Rational Functions — Complete Precalculus Guide
Domain, vertical asymptotes, holes, horizontal and slant asymptotes, intercepts, graphing strategy, end behavior, transformations of 1/x, and applications. Covers Stewart Precalculus Chapter 3 with worked examples and step-by-step solutions.
Quick Reference — The Essential Rules
Asymptotes vs. Holes
- Common factor cancels → hole
- Remaining denom zero → vertical asymptote
- Always factor first
Horizontal Asymptotes
- n < d → y = 0
- n = d → y = leading coeff ratio
- n > d → none (check for slant)
Intercepts
- Y-intercept: evaluate f(0)
- X-intercepts: set numerator = 0
- Use the simplified form
Topics on This Page
1. Definition and Domain
A rational function is any function that can be written as a ratio of two polynomials. If p(x) and q(x) are polynomials and q(x) is not identically zero, then f(x) = p(x) / q(x) is a rational function.
f(x) = p(x) / q(x)
p(x) and q(x) are polynomials · q(x) is not the zero polynomial
The domain of a rational function is the set of all real numbers for which the denominator is not zero. To find the domain:
Set the denominator q(x) equal to zero.
Solve for x. These are the excluded values.
Write the domain as all real numbers except those excluded values.
Example 1.1 — Find the domain of f(x) = (x + 3) / (x squared minus 9)
Set denominator = 0: x squared minus 9 = 0
Factor: (x minus 3)(x + 3) = 0 → x = 3 or x = minus 3
Domain: all real numbers except x = 3 and x = minus 3
Interval notation: (minus infinity, minus 3) union (minus 3, 3) union (3, infinity)
Example 1.2 — Find the domain of g(x) = (x squared + 1) / (x squared + 4)
Set denominator = 0: x squared + 4 = 0 → x squared = minus 4
No real solutions (x squared is always non-negative).
Domain: all real numbers, written as (minus infinity, infinity)
The denominator is always positive, so this function is defined everywhere.
Note on notation
In Stewart Precalculus, rational functions are typically assumed to be in their simplest form for the domain statement, but you must still identify holes and asymptotes separately after factoring. The domain excludes ALL x-values where the original denominator is zero — whether they produce holes or vertical asymptotes.
2. Vertical Asymptotes
A vertical asymptote is a vertical line x = a where the function grows without bound — f(x) approaches positive or negative infinity as x approaches a. This happens when the denominator approaches zero but the numerator does not.
The critical rule: factor and cancel first. Common factors that cancel give holes, not asymptotes. Only the factors remaining in the denominator after cancellation give vertical asymptotes.
Factor both numerator and denominator completely.
Cancel any common factors (these give holes, addressed next).
Set each remaining denominator factor equal to zero and solve. Each solution is a vertical asymptote.
Example 2.1 — f(x) = (x squared minus 4) / (x squared minus x minus 6)
Factor numerator: x squared minus 4 = (x minus 2)(x + 2)
Factor denominator: x squared minus x minus 6 = (x minus 3)(x + 2)
Common factor (x + 2) cancels → hole at x = minus 2
Remaining denominator factor: (x minus 3)
Set x minus 3 = 0 → x = 3
Vertical asymptote at x = 3. Hole at x = minus 2.
Most common mistake on vertical asymptotes
Students often skip factoring and just set the denominator equal to zero, then report all solutions as vertical asymptotes. This misses the critical step: canceled factors are holes, not asymptotes. On exams, this error costs full credit on the graphing problem.
3. Holes (Removable Discontinuities)
A hole, also called a removable discontinuity, occurs at x = a when the factor (x minus a) appears in both the numerator and the denominator and therefore cancels out. The original function is undefined at x = a, but the simplified function is defined there.
Graphically, a hole looks like an open circle — one missing point on an otherwise continuous curve. The graph approaches that point from both sides but never actually touches it.
How to Find a Hole
Factor numerator and denominator. Identify the common factor (x minus a) that cancels.
The x-coordinate of the hole is x = a (the value that makes the canceled factor zero).
To find the y-coordinate: substitute x = a into the SIMPLIFIED (reduced) rational function — NOT the original.
Write the hole as the ordered pair (a, y). Mark it as an open circle on the graph.
Example 3.1 — f(x) = (x squared minus 4) / (x minus 2)
Factor numerator: (x minus 2)(x + 2)
Denominator: (x minus 2)
Cancel (x minus 2): simplified form is x + 2, with x not equal to 2
Hole x-coordinate: x = 2
Hole y-coordinate: plug x = 2 into simplified form: 2 + 2 = 4
Hole at (2, 4). No vertical asymptote (denominator fully cancels).
Example 3.2 — f(x) = (x squared minus x minus 6) / (x squared minus 4)
Factor numerator: (x minus 3)(x + 2)
Factor denominator: (x minus 2)(x + 2)
Common factor (x + 2) cancels → simplified: (x minus 3) / (x minus 2), with x not equal to minus 2
Hole x-coordinate: x = minus 2
Hole y-coordinate: (minus 2 minus 3) / (minus 2 minus 2) = minus 5 / minus 4 = 5/4
Hole at (minus 2, 5/4). Vertical asymptote at x = 2.
4. Horizontal Asymptotes
A horizontal asymptote is a horizontal line y = k that the graph of f approaches as x goes to positive infinity or negative infinity. It describes the end behavior of the rational function — what the output looks like when the input gets very large in magnitude.
Let n = degree of the numerator and d = degree of the denominator. The three cases depend entirely on comparing n and d:
| Degree Comparison | Horizontal Asymptote | Example | Note |
|---|---|---|---|
| deg(num) less than deg(den) | y = 0 | (x + 1) / (x squared + 4) | Numerator degree 1, denominator degree 2 |
| deg(num) equals deg(den) | y = a / b (leading coefficient ratio) | (3x squared + 1) / (5x squared minus 2) | Horizontal asymptote y = 3/5 |
| deg(num) greater than deg(den) | None | (x squared minus 1) / (x minus 2) | Do long division; slant asymptote if deg diff = 1 |
Example 4.1 — Case 1: deg(num) < deg(den)
f(x) = (2x + 5) / (x squared + 1)
Numerator degree: 1. Denominator degree: 2.
Since 1 is less than 2, horizontal asymptote is y = 0.
As x grows large, the denominator dominates and f(x) approaches zero.
Example 4.2 — Case 2: deg(num) = deg(den)
f(x) = (3x squared minus 2) / (5x squared + 7)
Both have degree 2. Leading coefficient ratio: 3/5.
Horizontal asymptote: y = 3/5.
The constant terms and lower-degree terms become negligible.
Key distinction
Unlike vertical asymptotes, a rational function can cross its horizontal asymptote for some finite x-value. The horizontal asymptote only constrains the function as x grows very large — it does not prevent the graph from passing through that y-value at some point in the middle of the graph.
5. Slant (Oblique) Asymptotes
When the degree of the numerator is exactly one more than the degree of the denominator, there is no horizontal asymptote. Instead there is a slant asymptote (also called an oblique asymptote) — a non-horizontal line y = mx + b that the graph approaches as x goes to positive or negative infinity.
Method: Perform polynomial long division of the numerator by the denominator. The quotient (not the remainder) is the equation of the slant asymptote. The remainder term divided by the denominator approaches zero as x grows large, so it vanishes.
Example 5.1 — f(x) = (x squared minus 3x + 1) / (x + 2)
Degree of numerator is 2, degree of denominator is 1. Since 2 = 1 + 1, there is a slant asymptote.
Divide x squared minus 3x + 1 by x + 2:
x squared divided by x = x
x times (x + 2) = x squared + 2x
Subtract: (x squared minus 3x + 1) minus (x squared + 2x) = minus 5x + 1
minus 5x divided by x = minus 5
minus 5 times (x + 2) = minus 5x minus 10
Subtract: (minus 5x + 1) minus (minus 5x minus 10) = 11 (remainder)
Result: f(x) = (x minus 5) + 11 / (x + 2)
Slant asymptote: y = x minus 5
As x grows large, 11/(x + 2) approaches zero, so f(x) approaches y = x minus 5.
Example 5.2 — f(x) = (2x squared + x minus 3) / (x minus 1)
2x squared divided by x = 2x
2x times (x minus 1) = 2x squared minus 2x
Subtract: 2x squared + x minus 3 minus (2x squared minus 2x) = 3x minus 3
3x divided by x = 3
3 times (x minus 1) = 3x minus 3
Subtract: (3x minus 3) minus (3x minus 3) = 0 (no remainder)
Result: f(x) = 2x + 3 exactly (no remainder)
Slant asymptote: y = 2x + 3. Actually, f has a hole at x = 1 — check: (x minus 1)(2x + 3) = 2x squared + x minus 3. Yes, (x minus 1) is a factor of the numerator too.
When does a slant asymptote exist?
Only when deg(num) = deg(den) + 1. If the numerator is 2 or more degrees higher than the denominator, there is still no horizontal asymptote, but there is also no simple slant asymptote — the function grows like a polynomial, not a line.
6. X-Intercepts and Y-Intercepts
Y-Intercept
The y-intercept is the point where the graph crosses the vertical axis — it occurs at x = 0. Evaluate f(0), provided that 0 is in the domain (denominator not zero at x = 0).
f(x) = (3x + 6) / (x squared minus 9)
f(0) = (0 + 6) / (0 minus 9) = 6 / minus 9 = minus 2/3
Y-intercept: (0, minus 2/3)
X-Intercepts
X-intercepts occur where f(x) = 0, which means the numerator of the simplified form equals zero. Set the numerator equal to zero and solve. Verify each solution is in the domain (not a hole or asymptote location).
f(x) = (x squared minus 4) / (x squared minus 9)
Numerator = 0: x squared minus 4 = 0 → x = 2 or x = minus 2
X-intercepts: (2, 0) and (minus 2, 0)
Watch out: holes vs. x-intercepts
If a factor of the numerator also cancels with the denominator, it creates a hole — not an x-intercept. For example, if (x minus 2) cancels from both, then x = 2 is a hole, not a point where the graph touches the x-axis. Always work from the simplified form when finding x-intercepts.
7. Graphing Strategy — 8 Steps
Follow these steps in order. Each step gives you information needed by the next. Skipping steps is how graphing errors happen.
Find the domain
Set the denominator equal to zero and solve. Every solution is excluded from the domain. Write the domain in interval notation.
Factor and cancel
Factor numerator and denominator completely. Cancel common factors — each canceled factor (x minus a) produces a hole at x = a, not a vertical asymptote.
Identify holes
For each canceled factor, find the y-coordinate by substituting the hole x-value into the simplified function. Plot as an open circle.
Find vertical asymptotes
Set the remaining denominator (after cancellation) equal to zero. Each solution is a vertical asymptote. Draw a dashed vertical line.
Determine horizontal or slant asymptote
Compare numerator and denominator degrees. Apply the three-case rule for horizontal asymptotes, or do polynomial long division for a slant asymptote.
Find intercepts
Y-intercept: evaluate f(0), provided 0 is in the domain. X-intercepts: set the numerator of the simplified form equal to zero and solve. Verify each candidate is in the domain.
Build a sign chart
Plot vertical asymptotes and x-intercepts on a number line. Test a point in each interval to determine whether the function is positive or negative. This tells you which way each branch goes near the asymptotes.
Sketch the graph
Draw dashed lines for all asymptotes. Sketch branches in each interval consistent with the sign chart. Mark open circles at holes. Ensure each branch approaches its asymptote correctly.
8. End Behavior of Rational Functions
End behavior describes what happens to f(x) as x approaches positive infinity or negative infinity. For rational functions, end behavior is governed by the horizontal or slant asymptote.
Case A — Horizontal asymptote y = 0
When deg(num) is less than deg(den), both ends of the graph approach the x-axis.
f(x) = x / (x squared + 1)
As x goes to plus infinity: f(x) approaches 0 from above. As x goes to minus infinity: f(x) approaches 0 from below.
Case B — Horizontal asymptote y = L (nonzero)
When deg(num) equals deg(den), both ends approach the horizontal line y = L, where L is the ratio of leading coefficients.
f(x) = (2x squared + 3) / (x squared minus 1)
Leading coefficients: 2 and 1. As x goes to plus or minus infinity, f(x) approaches y = 2.
Case C — Slant asymptote
When deg(num) = deg(den) + 1, both ends of the graph approach a slanted line. The function grows without bound, but the distance from the line goes to zero.
f(x) = (x squared + 1) / (x minus 1) approaches y = x + 1 as x goes to infinity.
End behavior summary
The horizontal or slant asymptote is the "destination" of the function at the far left and far right of the graph. The vertical asymptotes and intercepts describe the behavior in the interior. Together they give you the complete shape of the graph without needing to plot dozens of individual points.
9. Transformations of f(x) = 1/x
The simplest rational function is f(x) = 1/x, called the reciprocal functionor the basic hyperbola. Its graph has two branches: one in the first quadrant and one in the third quadrant. All of the transformation rules for functions apply to it.
f(x) = 1 / x
Domain: all real numbers except 0 · Vertical asymptote: x = 0 · Horizontal asymptote: y = 0
Branches in quadrants I (x positive, y positive) and III (x negative, y negative)
| Function Form | Vertical Asymptote | Horizontal Asymptote | Transformation Description |
|---|---|---|---|
| 1 / x | x = 0 | y = 0 | Parent hyperbola, branches in quadrants I and III |
| 1 / (x minus h) | x = h | y = 0 | Shift right h units (left if h is negative) |
| 1 / x + k | x = 0 | y = k | Shift up k units (down if k is negative) |
| a / x | x = 0 | y = 0 | Vertical stretch by factor a; reflection if a is negative |
| a / (x minus h) + k | x = h | y = k | Full transformation; center of hyperbola at (h, k) |
Example 9.1 — Graph f(x) = 2 / (x minus 3) + 1
Start with parent 1/x. Apply transformations:
Horizontal shift right 3: vertical asymptote moves to x = 3
Vertical stretch by 2: branches stretch away from center
Vertical shift up 1: horizontal asymptote moves to y = 1
Center of the hyperbola: (3, 1). Asymptotes: x = 3 and y = 1.
Domain: all real numbers except x = 3. Range: all real numbers except y = 1.
Example 9.2 — Graph f(x) = minus 3 / (x + 1) minus 2
Horizontal shift left 1: vertical asymptote at x = minus 1
Vertical stretch by 3 and reflection (a = minus 3 is negative): branches flip to quadrants II and IV
Vertical shift down 2: horizontal asymptote moves to y = minus 2
Center: (minus 1, minus 2). Asymptotes: x = minus 1 and y = minus 2.
10. Worked Examples — Complete Analysis
f(x) = (x + 2) / (x squared minus 4)
Complete analysis: domain, holes, asymptotes, intercepts.
Step 1 — Factor
Denominator: x squared minus 4 = (x + 2)(x minus 2)
f(x) = (x + 2) / [(x + 2)(x minus 2)]
Step 2 — Cancel and find hole
Common factor (x + 2) cancels. Simplified: 1 / (x minus 2), with x not equal to minus 2.
Hole x-coordinate: x = minus 2
Hole y-coordinate: plug x = minus 2 into simplified: 1 / (minus 2 minus 2) = 1 / minus 4 = minus 1/4
Hole at (minus 2, minus 1/4)
Step 3 — Vertical asymptote
Remaining denominator: x minus 2 = 0 → x = 2
Vertical asymptote at x = 2
Step 4 — Horizontal asymptote
Simplified has deg(num) = 0, deg(den) = 1. Since 0 is less than 1:
Horizontal asymptote: y = 0
Step 5 — Intercepts
Y-intercept: f(0) = 1 / (0 minus 2) = minus 1/2 → point (0, minus 1/2)
X-intercepts: numerator of simplified = 1 (constant, never zero) → no x-intercepts
Summary
Domain: all reals except x = minus 2 and x = 2
Hole at (minus 2, minus 1/4) · Vertical asymptote: x = 2 · Horizontal asymptote: y = 0
Y-intercept: (0, minus 1/2) · No x-intercepts
f(x) = (2x squared + 3x minus 2) / (x squared minus 1)
Equal degrees — find horizontal asymptote and two x-intercepts.
Step 1 — Factor both
Numerator: 2x squared + 3x minus 2 = (2x minus 1)(x + 2)
Denominator: x squared minus 1 = (x minus 1)(x + 1)
No common factors → no holes
Step 2 — Vertical asymptotes
x minus 1 = 0 → x = 1; x + 1 = 0 → x = minus 1
Vertical asymptotes at x = 1 and x = minus 1
Step 3 — Horizontal asymptote
Both numerator and denominator have degree 2. Leading coefficients: 2 and 1.
Horizontal asymptote: y = 2
Step 4 — Intercepts
Y-intercept: f(0) = (0 + 0 minus 2) / (0 minus 1) = minus 2 / minus 1 = 2 → point (0, 2)
X-intercepts: set numerator = 0: (2x minus 1)(x + 2) = 0 → x = 1/2 or x = minus 2
Both values are in the domain (not holes or asymptotes).
X-intercepts at (1/2, 0) and (minus 2, 0)
f(x) = (x squared plus 2x minus 8) / (x minus 1)
Numerator degree is greater — find the slant asymptote by long division.
Step 1 — Check degrees and factor
deg(num) = 2, deg(den) = 1. Since 2 = 1 + 1, there is a slant asymptote.
Factor numerator: x squared + 2x minus 8 = (x + 4)(x minus 2)
No common factor with (x minus 1). No holes. Vertical asymptote at x = 1.
Step 2 — Long division
Divide x squared + 2x minus 8 by x minus 1:
x squared divided by x = x
x times (x minus 1) = x squared minus x
Subtract: (x squared + 2x minus 8) minus (x squared minus x) = 3x minus 8
3x divided by x = 3
3 times (x minus 1) = 3x minus 3
Subtract: (3x minus 8) minus (3x minus 3) = minus 5
f(x) = x + 3 + (minus 5) / (x minus 1)
Step 3 — Result
Slant asymptote: y = x + 3
Vertical asymptote: x = 1
Y-intercept: f(0) = (0 + 0 minus 8) / (0 minus 1) = minus 8 / minus 1 = 8 → point (0, 8)
X-intercepts: (x + 4)(x minus 2) = 0 → x = minus 4 or x = 2 → points (minus 4, 0) and (2, 0)
11. Common Mistakes to Avoid
These are the six most frequently-tested errors on rational function problems. Knowing them is as valuable as knowing the rules.
Mistake: Labeling a hole as a vertical asymptote
Fix: Always factor and cancel common factors FIRST. Canceled factors become holes. Only the remaining denominator factors become vertical asymptotes.
Mistake: Forgetting to plug into the simplified form for hole coordinates
Fix: To get the y-coordinate of a hole, substitute the x-value into the SIMPLIFIED rational function, not the original. The original is undefined there.
Mistake: Thinking a function cannot cross its horizontal asymptote
Fix: A rational function CAN cross its horizontal asymptote for finite x-values. The horizontal asymptote only describes end behavior as x approaches positive or negative infinity.
Mistake: Applying the horizontal asymptote rules to the original (un-simplified) function
Fix: Always compare the degrees of the SIMPLIFIED form after cancellation. The degree comparison should be done after removing common factors.
Mistake: Omitting the remainder when finding a slant asymptote
Fix: After polynomial long division, the slant asymptote is ONLY the quotient. Ignore the remainder term — it vanishes as x grows large.
Mistake: Setting the numerator equal to zero to find vertical asymptotes
Fix: Vertical asymptotes come from the denominator equaling zero (after cancellation). X-intercepts come from the numerator equaling zero.
12. Real-World Applications
Rational functions appear throughout science, engineering, and everyday problem-solving. Stewart Chapter 3 includes several application types worth recognizing.
Rate and Work Problems
When two workers complete a job at rates 1/a and 1/b jobs per hour, their combined rate is 1/a + 1/b = 1/T, where T is the time to finish together. Solving for T produces a rational equation.
Example: Two pipes fill a tank
Pipe A fills the tank in 6 hours. Pipe B fills it in 4 hours.
Combined rate: 1/6 + 1/4 = 2/12 + 3/12 = 5/12 tanks per hour
Time together: T = 12/5 = 2.4 hours
This is a rational equation because it involves reciprocals of time values.
Concentration Problems
When you add a substance to a mixture, the resulting concentration is a rational function. The concentration C depends on how much you add.
Example: Salt solution
A tank holds 100 liters of a 5 gram/liter salt solution (500 grams total).
You add x liters of pure water (0 g/L). New concentration:
C(x) = 500 / (100 + x)
As x approaches infinity, C(x) approaches 0. Horizontal asymptote at y = 0 makes physical sense.
Average Cost Functions
If a company has a fixed cost F plus variable cost per unit v, the total cost of producing x units is C(x) = F + vx. The average cost per unit is a rational function with a horizontal asymptote at y = v.
Example: Manufacturing
Fixed cost: $10,000. Variable cost: $5 per unit.
Total cost: C(x) = 10000 + 5x
Average cost: A(x) = (10000 + 5x) / x = 10000/x + 5
As x grows large, A(x) approaches $5. Producing more units spreads the fixed cost, reducing the average.
Vertical asymptote at x = 0 (cannot produce zero units). Slant-like structure with horizontal asymptote y = 5.
Physics — Lens Equation
In optics, the thin lens equation relates the focal length f, the object distance d sub o, and the image distance d sub i:
1/f = 1/d_o + 1/d_i
Solving for d sub i in terms of d sub o gives a rational function.
Example: focal length 10 cm, object at 30 cm from lens.
1/10 = 1/30 + 1/d_i → 1/d_i = 1/10 minus 1/30 = 3/30 minus 1/30 = 2/30 = 1/15
d_i = 15 cm
Electrical Circuits — Parallel Resistors
Two resistors connected in parallel obey a rational equation:
1/R = 1/R_1 + 1/R_2
Solving for R in terms of R sub 1 (treating R sub 2 as fixed) gives a rational function.
If R sub 2 = 6 ohms, then R = 6R sub 1 / (6 + R sub 1).
Horizontal asymptote at y = 6: as R sub 1 increases without bound, the parallel combination approaches R sub 2 = 6 ohms.
13. Partial Fractions Preview
Partial fraction decomposition is the reverse of combining rational expressions over a common denominator. Instead of adding fractions together, you split a rational function into a sum of simpler rational functions. This technique is essential in calculus (integration) and appears in some precalculus courses.
The key idea: if the denominator factors into linear factors, each factor gets its own fraction with an unknown constant in the numerator.
Simple Preview Example
Write (3x + 5) / [(x minus 1)(x + 2)] as a sum of partial fractions.
Set up: (3x + 5) / [(x minus 1)(x + 2)] = A / (x minus 1) + B / (x + 2)
Multiply both sides by (x minus 1)(x + 2):
3x + 5 = A(x + 2) + B(x minus 1)
Plug in x = 1: 3(1) + 5 = A(3) + B(0) → 8 = 3A → A = 8/3
Plug in x = minus 2: 3(minus 2) + 5 = A(0) + B(minus 3) → minus 1 = minus 3B → B = 1/3
Result: (8/3) / (x minus 1) + (1/3) / (x + 2)
Why does this matter for precalculus?
Partial fractions appear in advanced algebra topics, some textbooks include them in the systems of equations chapter, and they become central in calculus for integrating rational functions. Understanding rational function structure — especially factoring the denominator — is the prerequisite skill.
14. Practice Problems
Work through each problem completely before checking the answer summary below. For each function: find the domain, holes, asymptotes, and intercepts.
f(x) = (x squared minus 1) / (x squared minus x minus 2)
Find domain, holes, vertical asymptotes, horizontal asymptote, and intercepts.
Show answer
Factor num: (x minus 1)(x + 1). Factor den: (x minus 2)(x + 1).
Cancel (x + 1): hole at x = minus 1. Simplified: (x minus 1) / (x minus 2).
Hole y-coord: (minus 1 minus 1) / (minus 1 minus 2) = minus 2 / minus 3 = 2/3. Hole at (minus 1, 2/3).
Vertical asymptote: x = 2.
Horizontal asymptote: degrees equal (both 1), ratio = 1/1, so y = 1.
Y-intercept: f(0) = (0 minus 1) / (0 minus 2) = minus 1 / minus 2 = 1/2. Point (0, 1/2).
X-intercept: x minus 1 = 0 → x = 1. Point (1, 0).
g(x) = (3x squared minus 12) / (x squared + 2x minus 8)
Find all features including holes if any.
Show answer
Factor num: 3(x squared minus 4) = 3(x minus 2)(x + 2). Factor den: (x + 4)(x minus 2).
Cancel (x minus 2): hole at x = 2. Simplified: 3(x + 2) / (x + 4).
Hole y-coord: 3(2 + 2) / (2 + 4) = 12/6 = 2. Hole at (2, 2).
Vertical asymptote: x = minus 4.
Horizontal asymptote: degrees equal (both 1 after simplification), ratio = 3/1, so y = 3.
Y-intercept: g(0) = 3(0 + 2) / (0 + 4) = 6/4 = 3/2. Point (0, 3/2).
X-intercept: x + 2 = 0 → x = minus 2. Point (minus 2, 0).
h(x) = (x squared + x minus 6) / (x minus 1)
Find the slant asymptote using long division.
Show answer
deg(num) = 2, deg(den) = 1 → slant asymptote.
Factor num: (x + 3)(x minus 2). No common factor with (x minus 1). No holes.
Vertical asymptote: x = 1.
Long division: x squared + x minus 6 divided by x minus 1.
x squared divided by x = x. x(x minus 1) = x squared minus x. Subtract: 2x minus 6.
2x divided by x = 2. 2(x minus 1) = 2x minus 2. Subtract: minus 4 (remainder).
So h(x) = x + 2 + (minus 4) / (x minus 1). Slant asymptote: y = x + 2.
X-intercepts: (x + 3)(x minus 2) = 0 → x = minus 3 and x = 2.
Y-intercept: h(0) = (0 + 0 minus 6) / (0 minus 1) = minus 6 / minus 1 = 6. Point (0, 6).
Write f(x) = minus 2 / (x + 3) + 4 in terms of transformations of 1/x
Identify all asymptotes and the center of the hyperbola.
Show answer
This is f(x) = a/(x minus h) + k with a = minus 2, h = minus 3, k = 4.
Start with 1/x. Shift left 3: vertical asymptote at x = minus 3.
Multiply by minus 2: vertical stretch by 2, reflect (branches go to quadrants II and IV).
Shift up 4: horizontal asymptote moves to y = 4.
Center of hyperbola: (minus 3, 4).
Domain: all reals except x = minus 3. Range: all reals except y = 4.
Application: A chemical reaction produces x grams of a product.
The concentration function is C(x) = 5x / (x + 10) grams per liter. Find the horizontal asymptote and explain its meaning in context.
Show answer
Degrees of numerator and denominator are both 1. Leading coefficient ratio: 5/1 = 5.
Horizontal asymptote: y = 5 grams per liter.
Contextual meaning: as the amount of product increases without bound, the concentration approaches but never quite reaches 5 g/L. The 10 in the denominator represents a dilution factor or background volume. There is a physical upper limit on concentration for this system.
At x = 0: C(0) = 0 (no product, no concentration). At x = 10: C(10) = 50/20 = 2.5 g/L (halfway to the asymptote).
Related Precalculus Topics
Frequently Asked Questions
Do I need to factor first before finding any features?⌄
Yes — always. Factoring is the prerequisite for every other step. Without factoring, you cannot distinguish holes from vertical asymptotes, and you may incorrectly identify asymptotes from non-simplified denominators.
Can a rational function have more than one hole?⌄
Yes. Each distinct factor that appears in both the numerator and denominator cancels and becomes a hole. A rational function can theoretically have as many holes as there are common factors.
Can a rational function have both a horizontal asymptote and a slant asymptote?⌄
No — a rational function has at most one type of non-vertical end behavior. It has a horizontal asymptote when deg(num) is less than or equal to deg(den), and a slant asymptote only when deg(num) = deg(den) + 1. These cases are mutually exclusive.
How many vertical asymptotes can a rational function have?⌄
As many as the degree of the simplified denominator. A degree-n denominator has at most n real zeros, giving at most n vertical asymptotes. But some zeros may be complex (not real) or may cancel with numerator factors, so the actual number may be less.
What happens to the graph near a vertical asymptote?⌄
The graph goes to positive infinity or negative infinity as x approaches the vertical asymptote. You need to determine the sign from each side by testing nearby values. From one side the function may go to positive infinity and from the other to negative infinity, or both sides may go in the same direction — it depends on the specific function.
What is the difference between a rational function and a rational expression?⌄
A rational expression is an algebraic fraction p(x)/q(x) — it is an expression, not necessarily a function. A rational function is when you define f(x) = p(x)/q(x) and treat it as a rule mapping inputs to outputs with a specific domain. In practice, the terms are often used interchangeably in precalculus.
Ready to practice rational functions?
NailTheTest generates unlimited practice problems for rational functions, asymptotes, holes, and graphing — with instant step-by-step solutions.
Start Practicing Free