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Quizzes > High School Quizzes > Mathematics

Rational Functions Practice Test

Conquer rational functions with our practice quiz

Difficulty: Moderate
Grade: Grade 10
Study OutcomesCheat Sheet
Colorful paper art promoting a high school-level Rational Functions Challenge quiz.

What is the domain of the function f(x) = 1/(x - 3)?
All real numbers except x = 3
All real numbers except x = -3
x > 3
x < 3
The denominator becomes zero when x = 3, so that value must be excluded from the domain. Therefore, the domain is all real numbers except x = 3.
Which of the following is a vertical asymptote of the function f(x) = (2x + 1)/(x - 4)?
x = 4
x = -4
x = 2
x = -2
The vertical asymptote occurs where the denominator is zero. In this function, the denominator is zero when x = 4.
Identify the horizontal asymptote of f(x) = (3x² + 2)/(x² - 5).
y = 3
y = 1
y = 0
y = -3
Since the degrees of the numerator and denominator are the same, the horizontal asymptote is the ratio of the leading coefficients, which is 3/1 = 3.
Which rational function has a removable discontinuity (a hole) at x = 2?
f(x) = (x² - 4)/(x - 2)
f(x) = (x - 2)/(x² - 4)
f(x) = (x² - 4)/(x + 2)
f(x) = (x - 2)/(x - 4)
The function f(x) = (x² - 4)/(x - 2) factors to ((x - 2)(x + 2))/(x - 2), which cancels the common factor (x - 2) and creates a hole at x = 2.
What defines a rational function?
It is the ratio of two polynomials.
It is a polynomial function with a square root.
It is an exponential function.
It is a trigonometric function.
A rational function is defined as the ratio of two polynomials. This means it can be expressed in the form f(x) = P(x)/Q(x) where both P and Q are polynomials and Q(x) ≠ 0.
Determine the domain of f(x) = (x + 3)/(x² - 9).
All real numbers except x = 3 and x = -3
All real numbers
All real numbers except x = 3
All real numbers except x = -3
The denominator factors as (x - 3)(x + 3), so it is zero when x = 3 or x = -3. These values must be excluded from the domain.
Find the vertical asymptotes of f(x) = (2x - 1)/(x² - 4).
x = 2 and x = -2
x = 2 only
x = -2 only
No vertical asymptotes
The denominator factors as (x - 2)(x + 2), which equals zero when x = 2 or x = -2. Since there is no cancellation with the numerator, both values are vertical asymptotes.
Determine the horizontal asymptote of f(x) = (4x² - 1)/(2x² + 3).
y = 2
y = 0
y = 1/2
y = 4
The degrees of the numerator and denominator are equal, so the horizontal asymptote is the ratio of the leading coefficients, which yields 4/2 = 2.
Find the slant (oblique) asymptote of f(x) = (x² + x + 1)/(x - 1).
y = x + 2
y = x - 2
y = x + 1
y = 2x + 1
Performing polynomial division yields a quotient of x + 2, which is the equation of the slant asymptote. The remainder does not affect the asymptote.
Simplify the rational expression (x² - 9)/(x² + 3x).
(x - 3)/x
(x + 3)/x
(x - 3)/(x + 3)
(x + 3)/(x - 3)
Factoring the numerator as (x - 3)(x + 3) and the denominator as x(x + 3) shows that (x + 3) cancels, leaving (x - 3)/x. Remember to note the restrictions where x ≠ -3 and x ≠ 0.
For what value of x does f(x) = (x² + 2x)/(x + 2) have a hole?
x = -2
x = 0
x = 2
x = -1
The numerator factors as x(x + 2), and the denominator is (x + 2). Canceling the common factor (x + 2) creates a hole at x = -2, where the original function is undefined.
Solve the equation (1/(x - 1)) + (2/(x + 1)) = (3/(x² - 1)).
x = 4/3
x = 1
x = -1
x = 3
Multiplying both sides of the equation by the common denominator (x - 1)(x + 1) simplifies the equation to a linear form, yielding the solution x = 4/3 after solving.
Determine the end behavior of f(x) = (5x - 7)/(x + 2).
y = 5
y = x
y = 7
y = -2
Since the degrees of the numerator and denominator are equal (both degree 1), the end behavior is determined by the ratio of the leading coefficients, which is 5/1 = 5.
Which statement best characterizes a removable discontinuity in a rational function?
It occurs when a factor causing the denominator to be zero cancels out.
It represents a vertical asymptote of the function.
It is the same as a horizontal asymptote.
It is a point where the function is defined and continuous.
A removable discontinuity, or hole, happens when a factor in the numerator and denominator cancels, leaving the function undefined at that specific point. This differs from vertical asymptotes where cancellation does not occur.
If f(x) = (x + 4)/(x - 2), what is f(0)?
-2
2
4
0
Substituting x = 0 into the function gives f(0) = (0 + 4)/(0 - 2) = 4/(-2) = -2. This direct substitution computes the value of the function at x = 0.
Find all x-intercepts of f(x) = (x² - 4)/(x² - x - 6).
x = 2
x = -2
x = 2 and x = -2
x = 2, -2, and 3
Factor the numerator as (x - 2)(x + 2) and the denominator as (x - 3)(x + 2). The common factor (x + 2) cancels, so the only x-intercept is from the remaining factor, x - 2 = 0, giving x = 2.
Determine the oblique asymptote of f(x) = (2x² + 3x + 1)/(x - 2).
y = 2x + 7
y = 2x - 7
y = 2x
y = x + 7
Performing polynomial long division, the quotient is found to be 2x + 7, which represents the oblique asymptote of the function. The remainder is insignificant for determining the asymptote.
Simplify f(x) = (x³ - x)/(x² - 1).
f(x) = x
f(x) = x²
f(x) = x - 1
f(x) = 1/x
Factor the numerator as x(x² - 1) and the denominator as (x - 1)(x + 1). After canceling the common factor (x² - 1), the function simplifies to f(x) = x, though x ≠ 1 and x ≠ -1 must be considered.
Solve for x: (x/(x - 3)) = (2/(x + 3)).
No real solution
x = √23
x = -√23
x = 0
Cross-multiplying gives x(x + 3) = 2(x - 3), which simplifies to a quadratic with a negative discriminant. Therefore, there are no real solutions to this equation.
For f(x) = (2x² - 5x + 2)/(x - 2), which value must be excluded from the domain and why?
x = 2
x = -2
x = 0
x = 5/2
The denominator x - 2 becomes zero when x = 2, making the function undefined. Thus, x = 2 must be excluded from the domain to avoid division by zero.
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Study Outcomes

  1. Analyze domain restrictions of rational functions.
  2. Determine vertical and horizontal asymptotes.
  3. Apply algebraic techniques to simplify expressions.
  4. Solve rational equations to identify solutions.
  5. Interpret the graphical behavior of rational functions.
  6. Evaluate the impact of undefined points on function behavior.

Rational Functions Test Cheat Sheet

  1. Domain of Rational Functions - Your function's domain is all real x except where the denominator becomes zero. To avoid undefined spots, set the denominator equal to zero, solve for x, and exclude those values. This ensures your rational function only performs where it's allowed! OpenStax: Domain Essentials
  2. Vertical Asymptotes - These are the x-values where your function shoots off to infinity and the graph goes vertical. Factor the denominator, set each factor equal to zero, and those x-values are your asymptotes. Keep an eye on these to shape your curve's wild behavior! OpenStax: Vertical Asymptotes
  3. Horizontal and Slant Asymptotes - Compare degrees: if numerator degree is less than denominator, y=0 is your baseline; if degrees match, the asymptote is the ratio of leading coefficients; if numerator's degree is exactly one more, look for a slant asymptote via polynomial division. These rules give you the long‑run behavior of your function! OpenStax: Asymptote Guidelines
  4. Removable Discontinuities (Holes) - Sometimes factors cancel out and leave a "hole" in the graph at specific x-values. Identify any common numerator/denominator factors, cancel them, and mark the x-value of the canceled factor to show where the hole lives. It's like a secret gap in the graph that you can detect algebraically! OpenStax: Holes Explained
  5. Simplifying Rational Expressions - Break both numerator and denominator into their prime‑factor pieces, then cancel out matching factors like a pro. For example, (x²−1)/(x²−2x−3) factors to ((x+1)(x−1))/((x+1)(x−3)), and the (x+1) terms vanish. This snips the expression down to its simplest form! OpenStax: Simplification Tricks
  6. Multiplying Rational Expressions - First, factor everything and hunt for common factors. Multiply the numerators together and multiply the denominators together, then cancel any new common factors. It's like building with blocks: connect numerators, connect denominators, then remove extras! OpenStax: Multiply Rational Expressions
  7. Dividing Rational Expressions - Division means multiplying by the reciprocal of the second fraction. Flip the divisor, multiply across, then simplify by canceling factors. Always watch for restrictions from the original denominator to keep your work valid! OpenStax: Dividing Rational Expressions
  8. Adding and Subtracting Rational Expressions - Identify a common denominator (LCM of the denominators), rewrite each expression with that denominator, then add or subtract the numerators. Finally, factor and simplify the result. It's like finding a common language so the fractions can combine! OpenStax: Add/Subtract Rationals
  9. Graphing Rational Functions - Plot intercepts, asymptotes, holes, and a few strategic points to guide your sketch. Observe the behavior near vertical asymptotes and the end behavior dictated by horizontal/slant asymptotes. Then connect the dots in smooth curves - your graph will shout "I'm a rational function!" OpenStax: Graphing Guide
  10. Practice Problem‑Solving - Nothing beats consistent practice to master rational functions. Tackle a variety of problems: domain checks, asymptote analyses, simplification drills, and graphing challenges. Each solved problem adds confidence and cements your algebraic skills! Pearson: Practice Rational Functions
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