(x) \ge 1 > 0$ everywhere on its domain. [R1]
Because the derivative is strictly positive, $f(x)$ is strictly increasing on $(-\infty, -1)$ and on $(-1, \infty)$, so it has no stationary points / local extrema. [A1]

Exercise 9 · Quadratic Denominator Oblique Asymptote

Find the oblique asymptote and the point where the curve intersects it for $y = \frac{2x^3 - x^2 + 4x - 5}{x^2 + 2}$.

Show Full Solution

Divide numerator by $(x^2 + 2)$: [M1]
$2x^3 - x^2 + 4x - 5 = 2x(x^2 + 2) - (x^2 + 2) - 3 = (2x - 1)(x^2 + 2) - 3$. [M1]
$y = 2x - 1 - \frac{3}{x^2 + 2}$.
Oblique asymptote is $y = 2x - 1$. [A1]
Intersection with asymptote requires remainder term $= 0$:
$-\frac{3}{x^2 + 2} = 0$, which has no real solutions (since numerator is $-3 \neq 0$).
Hence the curve never intersects its oblique asymptote. [A1]

Exercise 10 · Full Curve Sketching & Range

For $f(x) = \frac{x^2 - 4}{x - 1}$:
(a) Find intercepts and all asymptotes.
(b) Show that $f(x)$ has no stationary points.
(c) State the range of $f(x)$.

Show Full Solution

(a) Intercepts & Asymptotes: [M1]
• $y$-intercept: $f(0) = 4$. $x$-intercepts: $x = \pm 2$. [A1]
• Vertical asymptote: $x = 1$.
• $f(x) = \frac{(x - 1)(x + 1) - 3}{x - 1} = x + 1 - \frac{3}{x - 1}$. Oblique asymptote: $y = x + 1$. [A1]

(b) Stationary points:
$f'(x) = 1 + \frac{3}{(x - 1)^2}$. Since $3/(x-1)^2 > 0$, $f'(x) > 0$ everywhere. No stationary points! [A1]

(c) Range:
As $x \to 1^+$, $f(x) \to -\infty$; as $x \to +\infty$, $f(x) \to +\infty$. Continuous on $(1, \infty) \implies$ takes all real values. Same on $(-\infty, 1)$.
Range is $\mathbb{R}$ (all real numbers). [A1]

IB Formula Booklet References (Topic 2)

Syllabus AHL 2.12: Rational Functions & Asymptotes

The IB Formula Booklet does not provide the oblique asymptote formula. The syllabus requires students to apply polynomial division and limit arguments:

$$f(x) = mx + c + \frac{R(x)}{Q(x)} \implies y = mx + c \text{ is the oblique asymptote if } \lim_{x \to \pm\infty} \frac{R(x)}{Q(x)} = 0$$

Key Formulas & Identities

1. Condition for Oblique Asymptotes

$$\deg(\text{Numerator}) = \deg(\text{Denominator}) + 1$$

2. Relative Position to Asymptote

$\frac{R(x)}{Q(x)} > 0 \implies$ curve lies above the asymptote.
$\frac{R(x)}{Q(x)} < 0 \implies$ curve lies below the asymptote.
$\frac{R(x)}{Q(x)} = 0 \implies$ point where curve crosses the asymptote.

3. Range via Quadratic Discriminant

$$y = \frac{ax^2+bx+c}{dx+e} \implies ax^2 + (b-dy)x + (c-ey) = 0 \implies \Delta_x \ge 0$$

Common Exam Pitfalls

Horizontal vs Oblique Confusion

A rational function CANNOT have both a horizontal asymptote and an oblique asymptote. If the degree of the numerator exceeds the denominator by 1, there is no horizontal asymptote.

"Curves Cannot Cross Asymptotes" Myth

Curves can never cross a vertical asymptote (where the function is undefined), but curves CAN and frequently do cross horizontal and oblique asymptotes at finite values of $x$.

Differentiating Rational Functions Efficiently

When finding turning points of $f(x) = \frac{P(x)}{Q(x)}$, decompose into quotient + remainder first: $f(x) = mx + c + R(x)Q(x)^{-1}$. Differentiating this sum is significantly faster and less error-prone than applying the quotient rule to the entire fraction!

Figure 2.2: Rational Function with Vertical and Oblique Asymptotes

1. The Global Anatomy of Rational Functions

2. Derivation of Oblique Asymptotes via Polynomial Long Division

3. Direction of Asymptotic Approach & Curve Intersections

4. Range of Rational Curves via Quadratic Discriminant Method

Worked Examples

Worked Example 1: Long Division & Asymptotic Approach Direction

Worked Example 2: Curve Crossing its Oblique Asymptote

Worked Example 3: Stationary Points & Second Derivative Test

Worked Example 4: Finding Range via the Quadratic Discriminant

Worked Example 5: Parameterized Rational Function

Practice Exercises (10 Tiered Questions)

IB Formula Booklet References (Topic 2)

Syllabus AHL 2.12: Rational Functions & Asymptotes

Key Formulas & Identities

1. Condition for Oblique Asymptotes

2. Relative Position to Asymptote

3. Range via Quadratic Discriminant

Common Exam Pitfalls

Horizontal vs Oblique Confusion

"Curves Cannot Cross Asymptotes" Myth

Differentiating Rational Functions Efficiently

Topic Diagnostic Test

Section A: Diagnostic Multiple Choice Questions

Section B: Examination-Style Free Response Questions