Rational Functions Practice- Precalculus Problems and Solutions

What Are Rational Functions?

A rational function is simply a ratio of two polynomials. That's it. If you can write it as f(x) = p(x)/q(x) where both p and q are polynomials and q(x) ≠ 0, you've got a rational function.

The denominator is where everything gets interesting. It's the reason rational functions have vertical asymptotes, holes, and restricted domains. Ignore the denominator at your own peril — that's where students lose marks.

The Essential Vocabulary You Need First

Before touching any practice problem, these terms must be locked in:

• Domain — all real numbers except where the denominator equals zero
• Vertical asymptote — occurs at x-values that make the denominator zero but can't be canceled
• Horizontal asymptote — the behavior of the function as x approaches ±∞
• Hole (removable discontinuity) — a point where both numerator and denominator share a factor that cancels
• x-intercept — where the numerator equals zero (and denominator doesn't)
• y-intercept — f(0), assuming 0 is in the domain

Getting Started: Solving Rational Functions

Step 1: Factor Everything

Factor the numerator and denominator completely. This reveals holes (canceled factors) and makes finding the domain trivial.

Step 2: Find the Domain

Set the denominator equal to zero. Solve for x. Those values are excluded from the domain. Write your domain in interval notation.

Step 3: Cancel Common Factors

Cross out factors that appear in both numerator and denominator. The simplified function tells you about holes. The original function tells you about the actual graph.

Step 4: Find Asymptotes

Vertical asymptotes happen at denominator zeros that don't cancel. Horizontal asymptotes depend on the degree of the numerator versus the denominator.

Step 5: Find Intercepts

Set x=0 for the y-intercept. Set the numerator equal to zero for x-intercepts. Check that neither falls in an excluded domain value.

Practice Problems with Solutions

Problem 1: Basic Domain and Simplification

Given: f(x) = (x² - 9)/(x² - x - 6)

Find: Domain, holes, and simplified form.

Solution:

Factor both parts:

(x² - 9) = (x + 3)(x - 3)

(x² - x - 6) = (x - 3)(x + 2)

Domain: x² - x - 6 ≠ 0 → x ≠ 3, x ≠ -2

Simplified form: f(x) = (x + 3)/(x + 2), with a hole at x = 3

The hole exists because (x - 3) canceled. At x = 3, the original function is undefined. The simplified function would give f(3) = 6/5, but that's not the actual value.

Problem 2: Finding Asymptotes

Given: f(x) = (3x²)/(x² - 4)

Find: All asymptotes.

Solution:

Vertical asymptotes: Set denominator to zero.

x² - 4 = 0 → x = ±2

Horizontal asymptote: Compare degrees. Both numerator and denominator are degree 2. The horizontal asymptote is the ratio of leading coefficients: y = 3/1 = 3

No holes exist — nothing cancels.

Problem 3: Graph Behavior

Given: f(x) = (2x)/(x + 1)

Find: Domain, intercepts, and asymptotes.

Solution:

Domain: x ≠ -1

Y-intercept: f(0) = 0 → passes through origin

X-intercept: 2x = 0 → x = 0

Vertical asymptote: x = -1

Horizontal asymptote: Both degree 1 → y = 2/1 = 2

Sketch this: the graph approaches y = 2 as x → ±∞, and shoots toward infinity near x = -1.

Problem 4: Adding Rational Expressions

Solve: 1/(x - 2) + 2/(x + 3)

Solution:

Find the LCD: (x - 2)(x + 3)

Rewrite each term:

1/(x - 2) = (x + 3)/[(x - 2)(x + 3)]

2/(x + 3) = 2(x - 2)/[(x - 2)(x + 3)]

Add the numerators:

(x + 3) + 2(x - 2) = x + 3 + 2x - 4 = 3x - 1

Answer: (3x - 1)/[(x - 2)(x + 3)]

Don't simplify unless asked. The factored form shows the domain restrictions immediately: x ≠ 2, x ≠ -3.

Problem 5: Solving Rational Equations

Solve: (x)/(x - 2) = 3/(x + 1)

Solution:

Cross-multiply:

x(x + 1) = 3(x - 2)

x² + x = 3x - 6

x² + x - 3x + 6 = 0

x² - 2x + 6 = 0

Use the quadratic formula:

x = [2 ± √(4 - 24)]/2 = [2 ± √(-20)]/2

No real solutions. The discriminant is negative. This happens — move on.

Always check that your solutions don't make any denominator zero. In this case, x ≠ 2 and x ≠ -1.

Horizontal Asymptote Rules

These rules cover 95% of cases you'll encounter:

When the numerator is exactly one degree higher, you get a slanted/oblique asymptote. Divide the polynomials using long division to find it.

Common Mistakes That Cost You Points

• Forgetting to exclude domain values — if x = 2 makes the denominator zero, it's not in the domain. Period.
• Canceling before factoring — you can't cancel across addition. Factor first.
• Missing holes — canceled factors create holes. Don't ignore them.
• Confusing holes with asymptotes — holes happen where factors cancel. Asymptotes happen where they don't.
• Forgetting to check solutions — plug answers back into the original equation.

Quick Reference: Asymptote vs Hole

Here's the simple test: factor the rational expression completely. If a factor appears in both numerator and denominator, and cancels, you have a hole at that x-value. If a factor only appears in the denominator, you have a vertical asymptote there.

Example: f(x) = (x² - 4)/(x² - x - 6) = [(x+2)(x-2)]/[(x-3)(x+2)]

The (x+2) cancels → hole at x = -2

The (x-3) doesn't cancel → vertical asymptote at x = 3

Final Tips

Graphing rational functions by hand? Start with intercepts, plot the asymptotes as dashed lines, and sketch the behavior near each. The graph approaches asymptotes but never crosses horizontal asymptotes at the ends.

For any rational function problem, the process is always the same: factor, find domain restrictions, simplify, then identify features. Once you internalize this sequence, every problem becomes mechanical.

That's the substance. Practice factoring until it's automatic — it's the foundation everything else builds on.