Unit 5 Review and Quiz
A free Precalculus lesson from the “Rational Functions” unit, with a worked example and practice problems including step-by-step solutions.
This checkpoint verifies rational-function behavior before exponential and logarithmic models. This lesson is part of Precalculus: Advanced Functions, so the emphasis is on interpreting behavior, choosing the right representation, and explaining the result clearly rather than memorizing isolated algebra moves.
What you'll learn
- Review restrictions, holes, asymptotes, graphing, and rational equations
- Choose the correct function, graph, or modeling tool from mixed prompts
- Explain why the selected method fits the problem
Worked example
Problem. What value of x is excluded from the domain of r(x) = (x + 1)/(x - 3)?
- A rational function is undefined where its denominator is 0.
- Set the denominator equal to 0: x - 3 = 0.
- Solving gives x = 3, so 3 is excluded from the domain.
Answer: 3
Practice problems
1. Unit review 1 (Rational Function Basics): What value of x is excluded from the domain of r(x) = (x + 1)/(x - 3)?
Show solution
- A rational function is undefined where its denominator is 0.
- Set the denominator equal to 0: x - 3 = 0.
- Solving gives x = 3, so 3 is excluded from the domain.
Answer: 3
2. Unit review 2 (Domain Restrictions and Holes): For r(x) = (3(x - 4))/(x - 4), cancel the shared (x - 4) factor and find the y-value of the hole.
Show solution
- The factor (x - 4) appears in both the numerator and the denominator, so cancel it.
- What remains is the constant 3, defined for every x except 4.
- The hole sits on that constant value, so its y-value is 3.
Answer: 3
3. Unit review 3 (Vertical Asymptotes): r(x) = 5/((x - 5)(x - 12)). Enter the LARGER of the two vertical-asymptote x-values.
Show solution
- Each denominator factor set to 0 gives a vertical asymptote.
- x - 5 = 0 gives x = 5; x - 12 = 0 gives x = 12.
- The larger value is x = 12.
Answer: 12
4. Unit review 4 (Horizontal and Slant Asymptotes): Find the slant asymptote of r(x) = (x^2 + 3x + 9)/(x + 1) by long division. Enter it as y = mx + b.
Show solution
- Degrees differ by one (2 vs 1), so there is a slant asymptote equal to the quotient.
- Dividing: x^2 + 3x + 9 = (x + 1)(x + 2) + 7, so the quotient is x + 2 with a nonzero remainder.
- The slant asymptote is y = x + 2.
Answer: y = x + 2
5. Unit review 5 (Graphing Rational Functions): Find the vertical asymptote x-value for r(x) = 4/(x - 3).
Show solution
- A vertical asymptote occurs where the uncanceled denominator is 0.
- Set x - 3 = 0, so x = 3; the numerator 4 does not cancel it.
- So the vertical asymptote is x = 3.
Answer: 3
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