Precalculus Sample Problems- Practice with Detailed Solutions

Precalculus Sample Problems: The Problems You Actually Need to Master

Most textbooks throw hundreds of problems at you. Most of them are garbage. This guide cuts through the noise and shows you the precalculus problems that actually matter, with solutions that don't leave you guessing.

These are the problem types that show up on exams. Master these patterns and you won't be caught off guard.

Functions and Their Graphs

Understanding functions is the entire foundation of precalculus. If this part confuses you, everything else falls apart.

Problem 1: Evaluating Functions

Given f(x) = 2x² - 3x + 1, find f(3) and f(-2).

Solution:

For f(3):

f(3) = 2(3)² - 3(3) + 1

f(3) = 2(9) - 9 + 1

f(3) = 18 - 9 + 1

f(3) = 10

For f(-2):

f(-2) = 2(-2)² - 3(-2) + 1

f(-2) = 2(4) + 6 + 1

f(-2) = 8 + 6 + 1

f(-2) = 15

Problem 2: Finding Domain and Range

Find the domain of f(x) = √(x - 4) / (x² - 9).

Solution:

Two restrictions here:

• The square root requires x - 4 ≥ 0, so x ≥ 4
• The denominator cannot be zero: x² - 9 ≠ 0, so x ≠ ±3

The condition x ≥ 4 automatically excludes x = -3. We only need to check if x = 3 falls in our domain.

Since 3 < 4, it's already excluded by the square root restriction.

Domain: [4, ∞)

Polynomial and Rational Functions

Problem 3: Factoring Polynomials

Factor completely: x³ - 6x² + 11x - 6

Solution:

Use the Rational Root Theorem. Possible rational roots are factors of 6: ±1, ±2, ±3, ±6.

Test x = 1:

1 - 6 + 11 - 6 = 0 ✓

So (x - 1) is a factor. Divide using synthetic division:

1 | 1 -6 11 -6

| 1 -5 6

─────────────

1 -5 6 0

Now factor the quotient x² - 5x + 6:

x² - 5x + 6 = (x - 2)(x - 3)

Complete factorization: (x - 1)(x - 2)(x - 3)

Problem 4: Vertical Asymptotes

Find the vertical asymptotes of R(x) = (x + 2) / (x² - 4x + 3)

Solution:

Factor the denominator:

x² - 4x + 3 = (x - 1)(x - 3)

Vertical asymptotes occur where the denominator equals zero AND the numerator doesn't also equal zero at those points.

At x = 1: numerator = 1 + 2 = 3 ≠ 0 ✓

At x = 3: numerator = 3 + 2 = 5 ≠ 0 ✓

Vertical asymptotes: x = 1 and x = 3

Trigonometry Problems

Problem 5: Converting Between Degrees and Radians

Convert 225° to radians and express the result in terms of π.

Solution:

Use the conversion factor: π radians = 180°

225° × (π/180°) = (225π/180)

Simplify by dividing by 45: (5π/4)

225° = 5π/4 radians

This lands in Quadrant III, which checks out.

Problem 6: Solving Trigonometric Equations

Solve for x in the interval [0, 2π): 2sin²(x) - sin(x) - 1 = 0

Solution:

Let u = sin(x). The equation becomes:

2u² - u - 1 = 0

Factor:

(2u + 1)(u - 1) = 0

So u = -1/2 or u = 1

Case 1: sin(x) = 1

x = π/2

Case 2: sin(x) = -1/2

x = 7π/6 and x = 11π/6 (Quadrant III and IV)

Solutions: x = π/2, 7π/6, 11π/6

Exponential and Logarithmic Functions

Problem 7: Solving Exponential Equations

Solve: 3^(2x+1) = 81

Solution:

Rewrite 81 as a power of 3:

81 = 3^4

So: 3^(2x+1) = 3^4

2x + 1 = 4

2x = 3

x = 3/2

Problem 8: Solving Logarithmic Equations

Solve: log₂(x + 3) + log₂(x - 1) = 3

Solution:

Combine using the product rule:

log₂[(x + 3)(x - 1)] = 3

Convert to exponential form:

(x + 3)(x - 1) = 2³

(x + 3)(x - 1) = 8

Expand:

x² + 3x - x - 3 = 8

x² + 2x - 3 = 8

x² + 2x - 11 = 0

Use the quadratic formula:

x = (-2 ± √(4 + 44)) / 2

x = (-2 ± √48) / 2

x = (-2 ± 4√3) / 2

x = -1 ± 2√3

Check both in the original equation:

x = -1 + 2√3 ≈ 2.46 → valid (both logs positive)

x = -1 - 2√3 ≈ -4.46 → invalid (x + 3 would be negative)

Solution: x = -1 + 2√3

Systems of Equations

Problem 9: Solving by Substitution

Solve the system:

2x + y = 7

x² + y = 4

Solution:

From the first equation: y = 7 - 2x

Substitute into the second:

x² + (7 - 2x) = 4

x² - 2x + 7 = 4

x² - 2x + 3 = 0

Factor:

(x - 1)(x - 3) = 0

x = 1 or x = 3

Find corresponding y values:

If x = 1: y = 7 - 2(1) = 5

If x = 3: y = 7 - 2(3) = 1

Solutions: (1, 5) and (3, 1)

Verify both in the original equations—they check out.

Practice Problem Comparison Table

Here's a breakdown of which problem types appear most frequently on assessments:

Getting Started: How to Practice Effectively

Don't just read through these solutions. That wastes your time.

• Cover the solutions first. Try each problem with the book closed. Struggle with it for 10-15 minutes before looking at the answer
• Identify your weak spots. If you consistently miss trig problems, drill trig. Don't waste time on what you already know
• Time yourself. Exams have time limits. Practice under pressure
• Check your algebra. Most precalculus errors come from algebra mistakes, not conceptual ones

Work through 10-15 problems daily. Review the ones you get wrong. Within two weeks, you'll see your speed and accuracy improve noticeably.

Common Mistakes to Avoid

• Forgetting to check for extraneous solutions in logarithmic equations
• Not simplifying fractions when finding asymptotes
• Confusing degrees and radians in trig problems
• Losing negative signs when distributing
• Forgetting the ± when taking square roots of both sides

These mistakes cost points on every exam. Hunt them down in your practice work.