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Graphs of Rational Functions

A free Algebra II lesson from the “Equations, Rational Functions, and Conics” unit, with a worked example and practice problems including step-by-step solutions.

A rational function is a ratio of polynomials. VERTICAL ASYMPTOTES happen where the denominator is zero AND the factor does not cancel with the numerator. HOLES happen where a factor cancels in both. HORIZONTAL ASYMPTOTES depend on degrees: if degree of numerator is less than degree of denominator, y = 0; if equal, y = leading-coefficient ratio; if greater, no horizontal asymptote.

What you'll learn

Why it matters: Drug concentrations over time, economic supply/demand limits, gear ratios, and physics relationships involving 1/x all show rational-function behavior including asymptotes.

Worked example

Problem. For f(x) = (x + 3) / ((x - 2)(x + 3)), find the vertical asymptote and the hole.

  1. The factor (x + 3) cancels between numerator and denominator -> hole at x = -3.
  2. Remaining denominator factor (x - 2) gives the vertical asymptote at x = 2.

Answer: VA: x = 2. Hole: x = -3.

Practice problems

1. f(x) = 1 / (x - 3). Enter the x-value of the vertical asymptote.

Show solution
  1. Warm-up: First identify exactly what the question is asking: f(x) = 1 / (x - 3). Enter the x-value of the vertical asymptote.
  2. For function notation, treat the value inside parentheses as the input and carefully substitute it into the rule.
  3. Denominator = 0 at x = 3.
  4. Check the result by substituting or estimating: the response should match 3 and make sense in the original problem.

Answer: 3

2. f(x) = 1 / (x + 5). Enter the x-value of the vertical asymptote.

Show solution
  1. Warm-up: First identify exactly what the question is asking: f(x) = 1 / (x + 5). Enter the x-value of the vertical asymptote.
  2. For function notation, treat the value inside parentheses as the input and carefully substitute it into the rule.
  3. x + 5 = 0 at x = -5.
  4. Check the result by substituting or estimating: the response should match -5 and make sense in the original problem.

Answer: -5

3. f(x) = (x - 2) / (x^2 - 4). Enter the x-value of the hole.

Show solution
  1. Warm-up: First identify exactly what the question is asking: f(x) = (x - 2) / (x^2 - 4). Enter the x-value of the hole.
  2. For function notation, treat the value inside parentheses as the input and carefully substitute it into the rule.
  3. x^2 - 4 = (x - 2)(x + 2).
  4. The (x - 2) cancels with the numerator -> hole at x = 2.
  5. Check the result by substituting or estimating: the response should match 2 and make sense in the original problem.

Answer: 2

4. Same f(x) = (x - 2) / (x^2 - 4). Enter the x-value of the vertical asymptote.

Show solution
  1. Core Practice: First identify exactly what the question is asking: Same f(x) = (x - 2) / (x^2 - 4). Enter the x-value of the vertical asymptote.
  2. For function notation, treat the value inside parentheses as the input and carefully substitute it into the rule.
  3. After cancelling, the remaining denominator factor is (x + 2).
  4. Check the result by substituting or estimating: the response should match -2 and make sense in the original problem.

Answer: -2

5. f(x) = 3x / x^2. Enter the y-value of the horizontal asymptote.

Show solution
  1. Core Practice: First identify exactly what the question is asking: f(x) = 3x / x^2. Enter the y-value of the horizontal asymptote.
  2. For function notation, treat the value inside parentheses as the input and carefully substitute it into the rule.
  3. Numerator degree 1 < denominator degree 2 -> HA at y = 0.
  4. Check the result by substituting or estimating: the response should match 0 and make sense in the original problem.

Answer: 0

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