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Rational Expressions and Functions: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 188 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. Before the halves are products . 15 points. Question 1 of 10.

The expression $\frac{81-x^2}{x^2-4x-45}$ is to be reduced, evaluated at two inputs, and then held against a shorter form that a classmate offers for it. Nothing here can be settled while either half is still written as a sum.
1. Part A.
  Reduce the expression to lowest terms, and give the domain of the version printed above.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Give the number the printed expression returns at $x = -9$ and the number your reduced form returns there. Then do the same at $x = 9$ .
  
  Carry your own answer forward Use whatever reduced form you reached in part A. The credit is for evaluating both forms at the two inputs and comparing them, not for landing on a particular pair of numbers.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate divides out $9-x$ against $x-9$ and reports $\dfrac{x+9}{x+5}$ . Choose one input the printed expression accepts, give what the classmate's form and the printed expression each return there, and say why the two orders of subtraction cannot be treated as one factor.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

$-\dfrac{x+9}{x+5}$ , and the printed expression is defined at every real number except $9$ and $-5$ .
- $\dfrac{-(x+9)}{x+5}$ and $\dfrac{x+9}{-(x+5)}$ are the same expression; $\dfrac{-x-9}{x+5}$ is too
Part B

At $x = -9$ both return $0$ . At $x = 9$ the printed expression returns $\dfrac{0}{0}$ , which is no number at all, while the reduced form returns $-\dfrac{9}{7}$ .

Part C

At $x = 0$ the classmate's form gives $\dfrac{9}{5}$ while the printed expression gives $\dfrac{81}{-45} = -\dfrac{9}{5}$ . The two orders differ by a factor of $-1$ , since $9-x = -(x-9)$ , so cancelling them against each other throws that factor away: it negates every value and so reverses the sign of every nonzero one.
Worked solution

Part A

Factor both halves. The numerator is a difference of squares and the denominator needs two numbers with product $-45$ and sum $-4$ , namely $-9$ and $5$ :

$81-x^2 = (9-x)(9+x), \qquad x^2-4x-45 = (x-9)(x+5)$

Read the domain off that printed denominator now, before anything is divided out: it is zero at $x = 9$ and at $x = -5$ , so those two inputs are excluded and no others are.

The two halves look as though they share nothing, but $9-x$ and $x-9$ are opposites:

$9-x = -(x-9)$

Rewriting the numerator with that in place exposes the common factor, which is nonzero everywhere the domain allows:

$\frac{-(x-9)(x+9)}{(x-9)(x+5)} = -\frac{x+9}{x+5} \qquad (x \neq 9, \; x \neq -5)$

Part B

At $x = -9$ the numerator of the printed expression vanishes while its denominator $(x-9)(x+5)$ does not:

$\frac{81-81}{(-18)(-4)} = 0$

and the reduced form gives $-\dfrac{-9+9}{-9+5} = 0$ as well. The two agree, as they must at any input the printed expression accepts.

At $x = 9$ the printed expression is $\frac{81-81}{81-36-45} = \frac{0}{0}$ , which names no number: every number multiplies $0$ back into $0$ , so nothing is pinned down. The reduced form has no such trouble there:

$-\frac{9+9}{9+5} = -\frac{18}{14} = -\frac{9}{7}$

Part C

Take any input in the domain, say $x = 0$ . The printed expression is

$\frac{81-0}{0-0-45} = \frac{81}{-45} = -\frac{9}{5},$

while the classmate's form gives $\frac{0+9}{0+5} = \frac{9}{5}$ . One input is enough to settle it: the two disagree, so the classmate's form is not the printed expression.

Watch out. The two forms have the same size at every input, which is why this error survives a careless check. Test the sign, not just the magnitude.

The reason is that $9-x$ and $x-9$ are not the same polynomial. They differ by a factor of $-1$ , and the cancellation law removes a factor common to the top and the bottom, not a factor that matches one of them only up to sign. Taking the $-1$ out first, as part A did, keeps it in the answer where it belongs.

In one line

$\dfrac{81-x^2}{x^2-4x-45} = -\dfrac{x+9}{x+5}$ for every real number except $9$ and $-5$ , the factor of $-1$ arriving from $9-x = -(x-9)$ . At $x = -9$ both forms return $0$ ; at $x = 9$ the printed expression returns $\dfrac{0}{0}$ , which is no number, while the reduced form returns $-\dfrac{9}{7}$ , so that is where the two part company. A classmate who cancels $9-x$ against $x-9$ reports $\dfrac{x+9}{x+5}$ , which at $x = 0$ gives $\dfrac{9}{5}$ against the printed expression's $-\dfrac{9}{5}$ : the two orders of subtraction differ by a factor of $-1$ , and that factor cannot be cancelled away.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Factors both halves completely, writing $81-x^2$ as a difference of squares and $x^2-4x-45$ as $(x-9)(x+5)$ . . Worth 2 points.

Rewrites $9-x$ as $-(x-9)$ so that the common factor is exposed, and carries the factor of $-1$ into the answer. . Worth 2 points.

States the domain from the printed denominator, excluding both $9$ and $-5$ . . Worth 1 point.

Part B 5 points

Evaluates both forms at $x = -9$ and reports $0$ from each. . Worth 2 points.

Evaluates both forms at $x = 9$ , identifying the printed expression's value as $\frac{0}{0}$ rather than as $0$ . . Worth 2 points.

Names $x = 9$ as the input at which the two forms part company, and $-9$ as an ordinary input of both. . Worth 1 point.

Part C 5 points

Evaluates the classmate's form and the printed expression at one allowed input and reports both numbers. . Worth 2 points.

Explains that $9-x = -(x-9)$ , so cancelling the two against each other discards a factor of $-1$ , which negates every value and reverses the sign of every nonzero value. . Worth 3 points. needs an explanation, not just an answer
2. One denominator for two fractions . 17 points. Question 2 of 10.

Two fractions are to be combined: $\frac{4}{3x^2+6x} + \frac{5}{2x^2-8}.$ Neither denominator is ready to be used as it stands, and the last part turns to a step a classmate takes on the way there.
1. Part A.
  Factor both denominators completely, give the least common denominator, and list every value of $x$ the sum excludes.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Rewrite each fraction over that denominator and combine them into a single fraction in lowest terms.
  
  Carry your own answer forward Build each fraction up over whatever least common denominator you gave in part A. The credit is for multiplying numerator and denominator by the same missing factors and then combining, not for reaching a particular fraction.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
3. Part C.
  A classmate rewrites the first fraction over the least common denominator by multiplying only its denominator by $2(x-2)$ , writing $\dfrac{4}{6x(x-2)(x+2)}$ . Say what that step does to the fraction, give one allowed input at which the classmate's fraction and $\dfrac{4}{3x^2+6x}$ disagree, and report what each returns there.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 6 points
The answer

Part A

$3x(x+2)$ and $2(x-2)(x+2)$ , so the least common denominator is $6x(x-2)(x+2)$ , and the sum excludes $x = 0$ , $x = 2$ and $x = -2$ .

Part B

$\dfrac{23x-16}{6x(x-2)(x+2)}$ , which is already in lowest terms.

Part C

It divides the fraction's value by $2(x-2)$ , since only a factor applied to both halves leaves a fraction alone. At $x = 1$ the printed fraction is $\dfrac{4}{9}$ while the classmate's is $-\dfrac{2}{9}$ .

Worked solution

Part A

Take the numerical factor out of each denominator, then factor what is left:

$3x^2+6x = 3x(x+2), \qquad 2x^2-8 = 2(x^2-4) = 2(x-2)(x+2)$

The distinct non-constant factors are $x$ , $x+2$ and $x-2$ , each reaching only the first power in a single denominator, and the coefficients contribute $\operatorname{lcm}(3,2) = 6$ :

$\text{LCD} = 6x(x-2)(x+2)$

The excluded values come from the printed denominators, and the factored LCD lists them all at once: $x = 0$ , $x = 2$ and $x = -2$ .

Part B

The first fraction is missing the factors $2(x-2)$ and the second is missing $3x$ . Multiply each fraction's numerator and denominator by what it lacks:

$\frac{4}{3x(x+2)} \cdot \frac{2(x-2)}{2(x-2)} = \frac{8x-16}{6x(x-2)(x+2)}, \qquad \frac{5}{2(x-2)(x+2)} \cdot \frac{3x}{3x} = \frac{15x}{6x(x-2)(x+2)}$

Now the pieces are the same size, so the counts add:

$\frac{(8x-16) + 15x}{6x(x-2)(x+2)} = \frac{23x-16}{6x(x-2)(x+2)}$

The numerator vanishes only at $x = \frac{16}{23}$ , which is none of $0$ , $2$ or $-2$ , so it shares no factor with the denominator and the fraction is in lowest terms.

Part C

Building a fraction up rests on multiplying by a disguised $1$ :

$\frac{C}{D} = \frac{C}{D} \cdot \frac{K}{K} = \frac{CK}{DK} \qquad (D \neq 0, \; K \neq 0)$

The classmate multiplied $D$ by $K = 2(x-2)$ and left $C$ alone, which is multiplying by $\frac{1}{K}$ rather than by $\frac{K}{K}$ . The value is therefore divided by $2(x-2)$ .

One input settles it. At $x = 1$ , which the sum allows,

$\frac{4}{3(1)^2+6(1)} = \frac{4}{9}, \qquad \frac{4}{6(1)(1-2)(1+2)} = \frac{4}{-18} = -\frac{2}{9}$

Watch out. Here the two even differ in sign, because $2(x-2)$ is negative at $x = 1$ . A wrong step that only shrinks a value on one side of the page can flip it on the other.

In one line

The denominators factor as $3x(x+2)$ and $2(x-2)(x+2)$ , so the least common denominator is $6x(x-2)(x+2)$ and the sum excludes $x = 0$ , $x = 2$ and $x = -2$ . Built up and combined, the sum is $\dfrac{23x-16}{6x(x-2)(x+2)}$ , in lowest terms because $23x-16$ vanishes only at $\frac{16}{23}$ . Multiplying only a denominator by $2(x-2)$ divides the fraction's value by $2(x-2)$ instead of leaving it alone: at $x = 1$ the printed fraction is $\dfrac{4}{9}$ and the classmate's is $-\dfrac{2}{9}$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Factors each denominator completely, including pulling out the numerical factors $3$ and $2$ . . Worth 2 points.

Builds the least common denominator as $6x(x-2)(x+2)$ , rather than the product of the two whole denominators, $6x(x+2)^2(x-2)$ . . Worth 2 points.

Lists all three excluded values, $0$ , $2$ and $-2$ . . Worth 1 point.

Part B 6 points

Multiplies each fraction's numerator and denominator by the factors it lacks, $2(x-2)$ and $3x$ respectively. . Worth 3 points.

Combines over the single denominator and collects the numerator to $23x-16$ . . Worth 2 points.

Checks that the numerator shares no factor with the denominator, so the fraction is in lowest terms. . Worth 1 point.

Part C 6 points

Identifies the step as multiplying the denominator alone, so the value is divided by $2(x-2)$ rather than left alone. . Worth 2 points.

Explains that only a factor applied to numerator and denominator alike is a disguised $1$ and preserves the value. . Worth 2 points. needs an explanation, not just an answer

Gives one allowed input with both values, for instance $\frac{4}{9}$ against $-\frac{2}{9}$ at $x = 1$ . . Worth 2 points.
3. Pouring in what is already there . 18 points. Question 3 of 10.

A vat holds $40$ litres of a mixture that is $25\%$ acid by volume. Pure acid is stirred in, $x$ litres of it, and nothing else is added or removed. Volumes are in litres, and $x$ is not negative.
1. Part A.
  Write the fraction of the mixture that is acid, once $x$ litres of pure acid have been stirred in, as a single rational expression in $x$ . State the value of $x$ your expression excludes, and say separately what the vat itself allows.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 6 points
2. Part B.
  Find how many litres of pure acid make the mixture $40\%$ acid, and check the answer against the vat.
  
  Carry your own answer forward Set whatever expression you wrote in part A equal to the target fraction and solve. The credit is for clearing the denominator and testing the candidate, not for a particular number of litres.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  A technician asks for a mixture that is $100\%$ acid. Solve the equation that asks for it, say what the answer means for the vat, and describe what happens to the acid fraction as more and more pure acid is poured in.
  
  Carry your own answer forward Set your part A expression equal to $1$ and clear it. The credit is for reading the statement the clearing leaves standing and for explaining the vat's behaviour, not for a particular equation.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
The answer

Part A

$\dfrac{10+x}{40+x}$ . The expression excludes $x = -40$ ; the vat allows only $x \ge 0$ , so the excluded value is never in reach.

Part B

$10$ litres. The vat then holds $20$ litres of acid in $50$ litres of mixture, and $\frac{20}{50} = 40\%$ .

Part C

Clearing gives $10 = 40$ , which is false, so there is no candidate at all and no volume works. Written as $1 - \dfrac{30}{40+x}$ , the fraction climbs toward $1$ without reaching it: the $30$ litres that are not acid never leave.

Worked solution

Part A

The vat starts with $25\%$ of $40$ litres of acid, that is $10$ litres, inside a total of $40$ litres. Pouring in $x$ litres of pure acid adds $x$ to the acid and $x$ to the total, since the acid is part of the mixture:

$\text{fraction that is acid} = \frac{10+x}{40+x}$

As an expression this is undefined only where $40 + x = 0$ , that is at $x = -40$ . That is a statement about the algebra. The vat imposes its own, separate condition: you cannot pour in a negative volume, so $x \ge 0$ , and every value the vat allows is one the expression accepts.

Part B

Set the fraction equal to $\frac{2}{5}$ and clear the denominator, which is nonzero throughout the allowed range:

$\frac{10+x}{40+x} = \frac{2}{5} \;\Longrightarrow\; 5(10+x) = 2(40+x) \;\Longrightarrow\; 50 + 5x = 80 + 2x \;\Longrightarrow\; x = 10$

The candidate $10$ is not the excluded value $-40$ , so it is a genuine solution, and it is not negative, so the vat accepts it too. Checking against the vat: $10 + 10 = 20$ litres of acid inside $40 + 10 = 50$ litres of mixture, and $\frac{20}{50} = 0.4$ .

Part C

Set the fraction equal to $1$ and clear:

$\frac{10+x}{40+x} = 1 \;\Longrightarrow\; 10 + x = 40 + x \;\Longrightarrow\; 10 = 40$

The variable cancels and a false statement is left, so the cleared equation has no roots at all. This is the second of the two ways an equation reaches an empty answer: nothing was produced for the domain to reject.

Dividing tells you what the vat is doing instead:

$\frac{10+x}{40+x} = 1 - \frac{30}{40+x}$

The correction $\frac{30}{40+x}$ is the $30$ litres of non-acid measured against the growing total. It shrinks as $x$ grows, but it is never zero, so the fraction rises toward $1$ and never arrives. The line $y = 1$ is a horizontal asymptote of this model, and the $30$ litres that will not leave the vat are the reason.

In one line

After $x$ litres of pure acid the mixture is $\dfrac{10+x}{40+x}$ acid, an expression excluding $x = -40$ while the vat separately allows only $x \ge 0$ . Setting it equal to $\frac{2}{5}$ gives $x = 10$ litres, leaving $20$ litres of acid in $50$ . Setting it equal to $1$ clears to $10 = 40$ , which is false, so there is no candidate and no volume of acid makes the mixture pure. The reason is visible in $1 - \dfrac{30}{40+x}$ : the $30$ litres that are not acid never leave the vat, so the fraction climbs toward the horizontal asymptote $y = 1$ without reaching it.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 6 points

Identifies the starting acid as $10$ litres from $25\%$ of $40$ . . Worth 2 points.

Adds $x$ to the total as well as to the acid, giving $\frac{10+x}{40+x}$ rather than $\frac{10+x}{40}$ . . Worth 2 points.

Names $x = -40$ as the excluded value and $x \ge 0$ as the separate condition the situation imposes. . Worth 2 points.

Part B 6 points

Sets the modelled fraction equal to $\frac{2}{5}$ and clears the denominator correctly. . Worth 2 points.

Solves to $x = 10$ and confirms the candidate is not an excluded value. . Worth 2 points.

Reports the answer in litres and checks it against the vat's contents, $20$ litres of acid in $50$ . . Worth 2 points.

Part C 6 points

Clears the equation, reads the statement the clearing leaves standing, and distinguishes producing no candidate from rejecting one. . Worth 2 points.

Rewrites the fraction as $1 - \frac{30}{40+x}$ , or argues equivalently, to show the fraction rising toward $1$ . . Worth 2 points.

Reads the result back to the vat: the non-acid volume never leaves, so $y = 1$ is approached and never met. . Worth 2 points.
4. Reading a chain from the left . 21 points. Question 4 of 10.

A chain of two divisions is written below. $\frac{x+1}{x-3} \div \frac{x-6}{x+2} \div \frac{x+7}{x-4}$ Read it left to right, as division is always read. Collect what it demands before rearranging any of it.
1. Part A.
  List every value of $x$ the chain excludes, naming for each one the polynomial whose vanishing forces it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 7 points
2. Part B.
  Carry the chain out and give the result in lowest terms. Say which of the restrictions the result's own denominator still shows.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 7 points
3. Part C.
  Now change the last divisor's numerator from $x+7$ to $x+2$ , leaving everything else alone. Give the new result in lowest terms, and rule on a classmate's claim that $x \neq -2$ may now be dropped because $x+2$ appears both above the bar and below it.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer

Part A

$x = 3$ from $x-3$ ; $x = -2$ from $x+2$ ; $x = 6$ from $x-6$ , the first divisor's numerator; $x = 4$ from $x-4$ ; and $x = -7$ from $x+7$ , the second divisor's numerator.

Part B

$\dfrac{(x+1)(x+2)(x-4)}{(x-3)(x-6)(x+7)}$ . Its denominator shows $x \neq 3$ , $x \neq 6$ and $x \neq -7$ ; the restrictions $x \neq -2$ and $x \neq 4$ appear nowhere in it.

Part C

The new result is $\dfrac{(x+1)(x-4)}{(x-3)(x-6)}$ , and the claim fails. Cancelling removes a factor from the page, never a value from the domain: $x = -2$ is forced twice over here, once by the first divisor's denominator and once by the second divisor's numerator.

Worked solution

Part A

Each division constrains three polynomials: the two denominators, and the divisor's numerator, because a divisor may not be zero. Taking the two divisions in turn,

$\frac{A}{B} \div \frac{C}{D} \quad \text{requires} \quad B \neq 0, \quad D \neq 0, \quad C \neq 0.$

The first division has $B = x-3$ , $D = x+2$ and $C = x-6$ , giving $x \neq 3$ , $x \neq -2$ and $x \neq 6$ . The second division divides that result by $\frac{x+7}{x-4}$ , so its own denominator gives $x \neq 4$ and its numerator gives $x \neq -7$ . Five conditions in all, and only the very first numerator, $x+1$ , is free to vanish.

Part B

Invert each divisor in turn and multiply; the dividend never moves.

$\frac{x+1}{x-3} \cdot \frac{x+2}{x-6} \cdot \frac{x-4}{x+7} = \frac{(x+1)(x+2)(x-4)}{(x-3)(x-6)(x+7)}$

Nothing cancels, since no factor appears both above and below. Even so, two restrictions have gone missing: $x+2$ and $x-4$ began life as denominators and the flips carried them upstairs, where a polynomial is free to vanish. The page now announces $3$ , $6$ and $-7$ and says nothing about $-2$ or $4$ , at which the result returns the ordinary values $\frac{(-1)(0)(-6)}{(-5)(-8)(5)} = 0$ and $\frac{(5)(6)(0)}{(1)(-2)(11)} = 0$ .

Part C

With the edit, the chain is $\frac{x+1}{x-3} \div \frac{x-6}{x+2} \div \frac{x+2}{x-4}$ . Flipping both divisors,

$\frac{x+1}{x-3} \cdot \frac{x+2}{x-6} \cdot \frac{x-4}{x+2} = \frac{(x+1)(x-4)}{(x-3)(x-6)}$

so $x+2$ does cancel. Now collect the conditions from the printed chain, exactly as in part A: $x \neq 3$ , $x \neq -2$ from the first divisor's denominator, $x \neq 6$ from its numerator, $x \neq 4$ from the second divisor's denominator, and $x \neq -2$ again from the second divisor's numerator.

So $x = -2$ arrives from two independent places, and cancelling addresses neither. A cancellation is a statement about the symbols on the page; the domain was fixed by the printed expression before any symbol moved. The reduced form returns $\frac{(-1)(-6)}{(-5)(-8)} = \frac{6}{40} = \frac{3}{20}$ at $x = -2$ , and the printed chain returns nothing there.

In one line

The chain excludes $x = 3$ and $x = -2$ from the first two denominators, $x = 6$ from the first divisor's numerator, $x = 4$ from the second divisor's denominator and $x = -7$ from the second divisor's numerator. Carried out it is $\dfrac{(x+1)(x+2)(x-4)}{(x-3)(x-6)(x+7)}$ , whose denominator displays only $3$ , $6$ and $-7$ , so $-2$ and $4$ survive nowhere but in writing. With $x+7$ replaced by $x+2$ the result reduces to $\dfrac{(x+1)(x-4)}{(x-3)(x-6)}$ , and $x \neq -2$ still stands: cancelling clears a factor from the page rather than a value from the domain, and here $-2$ is forced twice over.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 7 points

Collects the restrictions from both printed denominators of the first division, $x \neq 3$ and $x \neq -2$ . . Worth 2 points.

Collects $x \neq 6$ and $x \neq -7$ from the two divisors' numerators, naming the division symbol as the source. . Worth 3 points.

Collects $x \neq 4$ from the second divisor's denominator, and leaves $x+1$ unconstrained. . Worth 2 points.

Part B 7 points

Inverts each divisor and multiplies across, leaving the dividend where it is. . Worth 3 points.

Reports the product in lowest terms and notes that nothing cancels. . Worth 2 points.

Names $3$ , $6$ and $-7$ as the restrictions the answer still displays, and $-2$ and $4$ as the two it does not. . Worth 2 points.

Part C 7 points

Carries out the edited chain and reports $\frac{(x+1)(x-4)}{(x-3)(x-6)}$ , noting that $x+2$ cancels. . Worth 2 points.

States a verdict on the claim and supports it from what fixes the domain, namely the printed expression rather than what survives cancelling. . Worth 3 points. needs an explanation, not just an answer

Points out that $x = -2$ is forced from two separate places, the first divisor's denominator and the second divisor's numerator. . Worth 2 points.
5. Clearing, and then changing one number . 18 points. Question 5 of 10.

All three parts concern $\frac{x}{x-4} + \frac{5}{x+6} = \frac{40}{x^2+2x-24}.$ One of its three denominators is printed unfactored.
1. Part A.
  Give the least common denominator of the three fractions, and say which values of $x$ it forbids.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Clear the fractions with that denominator, solve what is left, and report the equation's solution set.
  
  Carry your own answer forward Clear with whatever LCD you named in part A, and test each candidate against the excluded values you listed there. The credit is for the clearing and the test, not for a particular solution set.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  Replace the $40$ on the right by a constant $k$ . Decide whether any value of $k$ would make every real candidate an excluded value, and give the argument for your decision.
  
  Carry your own answer forward Repeat your part B clearing with $k$ in place of $40$ , and compare the result with the excluded values you found in part A. The credit is for the comparison, not for a particular coefficient.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer

Part A

The least common denominator is $(x-4)(x+6)$ , since $x^2+2x-24$ factors as $(x-4)(x+6)$ . It is zero exactly at $x = 4$ and $x = -6$ , and those are the forbidden values.

Part B

The cleared equation is $x^2+11x-60 = 0$ , with candidates $x = -15$ and $x = 4$ . The candidate $4$ is an excluded value, so it is extraneous; the solution set is $\{-15\}$ .

Part C

No value of $k$ does it. Clearing always gives $x^2+11x-(20+k) = 0$ , whose two roots sum to $-11$ . Two roots drawn from $\{4, -6\}$ , repeats allowed, sum to $8$ , $-2$ or $-12$ , never $-11$ . So whenever real candidates exist, at least one of them is genuine.

Worked solution

Part A

Two numbers with product $-24$ and sum $2$ are $6$ and $-4$ , so

$x^2+2x-24 = (x-4)(x+6).$

The three denominators are therefore $x-4$ , $x+6$ and $(x-4)(x+6)$ . Setting each factor to zero gives the excluded values $x = 4$ and $x = -6$ , both of which the two linear denominators already display once the quadratic is factored. The least common denominator is $(x-4)(x+6)$ , which the third denominator already is, and being built from those factors it is zero at exactly the values the equation forbids.

Part B

Multiply each of the three terms by $(x-4)(x+6)$ and cancel:

$x(x+6) + 5(x-4) = 40$

Expand and collect:

$x^2 + 6x + 5x - 20 = 40 \;\Longrightarrow\; x^2 + 11x - 60 = 0 \;\Longrightarrow\; (x+15)(x-4) = 0$

The candidates are $x = -15$ and $x = 4$ . Test each against the excluded list, one number at a time: $4$ is on it, so it is extraneous however correctly it was derived; $-15$ is not, so it is genuine. Substituting $-15$ into the original as a check on the arithmetic, the left side is $\frac{-15}{-19} + \frac{5}{-9} = \frac{135-95}{171} = \frac{40}{171}$ and the right side is $\frac{40}{225-30-24} = \frac{40}{171}$ .

Part C

Clear the general equation exactly as in part B, with $k$ in place of $40$ :

$x(x+6) + 5(x-4) = k \;\Longrightarrow\; x^2 + 11x - (20+k) = 0$

Whatever $k$ is, the coefficient of $x^2$ is $1$ and the coefficient of $x$ is $11$ : the constant term is the only thing $k$ can move. For a monic quadratic the sum of the two roots is minus the middle coefficient, so

$r_1 + r_2 = -11 \qquad \text{for every } k.$

Now suppose every real candidate were excluded. Each root would then have to be $4$ or $-6$ , the two values part A forbids, and there are only three ways to choose two roots from that pair:

$4+4 = 8, \qquad 4+(-6) = -2, \qquad (-6)+(-6) = -12.$

None of the three is $-11$ , so no $k$ arranges it, repeated roots included.

Watch out. That is a statement about REAL candidates only, and it is not the same as saying the equation always has a solution. The discriminant here is $121 + 4(20+k) = 201 + 4k$ , so at $k = -51$ it is $-3$ and the cleared equation has no real roots at all. There is then nothing for the domain to reject and nothing to solve the equation either.

In one line

The quadratic denominator factors as $(x-4)(x+6)$ , so the excluded values are $4$ and $-6$ and the LCD is $(x-4)(x+6)$ . Clearing gives $x^2+11x-60 = 0$ with candidates $-15$ and $4$ ; the candidate $4$ is excluded and therefore extraneous, so the solution set is $\{-15\}$ . With $k$ in place of $40$ the clearing always gives $x^2+11x-(20+k) = 0$ , whose roots sum to $-11$ whatever $k$ is, while two roots drawn from the forbidden pair $\{4,-6\}$ can only sum to $8$ , $-2$ or $-12$ . So no $k$ makes every real candidate excluded: whenever real candidates exist, at least one is genuine. That is not a promise of a solution, since $k = -51$ leaves the discriminant $201+4k$ negative and no real candidates at all.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Factors $x^2+2x-24$ as $(x-4)(x+6)$ . . Worth 2 points.

Lists both excluded values, $4$ and $-6$ , before clearing or solving the equation. . Worth 2 points.

States the LCD as $(x-4)(x+6)$ rather than the product of all three denominators. . Worth 1 point.

Part B 6 points

Multiplies all three terms by the LCD and reaches $x(x+6) + 5(x-4) = 40$ . . Worth 2 points.

Solves the cleared quadratic to the candidates $-15$ and $4$ . . Worth 2 points.

Rejects $4$ as extraneous by comparing it with the original excluded values, and reports the solution set as $\{-15\}$ . . Worth 2 points.

Part C 7 points

Clears the general equation to $x^2+11x-(20+k) = 0$ and observes that $k$ moves only the constant term. . Worth 2 points.

Reads the root sum off the fixed middle coefficient as $-11$ , and lists the sums two roots drawn from $\{4,-6\}$ can have, repeats included. . Worth 2 points.

States a verdict on whether such a $k$ exists and supports it by comparing those sums with $-11$ , so that every $k$ is settled at once. . Worth 3 points. needs an explanation, not just an answer
6. Every feature of one rule . 19 points. Question 6 of 10.

The rule $f(x) = \frac{2x^2+2x-24}{x^2-9x+18}$ arrives unfactored, and every question below is settled by algebra rather than by plotting.
1. Part A.
  Factor the top and the bottom completely, state the domain of $f$ , and classify each excluded input, giving the reason from the count of copies.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 7 points
2. Part B.
  Give the coordinates of the hole, both intercepts, and the horizontal asymptote.
  
  Carry your own answer forward Evaluate whatever reduced rule you reached in part A. The credit is for taking the hole's height from the reduced expression and the intercepts from the domain, not for a particular set of coordinates.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  Describe what the graph does immediately on each side of $x = 6$ , and what it does far out to the right and far out to the left, giving the reason for each from the reduced rule.
  
  Carry your own answer forward Work from your own reduced rule from part A. The credit is for the reasoning you run on it, not for a particular pair of descriptions.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
The answer

Part A

$\dfrac{2(x+4)(x-3)}{(x-3)(x-6)}$ ; domain: every real except $3$ and $6$ . $x = 3$ is a hole (one copy above, one below) and $x = 6$ a vertical asymptote (none above, one below).

Part B

Hole at $\left(3, -\frac{14}{3}\right)$ ; $x$ -intercept $(-4, 0)$ ; $y$ -intercept $\left(0, -\frac{4}{3}\right)$ ; horizontal asymptote $y = 2$ .

Part C

Written as $2 + \dfrac{20}{x-6}$ : just left of $6$ the correction is large and negative, so the curve dives; just right of $6$ it is large and positive, so the curve climbs. Far to the right it approaches $y = 2$ from above, and far to the left from below.

Worked solution

Part A

Take the common factor $2$ out of the numerator first, then factor each quadratic:

$2x^2+2x-24 = 2(x^2+x-12) = 2(x+4)(x-3), \qquad x^2-9x+18 = (x-3)(x-6)$

The original denominator is zero at $x = 3$ and $x = 6$ , so the domain is every real number except those two.

Now count copies of each denominator factor. The factor $x-3$ appears once above and once below, so it cancels completely and nothing of it survives underneath: $x = 3$ is a hole. The factor $x-6$ appears once below and not at all above, so a copy survives and the values blow up there: $x = 6$ is a vertical asymptote. Reduced,

$f(x) = \frac{2(x+4)}{x-6} \qquad (x \neq 3, \; x \neq 6).$

Part B

Work from the reduced rule $\frac{2(x+4)}{x-6}$ , which agrees with $f$ everywhere $f$ is defined.

The hole's height is that expression at $x = 3$ :

$\frac{2(3+4)}{3-6} = \frac{14}{-3} = -\frac{14}{3}$

The $x$ -intercept is the zero of the reduced numerator that lies in the domain, $x = -4$ , giving $(-4,0)$ . The $y$ -intercept is $f(0) = \frac{2(4)}{-6} = -\frac{4}{3}$ , and $0$ is in the domain, so $\left(0,-\frac{4}{3}\right)$ is a point of the graph.

For the end behaviour the degrees tie at $2$ in the printed rule and at $1$ in the reduced one, and cancelling changes neither the comparison nor the ratio. The leading coefficients are $2$ and $1$ , so the horizontal asymptote is $y = 2$ .

Part C

Divide the reduced rule, which separates the part that matters far out from the part that matters near the asymptote:

$\frac{2(x+4)}{x-6} = \frac{2x+8}{x-6} = 2 + \frac{20}{x-6}$

The check is $2(x-6) + 20 = 2x + 8$ .

Near $x = 6$ the correction $\frac{20}{x-6}$ is what decides everything. Just to the left, $x-6$ is a small negative number, so the correction is large and negative and the curve dives toward the bottom of the page. Just to the right, $x-6$ is a small positive number, so the correction is large and positive and the curve climbs. This is the sign test of a sketch, run at the one place the sign can change.

Far out the correction dies away. For large positive $x$ it is small and positive, so the curve sits just above $y = 2$ ; for large negative $x$ it is small and negative, so the curve sits just below. The hole at $x = 3$ changes none of this: the factor that made it cancelled out of the formula.

In one line

$f(x) = \dfrac{2(x+4)(x-3)}{(x-3)(x-6)}$ , so the domain is every real number except $3$ and $6$ ; $x = 3$ is a hole because its factor cancels completely, and $x = 6$ is a vertical asymptote because a copy of $x-6$ survives below the bar. Reduced, $f(x) = \dfrac{2(x+4)}{x-6}$ , so the hole sits at $\left(3,-\frac{14}{3}\right)$ , the intercepts are $(-4,0)$ and $\left(0,-\frac{4}{3}\right)$ , and the tied degrees with leading coefficients $2$ and $1$ give the horizontal asymptote $y = 2$ . Written as $2 + \dfrac{20}{x-6}$ , the curve dives just left of $6$ and climbs just right of it, and settles onto $y = 2$ from above on the right and from below on the left.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 7 points

Factors both halves completely, taking the $2$ out of the numerator before factoring $x^2+x-12$ . . Worth 2 points.

States the domain from the original denominator, excluding $3$ and $6$ . . Worth 2 points.

Classifies $3$ as a hole and $6$ as a vertical asymptote by counting copies of each factor, not merely by noting whether the factor appears upstairs. . Worth 3 points.

Part B 6 points

Evaluates the reduced expression at the hole's input rather than the printed rule, reaching height $-\frac{14}{3}$ . . Worth 2 points.

Gives both intercepts as coordinate pairs, checking that each input is in the domain. . Worth 2 points.

Names $y = 2$ from the tied degrees and the ratio of leading coefficients, not from the constant terms. . Worth 2 points.

Part C 6 points

Rewrites the reduced rule as a constant plus a proper fraction, or argues equivalently from the signs of numerator and denominator. . Worth 2 points.

Describes both branches at $x = 6$ , distinguishing the two sides by the sign of $x-6$ . . Worth 2 points.

Says which side of $y = 2$ the curve lies on far to the right and far to the left, with the reason. . Worth 2 points.
7. A short reduction, and a domain that is not . 20 points. Question 7 of 10.

$\frac{\frac{1}{x}-\frac{1}{5}}{\frac{1}{x^2}-\frac{1}{25}}$ Reducing this expression is the quick half of the question. Deciding which real numbers it will accept takes longer, and a classmate's rule for doing that is on trial in the last part.
1. Part A.
  Simplify the complex fraction completely.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
2. Part B.
  List every value of $x$ the printed complex fraction excludes, naming the denominator responsible for each, and say which of them your answer still refuses.
  
  Carry your own answer forward Compare the exclusions with whatever answer you reached in part A. The credit is for hunting every denominator, the long bar included, and then reading your own answer's denominator.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 7 points
3. Part C.
  A classmate says that every value a complex fraction excludes is a zero of one of the small denominators printed inside it. Produce a counterexample from this expression, and state what a search that misses nothing has to look at.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 7 points
The answer
Part A

$\dfrac{5x}{x+5}$
- $\dfrac{5x}{5+x}$ is the same expression
Part B

$x = 0$ from the inner denominators $x$ and $x^2$ , and $x = 5$ and $x = -5$ from the bottom of the long bar, which may not be zero. Of these, $\dfrac{5x}{x+5}$ refuses only $-5$ .

Part C

$x = 5$ is the counterexample: it is a zero of neither $x$ nor $x^2$ , yet the bottom of the long bar vanishes there and one may not divide by zero. A complete search reads every inner denominator and then sets the whole bottom expression to zero as well.
Worked solution

Part A

Combine the top over $5x$ and the bottom over $25x^2$ :

$\frac{1}{x} - \frac{1}{5} = \frac{5-x}{5x}, \qquad \frac{1}{x^2} - \frac{1}{25} = \frac{25-x^2}{25x^2} = \frac{(5-x)(5+x)}{25x^2}$

The long bar is a division, so multiply the top by the bottom's reciprocal:

$\frac{5-x}{5x} \cdot \frac{25x^2}{(5-x)(5+x)} = \frac{25x^2(5-x)}{5x(5-x)(5+x)} = \frac{5x}{x+5}$

Part B

There are two kinds of denominator to hunt through. The inner ones, $x$ and $x^2$ , force $x \neq 0$ ; the constants $5$ and $25$ force nothing.

The long bar is the third. Its bottom is $\frac{25-x^2}{25x^2}$ , and a fraction with a nonzero denominator is zero exactly when its numerator is, so

$25 - x^2 = 0 \;\Longrightarrow\; x = 5 \text{ or } x = -5.$

Dividing by zero is undefined, so both are excluded as well. Three exclusions in all: $0$ , $5$ and $-5$ .

The answer $\frac{5x}{x+5}$ has one denominator, $x+5$ , so it refuses $-5$ on its own. At $x = 5$ it returns $\frac{25}{10} = \frac{5}{2}$ and at $x = 0$ it returns $0$ , both perfectly ordinary numbers, so those two restrictions survive only in writing.

Part C

The inner denominators are $x$ and $x^2$ , whose only zero is $0$ . The claim therefore predicts that $0$ is the only exclusion.

Test $x = 5$ against the printed expression:

$\frac{\frac{1}{5}-\frac{1}{5}}{\frac{1}{25}-\frac{1}{25}} = \frac{0}{0}$

which names no number, since every number multiplies $0$ back into $0$ . So $5$ is excluded, and it is a zero of no inner denominator. The claim is false.

Watch out. Both halves vanishing does not rescue the division. A quotient must name the one quantity that multiplies the divisor back to the dividend, and when both are zero every quantity does.

A search that misses nothing therefore has two stages: exclude the zeros of every denominator printed inside, and then set the entire bottom of the long bar equal to zero and exclude its solutions too.

In one line

The complex fraction simplifies to $\dfrac{5x}{x+5}$ . It excludes $x = 0$ , from the inner denominators $x$ and $x^2$ , and $x = 5$ and $x = -5$ , from the bottom of the long bar, which may not be zero; of those three the answer's own denominator refuses only $-5$ . The classmate's claim fails on $x = 5$ , which is a zero of no inner denominator and yet makes the printed expression read $\frac{0}{0}$ . A search that misses nothing excludes the zeros of every inner denominator and then the solutions of setting the whole bottom of the main bar to zero.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 6 points

Combines the numerator and the denominator each into a single fraction before dividing. . Worth 2 points.

Factors $25-x^2$ so that the common factor $5-x$ is visible, rather than leaving it as a difference. . Worth 2 points.

Multiplies by the reciprocal of the bottom and reduces to $\frac{5x}{x+5}$ . . Worth 2 points.

Part B 7 points

Gives $x \neq 0$ from the inner denominators $x$ and $x^2$ . . Worth 2 points.

Sets the bottom of the long bar to zero and obtains $x \neq 5$ and $x \neq -5$ , naming the main division as the source. . Worth 3 points.

Identifies $-5$ as the only exclusion the answer still enforces. . Worth 2 points.

Part C 7 points

Produces a specific excluded value that is a zero of no inner denominator, or argues that none exists. . Worth 2 points.

Shows why it is excluded, by evaluating the bottom of the long bar there or by setting that expression to zero. . Worth 3 points. needs an explanation, not just an answer

States the corrected search: every inner denominator, and then the whole bottom of the main bar. . Worth 2 points.
8. One more score, and then more . 19 points. Question 8 of 10.

A student's mean mark over $n$ tests is $78$ . One further test is taken, scoring $96$ , and the mean over all the tests is then $80$ . Every test carries the same weight, and $n$ counts whole tests.
1. Part A.
  Write an equation in $n$ that says what the new mean is. State the value of $n$ the equation excludes, and say separately what the situation itself requires of $n$ .
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 6 points
2. Part B.
  Solve the equation, report how many tests came before the last one, and give the total marks over all the tests taken so far.
  
  Carry your own answer forward Solve whichever equation you wrote in part A and test the candidate against the excluded value you named there. The credit is for the clearing and the two checks, not for a particular number of tests.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  Further tests are now taken, each scoring $96$ . Write the mean after $k$ of them as a rational expression in $k$ , solve for the $k$ that would make the mean exactly $85$ , and say which of the two filters, if either, rules that value out.
  
  Carry your own answer forward Start from the total and the number of tests you reported in part B. The credit is for building the new mean, solving it, and naming which filter rules on the root, not for a particular value of $k$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer

Part A

$\dfrac{78n+96}{n+1} = 80$ . The equation excludes $n = -1$ ; the situation requires $n$ to be a positive whole number, so the excluded value is never in reach.

Part B

$n = 8$ , so eight tests came before the last one, and the nine tests total $720$ marks.

Part C

The mean is $\dfrac{720+96k}{9+k}$ , and the equation gives $k = \dfrac{45}{11}$ . The domain filter passes it: the only excluded value is $k = -9$ , so it genuinely solves the equation. The situation filter rejects it, since $k$ counts whole tests, so no permitted number of further tests makes the mean exactly $85$ .

Worked solution

Part A

A mean of $78$ over $n$ tests means the marks total $78n$ . Adding one test scoring $96$ makes the total $78n + 96$ over $n+1$ tests, and that mean is $80$ :

$\frac{78n+96}{n+1} = 80$

As an equation this is undefined where $n + 1 = 0$ , that is at $n = -1$ , and that is the whole of its excluded list. The situation adds a condition of a different kind: $n$ counts tests already taken, so it must be a positive whole number. The two filters are separate, and here everything the situation allows is something the equation can evaluate.

Part B

Multiply both sides by $n+1$ , which is nonzero for every $n$ the situation allows:

$78n + 96 = 80(n+1) \;\Longrightarrow\; 78n + 96 = 80n + 80 \;\Longrightarrow\; 16 = 2n \;\Longrightarrow\; n = 8$

The candidate $8$ is not the excluded value $-1$ , so it is genuine, and it is a positive whole number, so the situation accepts it too. Eight tests came before the last one, and there are nine tests in all, totalling $9 \times 80 = 720$ marks. Checking the story, $78 \times 8 = 624$ , and $624 + 96 = 720$ .

Part C

After $k$ further tests at $96$ each, the marks total $720 + 96k$ over $9 + k$ tests:

$\frac{720+96k}{9+k} = 85 \;\Longrightarrow\; 720 + 96k = 765 + 85k \;\Longrightarrow\; 11k = 45 \;\Longrightarrow\; k = \frac{45}{11}$

Run the two filters in order, and say which is which. The domain filter passes it: the only excluded value is $k = -9$ , and $\frac{45}{11}$ is not it, so this root genuinely solves the equation, as substituting it back confirms. The situation filter rejects it: $k$ counts whole tests, and no student sits $\frac{45}{11}$ of a test.

Watch out. This rejection is not an extraneous root. An extraneous root fails the equation's own domain; this one satisfies the equation exactly and fails the story instead.

So the answer to the question as asked is that no permitted value of $k$ exists: the mean can be brought to exactly $85$ by no whole number of further tests at $96$ .

In one line

The mean after the extra test is $\dfrac{78n+96}{n+1} = 80$ , an equation excluding $n = -1$ while the situation separately requires a positive whole number. Solving gives $n = 8$ , so nine tests have been taken and they total $720$ marks. After $k$ further tests at $96$ the mean is $\dfrac{720+96k}{9+k}$ , and setting that equal to $85$ gives $k = \dfrac{45}{11}$ , a genuine root of the equation, since the only excluded value is $k = -9$ . The situation rejects it all the same, because $k$ counts whole tests, so no permitted number of further tests brings the mean to exactly $85$ . That rejection is not an extraneous root: the value satisfies the equation and fails the story.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 6 points

Turns the mean of $78$ over $n$ tests into a total of $78n$ marks. . Worth 2 points.

Writes the new mean over $n+1$ tests, not over $n$ , and sets it equal to $80$ . . Worth 2 points.

Names $n = -1$ as the excluded value and, separately, the whole-number requirement the situation imposes. . Worth 2 points.

Part B 6 points

Clears the denominator and solves to $n = 8$ . . Worth 2 points.

Confirms the candidate against both the excluded value and the whole-number requirement. . Worth 2 points.

Reports the answer as a count of tests and gives the total of $720$ marks over nine tests. . Worth 2 points.

Part C 7 points

Writes the mean after $k$ further tests as $\frac{720+96k}{9+k}$ , adding $k$ to the count as well as $96k$ to the total. . Worth 2 points.

Solves the equation to $k = \frac{45}{11}$ . . Worth 2 points.

Names the situation, not the domain, as what rejects $\frac{45}{11}$ , and explains why that is not an extraneous root. . Worth 3 points. needs an explanation, not just an answer
9. One rule, divided two ways . 20 points. Question 9 of 10.

$f(x) = \frac{2x^3+x^2-13x+6}{x^2-4}$ The numerator's degree is one more than the denominator's, so a slanted line is in prospect. Whether the curve ever touches it is a separate question.
1. Part A.
  Factor the numerator and the denominator completely, state the domain of $f$ , and classify each excluded input.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 7 points
2. Part B.
  Divide the reduced rule, report $f$ as a quotient plus a remainder term, and name the line the curve settles onto far from the origin.
  
  Carry your own answer forward Divide whichever reduced rule you reached in part A. The credit is for the division you carry out and the line you read off your own quotient, not for a particular line.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
3. Part C.
  Decide whether the graph of $f$ ever meets the line you named in part B. Then divide the PRINTED rule by $x^2-4$ , and say what the remainder you get there does and does not settle about crossings.
  
  Carry your own answer forward Work from the decomposition you wrote in part B and the line you named there. The credit is for the argument you run on your own remainders, not for a particular verdict.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
The answer

Part A

$f(x) = \dfrac{(x-2)(2x-1)(x+3)}{(x-2)(x+2)}$ , with domain every real number except $2$ and $-2$ . The factor $x-2$ cancels, so $x = 2$ is a hole; $x = -2$ is a vertical asymptote.

Part B

$f(x) = 2x + 1 - \dfrac{5}{x+2}$ for $x \neq 2$ , so the oblique asymptote is $y = 2x+1$ .

Part C

The graph never meets the line: the gap is $-\dfrac{5}{x+2}$ , which is never $0$ . Dividing the printed rule gives the remainder $-5(x-2)$ , which does vanish, at $x = 2$ . That settles nothing, because $2$ is the hole: a remainder's zero produces a crossing only at an input the function accepts.

Worked solution

Part A

The denominator is a difference of squares, $x^2-4 = (x-2)(x+2)$ , so the excluded inputs are $2$ and $-2$ .

For the numerator, test the denominator's zeros first, since a shared factor is exactly what decides the classification. At $x = 2$ ,

$2(8) + 4 - 26 + 6 = 0,$

so by the Factor Theorem $x-2$ divides it. Synthetic division by $2$ on the coefficients $2, 1, -13, 6$ gives $2, 5, -3$ with remainder $0$ , so

$2x^3+x^2-13x+6 = (x-2)(2x^2+5x-3) = (x-2)(2x-1)(x+3).$

At $x = -2$ the numerator is $-16 + 4 + 26 + 6 = 20$ , not zero, so $x+2$ is not a factor. Counting copies: $x-2$ appears once above and once below and cancels completely, so $x = 2$ is a hole; $x+2$ appears once below and never above, so $x = -2$ is a vertical asymptote.

Part B

Work from the reduced rule, which agrees with $f$ everywhere $f$ is defined:

$f(x) = \frac{(2x-1)(x+3)}{x+2} = \frac{2x^2+5x-3}{x+2} \qquad (x \neq 2, \; x \neq -2)$

Divide $2x^2+5x-3$ by $x+2$ . The first step takes $2x(x+2) = 2x^2+4x$ and leaves $x-3$ ; the second takes $1 \cdot (x+2) = x+2$ and leaves $-5$ . Multiplying back checks it: $(2x+1)(x+2) - 5 = 2x^2+5x+2-5 = 2x^2+5x-3$ . So

$f(x) = 2x + 1 - \frac{5}{x+2}.$

The correction is a proper fraction, so it dies away far out and the curve settles onto the line $y = 2x+1$ .

Part C

The vertical gap between the curve and the line is what part B separated out:

$f(x) - (2x+1) = -\frac{5}{x+2}$

A fraction with a nonzero constant on top is never zero, so the gap never closes and the graph never meets the line.

Now divide the printed rule instead of the reduced one. Dividing $2x^3+x^2-13x+6$ by $x^2-4$ gives the quotient $2x+1$ and a remainder that is no longer constant:

$2x^3+x^2-13x+6 = (2x+1)(x^2-4) + (-5x+10), \qquad -5x+10 = -5(x-2)$

That remainder does have a zero, at $x = 2$ , and reading it as a crossing is exactly the trap. The value $x = 2$ is the hole found in part A: the function has no value there, so the graph has no point above it and there is nothing to do the meeting. The two divisions do not disagree; they are two decompositions of the same function, and the second one's extra zero sits precisely where the function has been punched out.

Watch out. A remainder's zero is a candidate for a crossing, not a crossing. It counts only when it is an input the function accepts.

In one line

$f(x) = \dfrac{(x-2)(2x-1)(x+3)}{(x-2)(x+2)}$ , so the domain excludes $2$ and $-2$ ; the factor $x-2$ cancels completely, making $x = 2$ a hole, while $x = -2$ is a vertical asymptote. Dividing the reduced rule gives $f(x) = 2x+1-\dfrac{5}{x+2}$ , so the oblique asymptote is $y = 2x+1$ , and the gap to it is a nonzero constant over $x+2$ , which never closes: the graph never meets its asymptote. Dividing the PRINTED rule by $x^2-4$ instead gives the same quotient with the remainder $-5(x-2)$ , which does vanish, at $x = 2$ . That produces no crossing either, because $2$ is the hole: a remainder's zero is a candidate for a meeting only at an input the function accepts, and this one sits exactly where the graph has been punched out.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 7 points

Factors the denominator and uses its zeros to find the numerator's factor $x-2$ , rather than hunting for roots at random. . Worth 2 points.

Completes the numerator's factorisation to $(x-2)(2x-1)(x+3)$ . . Worth 3 points.

States the domain from the original denominator and classifies $2$ as a hole and $-2$ as a vertical asymptote. . Worth 2 points.

Part B 6 points

Divides to a quotient and a proper remainder, reaching $2x+1-\frac{5}{x+2}$ . . Worth 3 points.

Checks the division by multiplying back. . Worth 1 point.

Names $y = 2x+1$ as the oblique asymptote, on the ground that the correction term dies away. . Worth 2 points.

Part C 7 points

States a verdict on whether the curve meets the line, identifying the gap as the fraction term of part B's decomposition. . Worth 2 points.

Divides the printed rule by $x^2-4$ and reports the remainder $-5(x-2)$ together with its zero. . Worth 2 points.

Explains why that zero produces no crossing, on the ground that a remainder's zero counts only at an input inside the domain and $2$ is the hole. . Worth 3 points. needs an explanation, not just an answer
10. A quotient with a graph . 21 points. Question 10 of 10.

$f(x) = \frac{x^2-7x}{x^2+2x-8} \div \frac{x^2-9x+14}{x^2+8x+16}$ A quotient of rational expressions is a rational function, so it has a graph. What that graph is missing is decided before anything is cancelled.
1. Part A.
  Carry the division out and give the result in lowest terms. List every value of $x$ the printed quotient excludes, naming the polynomial responsible for each.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 7 points
2. Part B.
  Give the coordinates of every point the graph is missing, name the vertical asymptote, and give the horizontal asymptote.
  
  Carry your own answer forward Evaluate whichever reduced rule you reached in part A at whichever excluded inputs cancelled there. The credit is for taking each height from the reduced rule and for sorting the cancelled factors from the surviving one.
  
  Solve and show your work Write each step out, and end with the value and its units. 7 points
3. Part C.
  A classmate reports the graph's $x$ -intercepts as the zeros of the reduced numerator, $0$ and $-4$ . Say what is wrong with that, and state a rule for reading the $x$ -intercepts off a quotient like this one that would work on any example.
  
  Carry your own answer forward Compare the zeros of your own reduced numerator with the excluded list you wrote in part A. The credit is for the comparison and the corrected rule, not for a particular intercept.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 7 points
The answer

Part A

$\dfrac{x(x+4)}{(x-2)^2}$ , excluding $x = -4$ and $x = 2$ from the dividend's denominator, $x = -4$ from the divisor's denominator, and $x = 7$ and $x = 2$ from the divisor's numerator.

Part B

Holes at $(-4, 0)$ and $\left(7, \frac{77}{25}\right)$ ; the vertical asymptote is $x = 2$ ; the horizontal asymptote is $y = 1$ .

Part C

Only $(0,0)$ is an intercept. The input $-4$ is excluded by the printed quotient, so the graph has no point above it: the hole sits on the axis rather than crossing it. The rule is to take the zeros of the reduced numerator and keep only those the ORIGINAL expression admits.

Worked solution

Part A

Factor all four polynomials first:

$\frac{x(x-7)}{(x+4)(x-2)} \div \frac{(x-7)(x-2)}{(x+4)^2}$

Collect the three sets of conditions from this form, before flipping. The dividend's denominator gives $x \neq -4$ and $x \neq 2$ ; the divisor's denominator gives $x \neq -4$ ; and the divisor's numerator gives $x \neq 7$ and $x \neq 2$ , because a divisor may not be zero. Three distinct values: $-4$ , $2$ and $7$ .

Now flip and multiply:

$\frac{x(x-7)}{(x+4)(x-2)} \cdot \frac{(x+4)^2}{(x-7)(x-2)} = \frac{x(x-7)(x+4)^2}{(x+4)(x-2)^2(x-7)} = \frac{x(x+4)}{(x-2)^2}$

Part B

Of the three excluded inputs, two had their factors cancel completely and one did not. Evaluate the reduced rule at each cancelled one for its height:

$\frac{(-4)(-4+4)}{(-4-2)^2} = \frac{0}{(-6)^2} = 0, \qquad \frac{7(7+4)}{(7-2)^2} = \frac{77}{25}$

So the holes sit at $(-4,0)$ and $\left(7,\frac{77}{25}\right)$ . The factor $x-2$ survives below the bar, twice over, so $x = 2$ is a vertical asymptote.

For the end behaviour, the reduced rule has degree $2$ above and degree $2$ below with leading coefficients $1$ and $1$ , so the horizontal asymptote is $y = 1$ .

Part C

The reduced numerator $x(x+4)$ does vanish at $0$ and at $-4$ , so the classmate's arithmetic is right and the conclusion is not.

An $x$ -intercept is a point of the graph whose height is zero, and a graph has a point above an input only when the function is defined there. Part A found $-4$ excluded, forced twice over: it is a zero of the dividend's denominator and of the divisor's denominator alike. So the graph has no point above $x = -4$ at all. What sits there is the hole from part B, at height $0$ , which is a missing point of the axis rather than a crossing of it.

Watch out. A hole whose height happens to be zero is the easiest one to draw as an intercept, because the reduced rule agrees with the axis there.

The input $0$ is a different matter: it is on none of the excluded lists, and the reduced rule gives

$\frac{0(0+4)}{(0-2)^2} = \frac{0}{4} = 0,$

so $(0,0)$ is a genuine intercept.

The rule, stated so that it works anywhere: reduce, take the zeros of the reduced numerator, and then discard every one of them that the original expression excludes.

In one line

The quotient reduces to $\dfrac{x(x+4)}{(x-2)^2}$ and excludes $x = -4$ , $x = 2$ and $x = 7$ : $-4$ and $2$ from the dividend's denominator, $-4$ again from the divisor's denominator, and $7$ and $2$ from the divisor's numerator, which may not vanish. The graph therefore has holes at $(-4,0)$ and $\left(7,\frac{77}{25}\right)$ , a vertical asymptote at $x = 2$ and the horizontal asymptote $y = 1$ . Its only $x$ -intercept is $(0,0)$ : the reduced numerator also vanishes at $-4$ , but that input is outside the domain, so what sits on the axis there is a missing point rather than a crossing. The rule is to take the zeros of the reduced numerator and keep only those the original expression admits.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 7 points

Factors all four polynomials before flipping anything. . Worth 2 points.

Collects $x \neq 7$ and $x \neq 2$ from the divisor's numerator alongside the denominators' restrictions. . Worth 3 points.

Reaches $\frac{x(x+4)}{(x-2)^2}$ in lowest terms. . Worth 2 points.

Part B 7 points

Identifies $-4$ and $7$ as the cancelled inputs and $2$ as the surviving one. . Worth 2 points.

Evaluates the reduced rule at both cancelled inputs and reports the holes as coordinate pairs. . Worth 3 points.

Names $y = 1$ from the tied degrees and the ratio of leading coefficients. . Worth 2 points.

Part C 7 points

Rules on each of the classmate's two candidates against the printed quotient's excluded list, and says what a graph must have above an input before that input can be an intercept. . Worth 3 points. needs an explanation, not just an answer

Reports which of the two candidates survives that test, with the evaluation supporting it. . Worth 2 points.

States the general rule: the zeros of the reduced numerator that lie in the ORIGINAL domain. . Worth 2 points.

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. Before the halves are products 15 points

Part A 5

Factors both halves completely, writing $81-x^2$ as a difference of squares and $x^2-4x-45$ as $(x-9)(x+5)$ . 2
Rewrites $9-x$ as $-(x-9)$ so that the common factor is exposed, and carries the factor of $-1$ into the answer. 2
States the domain from the printed denominator, excluding both $9$ and $-5$ . 1

Part B 5

Evaluates both forms at $x = -9$ and reports $0$ from each. 2
Evaluates both forms at $x = 9$ , identifying the printed expression's value as $\frac{0}{0}$ rather than as $0$ . 2
Names $x = 9$ as the input at which the two forms part company, and $-9$ as an ordinary input of both. 1

Part C 5

Evaluates the classmate's form and the printed expression at one allowed input and reports both numbers. 2
Explains that $9-x = -(x-9)$ , so cancelling the two against each other discards a factor of $-1$ , which negates every value and reverses the sign of every nonzero value. 3

2. One denominator for two fractions 17 points

Part A 5

Factors each denominator completely, including pulling out the numerical factors $3$ and $2$ . 2
Builds the least common denominator as $6x(x-2)(x+2)$ , rather than the product of the two whole denominators, $6x(x+2)^2(x-2)$ . 2
Lists all three excluded values, $0$ , $2$ and $-2$ . 1

Part B 6

Multiplies each fraction's numerator and denominator by the factors it lacks, $2(x-2)$ and $3x$ respectively. 3
Combines over the single denominator and collects the numerator to $23x-16$ . 2
Checks that the numerator shares no factor with the denominator, so the fraction is in lowest terms. 1

Part C 6

Identifies the step as multiplying the denominator alone, so the value is divided by $2(x-2)$ rather than left alone. 2
Explains that only a factor applied to numerator and denominator alike is a disguised $1$ and preserves the value. 2
Gives one allowed input with both values, for instance $\frac{4}{9}$ against $-\frac{2}{9}$ at $x = 1$ . 2

3. Pouring in what is already there 18 points

Part A 6

Identifies the starting acid as $10$ litres from $25\%$ of $40$ . 2
Adds $x$ to the total as well as to the acid, giving $\frac{10+x}{40+x}$ rather than $\frac{10+x}{40}$ . 2
Names $x = -40$ as the excluded value and $x \ge 0$ as the separate condition the situation imposes. 2

Part B 6

Sets the modelled fraction equal to $\frac{2}{5}$ and clears the denominator correctly. 2
Solves to $x = 10$ and confirms the candidate is not an excluded value. 2
Reports the answer in litres and checks it against the vat's contents, $20$ litres of acid in $50$ . 2

Part C 6

Clears the equation, reads the statement the clearing leaves standing, and distinguishes producing no candidate from rejecting one. 2
Rewrites the fraction as $1 - \frac{30}{40+x}$ , or argues equivalently, to show the fraction rising toward $1$ . 2
Reads the result back to the vat: the non-acid volume never leaves, so $y = 1$ is approached and never met. 2

4. Reading a chain from the left 21 points

Part A 7

Collects the restrictions from both printed denominators of the first division, $x \neq 3$ and $x \neq -2$ . 2
Collects $x \neq 6$ and $x \neq -7$ from the two divisors' numerators, naming the division symbol as the source. 3
Collects $x \neq 4$ from the second divisor's denominator, and leaves $x+1$ unconstrained. 2

Part B 7

Inverts each divisor and multiplies across, leaving the dividend where it is. 3
Reports the product in lowest terms and notes that nothing cancels. 2
Names $3$ , $6$ and $-7$ as the restrictions the answer still displays, and $-2$ and $4$ as the two it does not. 2

Part C 7

Carries out the edited chain and reports $\frac{(x+1)(x-4)}{(x-3)(x-6)}$ , noting that $x+2$ cancels. 2
States a verdict on the claim and supports it from what fixes the domain, namely the printed expression rather than what survives cancelling. 3
Points out that $x = -2$ is forced from two separate places, the first divisor's denominator and the second divisor's numerator. 2

5. Clearing, and then changing one number 18 points

Part A 5

Factors $x^2+2x-24$ as $(x-4)(x+6)$ . 2
Lists both excluded values, $4$ and $-6$ , before clearing or solving the equation. 2
States the LCD as $(x-4)(x+6)$ rather than the product of all three denominators. 1

Part B 6

Multiplies all three terms by the LCD and reaches $x(x+6) + 5(x-4) = 40$ . 2
Solves the cleared quadratic to the candidates $-15$ and $4$ . 2
Rejects $4$ as extraneous by comparing it with the original excluded values, and reports the solution set as $\{-15\}$ . 2

Part C 7

Clears the general equation to $x^2+11x-(20+k) = 0$ and observes that $k$ moves only the constant term. 2
Reads the root sum off the fixed middle coefficient as $-11$ , and lists the sums two roots drawn from $\{4,-6\}$ can have, repeats included. 2
States a verdict on whether such a $k$ exists and supports it by comparing those sums with $-11$ , so that every $k$ is settled at once. 3

6. Every feature of one rule 19 points

Part A 7

Factors both halves completely, taking the $2$ out of the numerator before factoring $x^2+x-12$ . 2
States the domain from the original denominator, excluding $3$ and $6$ . 2
Classifies $3$ as a hole and $6$ as a vertical asymptote by counting copies of each factor, not merely by noting whether the factor appears upstairs. 3

Part B 6

Evaluates the reduced expression at the hole's input rather than the printed rule, reaching height $-\frac{14}{3}$ . 2
Gives both intercepts as coordinate pairs, checking that each input is in the domain. 2
Names $y = 2$ from the tied degrees and the ratio of leading coefficients, not from the constant terms. 2

Part C 6

Rewrites the reduced rule as a constant plus a proper fraction, or argues equivalently from the signs of numerator and denominator. 2
Describes both branches at $x = 6$ , distinguishing the two sides by the sign of $x-6$ . 2
Says which side of $y = 2$ the curve lies on far to the right and far to the left, with the reason. 2

7. A short reduction, and a domain that is not 20 points

Part A 6

Combines the numerator and the denominator each into a single fraction before dividing. 2
Factors $25-x^2$ so that the common factor $5-x$ is visible, rather than leaving it as a difference. 2
Multiplies by the reciprocal of the bottom and reduces to $\frac{5x}{x+5}$ . 2

Part B 7

Gives $x \neq 0$ from the inner denominators $x$ and $x^2$ . 2
Sets the bottom of the long bar to zero and obtains $x \neq 5$ and $x \neq -5$ , naming the main division as the source. 3
Identifies $-5$ as the only exclusion the answer still enforces. 2

Part C 7

Produces a specific excluded value that is a zero of no inner denominator, or argues that none exists. 2
Shows why it is excluded, by evaluating the bottom of the long bar there or by setting that expression to zero. 3
States the corrected search: every inner denominator, and then the whole bottom of the main bar. 2

8. One more score, and then more 19 points

Part A 6

Turns the mean of $78$ over $n$ tests into a total of $78n$ marks. 2
Writes the new mean over $n+1$ tests, not over $n$ , and sets it equal to $80$ . 2
Names $n = -1$ as the excluded value and, separately, the whole-number requirement the situation imposes. 2

Part B 6

Clears the denominator and solves to $n = 8$ . 2
Confirms the candidate against both the excluded value and the whole-number requirement. 2
Reports the answer as a count of tests and gives the total of $720$ marks over nine tests. 2

Part C 7

Writes the mean after $k$ further tests as $\frac{720+96k}{9+k}$ , adding $k$ to the count as well as $96k$ to the total. 2
Solves the equation to $k = \frac{45}{11}$ . 2
Names the situation, not the domain, as what rejects $\frac{45}{11}$ , and explains why that is not an extraneous root. 3

9. One rule, divided two ways 20 points

Part A 7

Factors the denominator and uses its zeros to find the numerator's factor $x-2$ , rather than hunting for roots at random. 2
Completes the numerator's factorisation to $(x-2)(2x-1)(x+3)$ . 3
States the domain from the original denominator and classifies $2$ as a hole and $-2$ as a vertical asymptote. 2

Part B 6

Divides to a quotient and a proper remainder, reaching $2x+1-\frac{5}{x+2}$ . 3
Checks the division by multiplying back. 1
Names $y = 2x+1$ as the oblique asymptote, on the ground that the correction term dies away. 2

Part C 7

States a verdict on whether the curve meets the line, identifying the gap as the fraction term of part B's decomposition. 2
Divides the printed rule by $x^2-4$ and reports the remainder $-5(x-2)$ together with its zero. 2
Explains why that zero produces no crossing, on the ground that a remainder's zero counts only at an input inside the domain and $2$ is the hole. 3

10. A quotient with a graph 21 points

Part A 7

Factors all four polynomials before flipping anything. 2
Collects $x \neq 7$ and $x \neq 2$ from the divisor's numerator alongside the denominators' restrictions. 3
Reaches $\frac{x(x+4)}{(x-2)^2}$ in lowest terms. 2

Part B 7

Identifies $-4$ and $7$ as the cancelled inputs and $2$ as the surviving one. 2
Evaluates the reduced rule at both cancelled inputs and reports the holes as coordinate pairs. 3
Names $y = 1$ from the tied degrees and the ratio of leading coefficients. 2

Part C 7

Rules on each of the classmate's two candidates against the printed quotient's excluded list, and says what a graph must have above an input before that input can be an intercept. 3
Reports which of the two candidates survives that test, with the evaluation supporting it. 2
States the general rule: the zeros of the reduced numerator that lie in the ORIGINAL domain. 2

Multiple choice

Free response