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Honors Algebra II Unit 7 Review: Rational Expressions and Functions

Unit 7 is about fractions made of polynomials: how to graph them, simplify them, combine them, solve equations with them, and break them back apart. By the end you should be able to sketch a rational function from its factored form, solve a rational equation while catching extraneous solutions, and decompose a fraction into partial fractions.

Read the overview first, then work through each topic's notes and try the check-yourself question before looking at the answer.

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What is Honors Algebra II unit 7?

A rational expression is one polynomial divided by another, like x+3x−2\frac{x+3}{x-2}. This unit treats these expressions as functions to graph, as algebra to simplify and solve, and as fractions you can take apart. It builds directly on factoring and polynomial division from earlier units.

Honors Algebra II Unit 7 covers rational expressions and functions. You learn to find the domain of a rational function, locate vertical asymptotes, holes, and horizontal or slant asymptotes, and use intercepts and sign charts to sketch the graph. You also simplify, multiply, divide, add, and subtract rational expressions, solve rational equations by clearing denominators, and check for extraneous solutions. The unit ends with partial fractions decomposition, which rewrites one complicated fraction as a sum of simpler ones and gets used heavily in calculus.

The denominator decides where a rational function breaks.

Any x-value that makes the denominator zero is excluded from the domain. After canceling common factors, check the remaining denominator. An excluded x-value gives a hole if the remaining denominator is nonzero there and a vertical asymptote if it is zero there.

Rational expressions follow the same rules as number fractions.

You multiply straight across, divide by multiplying by the reciprocal, and add or subtract only after getting a common denominator. Factor polynomials to find common factors and the least common denominator (LCD). Keep the values that make an original denominator zero excluded.

Clearing denominators can create answers that don't actually work.

Multiplying both sides of a rational equation by the LCD turns it into a polynomial equation, which is easier to solve. That step can introduce extraneous solutions, values that make an original denominator zero, so every answer has to be checked in the original equation.

Factor first, and keep track of what can't be zero.

Factoring helps with graphing, simplifying, solving, and decomposing rational expressions. Keep the excluded values, then check which factors remain after cancellation to identify holes and vertical asymptotes. For partial fractions, divide first if the numerator's degree is at least the denominator's degree.

Honors Algebra II unit 7 topics

Unit 7 review notes

7.1

Rational Functions and Their Graphs

A rational function is a ratio of two polynomials, f(x)=P(x)Q(x)f(x)=\frac{P(x)}{Q(x)}. Its domain is every real number except the zeros of Q(x)Q(x). Factor and cancel common factors to determine what happens at each excluded value. If the remaining denominator is nonzero there, the graph has a hole, which is a single missing point. If the remaining denominator is zero there, the graph has a vertical asymptote, where the graph heads toward ±∞\pm\infty.

End behavior comes from comparing degrees. The table covers horizontal and slant asymptotes. If the numerator's degree is more than one greater, long division can give a polynomial asymptote instead. A graph can cross a horizontal asymptote in the middle of the graph, but it can never cross a vertical asymptote.

Worked example. Analyze f(x)=x2−1x2+x−2f(x)=\frac{x^2-1}{x^2+x-2}. Factoring gives (x−1)(x+1)(x−1)(x+2)\frac{(x-1)(x+1)}{(x-1)(x+2)}, so (x−1)(x-1) cancels and the simplified form is x+1x+2\frac{x+1}{x+2}. The hole is at x=1x=1. Plugging x=1x=1 into the simplified form gives y=23y=\frac{2}{3}, so the hole is at (1,23)(1,\tfrac{2}{3}). The leftover denominator gives a vertical asymptote at x=−2x=-2. Both degrees are 2 with leading coefficients 1 and 1, so the horizontal asymptote is y=1y=1. Setting x+1=0x+1=0 gives the x-intercept (−1,0)(-1,0), and f(0)=12f(0)=\frac{1}{2} gives the y-intercept (0,12)(0,\tfrac{1}{2}). A sign chart using x=−2x=-2 and x=−1x=-1 shows where the graph sits above or below the x-axis.

  • Vertical asymptote: A vertical line x=ax=a where the graph heads toward ±∞\pm\infty, caused by a denominator factor that does not cancel.
  • Hole: A single missing point at an excluded x-value where the denominator is nonzero after all common factors are canceled.
  • Slant (oblique) asymptote: A non-horizontal line the graph approaches at the far left and right, found by long division when the numerator's degree is exactly one more than the denominator's.

What are the hole and vertical asymptote of g(x)=x2−4x2−x−2g(x)=\frac{x^2-4}{x^2-x-2}? (Answer: It factors to (x−2)(x+2)(x−2)(x+1)\frac{(x-2)(x+2)}{(x-2)(x+1)}, so there is a hole at (2,43)(2,\tfrac{4}{3}) and a vertical asymptote at x=−1x=-1.)

Degree comparisonEnd behaviorExample
Numerator degree less than denominator degreeHorizontal asymptote $y=0$$\frac{3x}{x^2+1}$ has $y=0$
Degrees equalHorizontal asymptote $y=\frac{a}{b}$ (ratio of leading coefficients)$\frac{4x^2-1}{2x^2+x}$ has $y=2$
Numerator degree is one moreSlant asymptote from the quotient of long division$\frac{2x^2+3x-1}{x-4}$ has $y=2x+11$
7.2

Rational Expressions and Equations

To simplify a rational expression, factor the numerator and denominator completely and cancel common factors. For example, x2−9x2−5x+6=(x+3)(x−3)(x−2)(x−3)=x+3x−2\frac{x^2-9}{x^2-5x+6}=\frac{(x+3)(x-3)}{(x-2)(x-3)}=\frac{x+3}{x-2}, with x≠2x\neq 2 and x≠3x\neq 3. The restriction x≠3x\neq 3 stays even though that factor canceled.

Multiply by multiplying across, and divide by multiplying by the reciprocal. When dividing, also exclude values that make the divisor zero. To add or subtract, factor each denominator, build the least common denominator (LCD), rewrite each fraction over the LCD, and combine numerators. A complex fraction (a fraction inside a fraction) simplifies when you multiply its top and bottom by the LCD of all the small denominators.

To solve a rational equation, multiply both sides by the LCD, solve the polynomial equation that results, then check each answer in the original equation. A proposed solution that makes an original denominator zero is extraneous.

Worked example. Solve 2x−1+3x+1=1\frac{2}{x-1}+\frac{3}{x+1}=1. Multiply by (x−1)(x+1)(x-1)(x+1) to get 2(x+1)+3(x−1)=x2−12(x+1)+3(x-1)=x^2-1, so 5x−1=x2−15x-1=x^2-1. That becomes x2−5x=0x^2-5x=0, or x(x−5)=0x(x-5)=0, so x=0x=0 or x=5x=5. Checking gives −2+3=1-2+3=1 and 12+12=1\frac{1}{2}+\frac{1}{2}=1, so both work.

In work problems, rates add. If pipes fill a tank in 4 and 6 hours, t4+t6=1\frac{t}{4}+\frac{t}{6}=1 gives t=125t=\frac{12}{5}, or 2.4 hours.

  • Least common denominator (LCD): For integer-coefficient denominators, use the least common multiple of the numerical coefficients and multiply by each distinct polynomial factor raised to the highest power that appears.
  • Extraneous solution: A value that solves the cleared polynomial equation but fails the original rational equation, usually because it makes a denominator zero.
  • Complex fraction: A fraction whose numerator, denominator, or both contain fractions.

Check yourself: Solve xx−3=3x−3+2\frac{x}{x-3}=\frac{3}{x-3}+2. (Answer: Multiplying by x−3x-3 gives x=3+2(x−3)x=3+2(x-3), so x=3x=3. But x=3x=3 makes the denominator zero, so it is extraneous and the equation has no solution.)

7.3

Partial Fractions Decomposition

Partial fractions decomposition runs adding fractions in reverse. Instead of combining 2x+1+3x−2\frac{2}{x+1}+\frac{3}{x-2} into one fraction, you start with the single fraction and recover the pieces. You will use this later in calculus to integrate rational functions.

Two conditions come first. The fraction being decomposed must have a numerator of lower degree than its denominator. If it doesn't, use long division, keep the polynomial quotient, and decompose the remainder divided by the original denominator. The denominator must be factored completely into linear factors and irreducible quadratics (quadratics with negative discriminant, like x2+1x^2+1).

The template depends on the factor type (see the table). Every repeated factor needs one fraction for each power, and every irreducible quadratic gets a numerator of Ax+BAx+B, not a single constant. After writing the template, multiply through by the full denominator and solve for the unknowns. Plugging in the zeros of the linear factors is the fastest method, and matching coefficients handles whatever is left.

Worked example. Decompose 5x−1(x+1)(x−2)\frac{5x-1}{(x+1)(x-2)}. Write Ax+1+Bx−2\frac{A}{x+1}+\frac{B}{x-2} and multiply by (x+1)(x−2)(x+1)(x-2) to get 5x−1=A(x−2)+B(x+1)5x-1=A(x-2)+B(x+1). Setting x=2x=2 gives 9=3B9=3B, so B=3B=3. Setting x=−1x=-1 gives −6=−3A-6=-3A, so A=2A=2. The answer is 2x+1+3x−2\frac{2}{x+1}+\frac{3}{x-2}. Check it by adding: 2(x−2)+3(x+1)=5x−12(x-2)+3(x+1)=5x-1.

  • Irreducible quadratic: A quadratic factor that cannot be factored over the real numbers because its discriminant b2−4acb^2-4ac is negative.
  • Proper rational expression: A rational expression whose numerator has a lower degree than its denominator. An improper expression must be divided first.
  • Matching coefficients: Setting the coefficients of each power of xx equal on both sides of an identity to build equations for the unknowns.

Write the template (don't solve) for x+4(x−3)2(x2+2)\frac{x+4}{(x-3)^2(x^2+2)}. (Answer: Ax−3+B(x−3)2+Cx+Dx2+2\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{x^2+2}.)

Denominator factorFractions to writeExample template
Distinct linear, like $(x-1)$One fraction with a constant on top$\frac{A}{x-1}$
Repeated linear, like $(x-1)^2$One fraction for each power, constants on top$\frac{A}{x-1}+\frac{B}{(x-1)^2}$
Irreducible quadratic, like $(x^2+1)$One fraction with a linear numerator$\frac{Ax+B}{x^2+1}$

Key terms

TermDefinition
Rational expressionA fraction whose numerator and denominator are polynomials, like $\frac{x^2-1}{x+4}$.
Rational functionA function of the form $f(x)=\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial.
Excluded valueAn x-value that makes a denominator zero, so it is left out of the domain.
Vertical asymptoteA line $x=a$ that the graph approaches as the output grows toward $\pm\infty$, caused by a denominator factor that doesn't cancel.
Horizontal asymptoteA line $y=b$ that the graph approaches as $x\to\pm\infty$, found by comparing the degrees of the numerator and denominator.
HoleA removable point of discontinuity at an excluded x-value where the denominator is nonzero after all common factors are canceled.
Sign chartA number line split at the zeros and excluded values that shows where a function is positive or negative. Holes and vertical asymptotes are excluded from the domain.
Extraneous solutionA result from the cleared equation that doesn't satisfy the original rational equation.
Partial fractions decompositionRewriting a rational expression using simpler fractions based on factors of its denominator. If the expression is improper, the result also includes a polynomial quotient.

Mistakes to avoid in Rational Expressions and Functions

Canceling terms instead of factors

In x+3x\frac{x+3}{x}, the xx is a term being added, not a factor, so you can't cancel it to get 3. You can only cancel something that multiplies the entire numerator and the entire denominator. Factor completely first, then cancel whole factors like (x−3)(x-3).

Skipping the check for extraneous solutions

Multiplying by the LCD removes the denominators, so the cleared equation no longer knows which values were forbidden. Before you solve, list the excluded values. Then throw out any answer on that list, and check every answer in the original equation.

Thinking a graph can never cross a horizontal asymptote

A horizontal asymptote only describes what happens as x→±∞x\to\pm\infty, so the graph can cross it for moderate x-values. For example, xx2+1\frac{x}{x^2+1} crosses its asymptote y=0y=0 at the origin. Vertical asymptotes are different, because the function is undefined there and the graph can never touch them.

Finding a hole's y-value with the original function

Plugging the hole's x-value into the original function gives 00\frac{0}{0}, which is undefined. Cancel the common factor first, then plug the x-value into the simplified function. For x2−4x−2\frac{x^2-4}{x-2}, use x+2x+2 at x=2x=2 to get the hole at (2,4)(2,4).

Study unit 7 other ways

Ready to review Unit 7?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.