Chapter test · nothing is marked until you submit

Special 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, 85 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. A square root, its domain, and an equation to solve . 10 points. Question 1 of 10.

Consider the radical function $h(x) = \sqrt{2x + 3}$ and the equation $h(x) = x$ that it leads to.
1. Part A.
  State the domain of $h$ , and evaluate $h(11)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve $\sqrt{2x + 3} = x$ . Report every solution, and reject any candidate that does not satisfy the original equation.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Squaring both sides is an irreversible step. Explain, in general, why squaring an equation can create a solution the original does not have, and why substituting back into the original equation is the only reliable check.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

Domain $x \ge -\tfrac{3}{2}$ ; $h(11) = 5$ .
Part B

$x = 3$ only; the candidate $x = -1$ is extraneous.
- Both $3$ and $-1$ solve the squared equation, but only $x = 3$ satisfies $\sqrt{2x + 3} = x$ ; reporting $x = -1$ keeps a root the original rejects
Part C

Squaring is not reversible: $A = B$ forces $A^2 = B^2$ , but $A^2 = B^2$ only means $A = B$ or $A = -B$ . So the squared equation also carries solutions of $\sqrt{2x + 3} = -x$ , and such a candidate passes it while failing the original. Only the original, which keeps the sign, rejects the intruder.
Worked solution

Part A

An even radicand cannot be negative, and the root is the principal value.

$2x + 3 \ge 0 \Longrightarrow x \ge -\tfrac{3}{2}, \qquad h(11) = \sqrt{2(11) + 3} = \sqrt{25} = 5$

Part B

Square both sides, then factor.

$\sqrt{2x + 3} = x \Longrightarrow 2x + 3 = x^2 \Longrightarrow x^2 - 2x - 3 = 0 \Longrightarrow (x - 3)(x + 1) = 0$

The candidates are $x = 3$ and $x = -1$ . Testing in the original: $\sqrt{2(3) + 3} = 3$ holds, but $\sqrt{2(-1) + 3} = 1 \ne -1$ , so $x = -1$ is extraneous. Only $x = 3$ survives.

Part C

From $A = B$ it always follows that $A^2 = B^2$ , so no genuine solution is lost. The converse fails, because a square erases a sign.

$A^2 = B^2 \quad\text{means}\quad A = B \ \text{ or } \ A = -B$

Squaring therefore folds in the solutions of $\sqrt{2x + 3} = -x$ , and $x = -1$ belongs to that second family, since $\sqrt{1} = 1 = -(-1)$ . The squared equation cannot tell the families apart, so only the original equation, which still carries the sign, can reject the intruder.

In one line

The domain of $h$ is $x \ge -\tfrac{3}{2}$ and $h(11) = 5$ . The equation $\sqrt{2x + 3} = x$ has the single solution $x = 3$ ; its companion candidate $x = -1$ solves the squared equation only, because squaring cannot distinguish $A = B$ from $A = -B$ , so every candidate must be checked in the original equation.

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 3 points

Sets the radicand $\ge 0$ and solves for the domain, and evaluates $h(11)$ as the principal root. . Worth 2 points.

States the domain as an inequality and reports $h(11)$ as a single nonnegative value. . Worth 1 point.

Part B 4 points

Squares both sides and reaches the quadratic $x^2 - 2x - 3 = 0$ . . Worth 1 point.

Factors and finds both candidates $x = 3$ and $x = -1$ . . Worth 2 points.

Tests each candidate in the original equation and rejects the one that fails. . Worth 1 point.

Part C 3 points

Argues that $A = B$ forces $A^2 = B^2$ but not conversely, so squaring can admit a root of $A = -B$ . . Worth 2 points. needs an explanation, not just an answer

Concludes that only substitution into the original equation can reject the extraneous candidate. . Worth 1 point.
2. An absolute-value equation and an inequality . 9 points. Question 2 of 10.

This question works with the absolute value expression $\lvert 3x - 6 \rvert$ .
1. Part A.
  Solve $\lvert 3x - 6 \rvert = 9$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Solve the inequality $\lvert 3x - 6 \rvert < 9$ , and write the solution as a single interval.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Part A solved $\lvert 3x - 6 \rvert = 9$ . Explain how the number of solutions of $\lvert 3x - 6 \rvert = c$ depends on the sign of the constant $c$ , giving the count in each of the three cases and the reason for it.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer
Part A

$x = 5$ or $x = -1$ .
- $x = -1$ or $x = 5$ in either order; a single answer such as $x = 5$ alone drops the negative case
Part B

$-1 < x < 5$ , the interval $(-1, 5)$ .
- $(-1, 5)$ is the same as $-1 < x < 5$ ; writing it as two separate rays would be the shape a "greater than" inequality gives, not this one
Part C

It depends only on the sign of $c$ : if $c > 0$ there are two solutions, because two numbers sit $c$ units from zero; if $c = 0$ there is one, since only $0$ is at distance zero; and if $c < 0$ there are none, because an absolute value is never negative and cannot equal a negative number.
Worked solution

Part A

The right side is positive, so the inside sits $9$ units from zero on either side.

$3x - 6 = 9 \ \text{ or } \ 3x - 6 = -9 \Longrightarrow x = 5 \ \text{ or } \ x = -1$

Part B

A "less than" absolute value traps the inside in a band.

$-9 < 3x - 6 < 9 \Longrightarrow -3 < 3x < 15 \Longrightarrow -1 < x < 5$

Part C

The bars measure distance from zero, which is never negative, so the count turns entirely on the sign of $c$ .

$c > 0: \ \text{two}; \qquad c = 0: \ \text{one}; \qquad c < 0: \ \text{none}$

For $c > 0$ , exactly two numbers are $c$ units from zero, namely $c$ and $-c$ . For $c = 0$ , only $0$ is at distance zero, so $3x - 6 = 0$ has the one solution $x = 2$ . For $c < 0$ , no number is a negative distance from zero, so there is nothing to find.

In one line

$\lvert 3x - 6 \rvert = 9$ gives $x = 5$ or $x = -1$ , and $\lvert 3x - 6 \rvert < 9$ gives the interval $(-1, 5)$ . The equation $\lvert 3x - 6 \rvert = c$ has two solutions when $c > 0$ , one when $c = 0$ , and none when $c < 0$ , because an absolute value is a nonnegative distance.

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 3 points

Recognizes a positive right side and sets the inside equal to both $9$ and $-9$ . . Worth 2 points.

Solves both linear equations and reports both values. . Worth 1 point.

Part B 3 points

Unwraps the "less than" absolute value into a single double inequality. . Worth 1 point.

Adds $6$ and divides by $3$ across all three parts correctly. . Worth 1 point.

Reports one interval rather than two rays. . Worth 1 point.

Part C 3 points

Grounds the three cases in an absolute value being a nonnegative distance, not in a rule about signs. . Worth 2 points. needs an explanation, not just an answer

States the correct count (two, one, none) for $c$ positive, zero, and negative. . Worth 1 point.
3. Evaluating floors, ceilings, and fractional parts . 9 points. Question 3 of 10.

This question evaluates the floor, the ceiling, and the fractional part $\{x\} = x - \lfloor x \rfloor$ , taking care on negative inputs.
1. Part A.
  Evaluate $\lfloor -3.4 \rfloor$ , $\lceil -3.4 \rceil$ , and $\{-3.4\}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Find every $x$ with $\lceil x \rceil = -2$ , and every $x$ with $\lfloor x \rfloor = 4$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Interpret the value $\{-3.4\}$ from part A: what does it measure about $-3.4$ , and why can a fractional part never be negative?
  
  Carry your own answer forward Interpret whichever value you found for $\{-3.4\}$ in part A, even if it was not the expected one; the credit is for saying what a fractional part measures and why it stays in its range, not for a particular decimal.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

$\lfloor -3.4 \rfloor = -4$ , $\lceil -3.4 \rceil = -3$ , and $\{-3.4\} = 0.6$ .

Part B

$\lceil x \rceil = -2$ for $-3 < x \le -2$ ; $\lfloor x \rfloor = 4$ for $4 \le x < 5$ .

Part C

The fractional part $0.6$ measures how far $-3.4$ sits above its floor $-4$ , the leftover after removing the whole step below it. It can never be negative because the floor is always at or below $x$ , so $x - \lfloor x \rfloor \ge 0$ , and it stays below $1$ because the next integer up exceeds $x$ .

Worked solution

Part A

The input lies between $-4$ and $-3$ , so the floor is the integer to its left and the ceiling the integer to its right.

$\lfloor -3.4 \rfloor = -4, \qquad \lceil -3.4 \rceil = -3, \qquad \{-3.4\} = -3.4 - (-4) = 0.6$

Part B

Read each solution off its defining inequality.

$\lceil x \rceil = -2: \ -3 < x \le -2, \qquad \lfloor x \rfloor = 4: \ 4 \le x < 5$

Part C

By definition $\{x\} = x - \lfloor x \rfloor$ , and $\{-3.4\} = 0.6$ is the distance from $-3.4$ up to the step it stands on, whose floor value is $-4$ . From $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$ , subtracting the floor gives

$0 \le x - \lfloor x \rfloor < 1,$

so the leftover is always at least $0$ and less than $1$ , never negative.

In one line

$\lfloor -3.4 \rfloor = -4$ , $\lceil -3.4 \rceil = -3$ , and $\{-3.4\} = 0.6$ . The ceiling equation holds on $-3 < x \le -2$ and the floor equation on $4 \le x < 5$ . The fractional part $0.6$ is how far $-3.4$ rises above its floor $-4$ ; it is never negative because the floor never exceeds $x$ .

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 3 points

Rounds the floor down to $-4$ and the ceiling to $-3$ , not toward zero. . Worth 2 points.

Computes the fractional part from $x - \lfloor x \rfloor$ , landing in $[0, 1)$ . . Worth 1 point.

Part B 3 points

Gives each solution as the correct half-open interval. . Worth 2 points.

Places the closed end on the correct side: the right for the ceiling, the left for the floor. . Worth 1 point.

Part C 3 points

Explains the fractional part as the distance from the input up to its floor, the leftover above the step. . Worth 2 points.

Argues from $\lfloor x \rfloor \le x$ that a fractional part is never negative. . Worth 1 point. needs an explanation, not just an answer
4. Reading the features of a rational function . 10 points. Question 4 of 10.

Consider the rational function $f(x) = \dfrac{x^2 - 4}{x^2 - 5x + 6}$ .
1. Part A.
  State the domain of $f$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
2. Part B.
  Identify the vertical asymptote and the hole of $f$ , giving the hole's coordinates.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Find the horizontal asymptote of $f$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
4. Part D.
  One line of working states, as a general rule, that every zero of a rational function's denominator gives a vertical asymptote. Explain why that rule is wrong, and name the feature such a zero produces when it fails.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
The answer

Part A

All real numbers except $x = 2$ and $x = 3$ .
Part B

Vertical asymptote $x = 3$ ; hole at $(2, -4)$ .
- The hole is the point $(2, -4)$ , found from the reduced form at $x = 2$ ; naming $x = 2$ as an asymptote instead would miss the cancellation
Part C

$y = 1$ .

Part D

The rule ignores cancellation. A denominator zero is a vertical asymptote only if its factor does not also divide the numerator. When the factor cancels, the shrinking denominator is matched by a shrinking numerator, nothing blows up, and that input is a hole (a single missing point) instead of an asymptote.
Worked solution

Part A

Only the denominator can restrict the domain, so factor it and set each factor to zero.

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \Longrightarrow x = 2 \ \text{ or } \ x = 3$

Part B

Factor top and bottom and cancel.

$f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x - 3)} = \frac{x + 2}{x - 3} \quad (x \ne 2)$

The factor $x - 2$ cancels, so $x = 2$ is a hole at height $\frac{2 + 2}{2 - 3} = -4$ , the point $(2, -4)$ . The factor $x - 3$ does not cancel, so $x = 3$ is a vertical asymptote.

Part C

The numerator and denominator have equal degree, so the horizontal asymptote is the ratio of leading coefficients.

$y = \frac{1}{1} = 1$

Part D

A denominator zero blows the graph up only when its factor survives after cancelling. If the same factor divides the numerator, it cancels, the output stays finite, and the input is a removable hole rather than a vertical asymptote.

$\text{shared factor} \Rightarrow \text{hole}, \qquad \text{unshared factor} \Rightarrow \text{vertical asymptote}$

So the rule fails exactly when a denominator factor is shared with the numerator, and what it produces there is a hole.

In one line

The domain of $f$ excludes $x = 2$ and $x = 3$ . It has a vertical asymptote at $x = 3$ , a hole at $(2, -4)$ where the factor $x - 2$ cancels, and a horizontal asymptote $y = 1$ from the equal degrees. The claim of an asymptote at $x = 2$ overlooks that cancellation, which makes $x = 2$ a hole instead.

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 2 points

Factors the denominator to find its zeros. . Worth 1 point.

Excludes both $x = 2$ and $x = 3$ from the domain. . Worth 1 point.

Part B 3 points

Factors and cancels the common factor $x - 2$ . . Worth 1 point.

Locates the vertical asymptote at the non-cancelling zero $x = 3$ . . Worth 1 point.

Evaluates the reduced form at $x = 2$ to place the hole at $(2, -4)$ . . Worth 1 point.

Part C 2 points

Compares the degrees of top and bottom and finds them equal. . Worth 1 point.

Reports the asymptote as the ratio of leading coefficients, $y = 1$ . . Worth 1 point.

Part D 3 points

Explains that a denominator zero is an asymptote only when its factor does not cancel with the numerator. . Worth 2 points. needs an explanation, not just an answer

Names the feature produced when the factor cancels: a hole. . Worth 1 point.
5. Average cost per item . 7 points. Question 5 of 10.

A workshop's total cost to make $n$ items is a fixed $500$ dollars plus $4$ dollars per item, so the average cost per item is $C(n) = \dfrac{500 + 4n}{n}$ dollars, for $n \ge 1$ .
1. Part A.
  Compute the average cost per item when $n = 100$ items are made.
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
2. Part B.
  Find the horizontal asymptote of $C(n)$ as $n$ grows large.
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
3. Part C.
  Interpret the horizontal asymptote in the language of the workshop: what does it say about the average cost per item as production grows, and why does the fixed cost stop mattering?
  
  Carry your own answer forward Interpret whichever horizontal asymptote you found in part B; the credit is for connecting that limiting value to the average cost and explaining why the fixed cost's share per item shrinks, not for a particular number.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
The answer

Part A

$C(100) = 9$ dollars per item.
Part B

$y = 4$ .
- $y = 4$ ; equivalently $C(n) = 4 + \dfrac{500}{n}$ , which settles toward $4$ as $n$ grows
Part C

As the workshop makes more items, the average cost per item settles toward $4$ dollars, the per-item variable cost. The fixed $500$ dollars is spread over more items, so its share per item shrinks toward nothing, leaving only the $4$ dollars each item costs on its own.
Worked solution

Part A

$C(100) = \frac{500 + 4(100)}{100} = \frac{900}{100} = 9 \text{ dollars per item}$

Part B

Rewrite the average cost as a base plus a shrinking term, or compare degrees.

$C(n) = \frac{4n + 500}{n} = 4 + \frac{500}{n}$

The degrees of top and bottom are equal, so the horizontal asymptote is the ratio of leading coefficients, $y = \frac{4}{1} = 4$ .

Part C

The asymptote $y = 4$ is the value the average cost approaches without reaching. In context, making more items spreads the fixed $500$ dollars over a larger batch, so the fixed share per item, $\dfrac{500}{n}$ , falls toward $0$ as $n$ grows.

$C(n) = 4 + \frac{500}{n}$

What remains is the $4$ dollars of variable cost that every item carries no matter how many are made.

In one line

$C(100) = 9$ dollars per item, and the average cost has horizontal asymptote $y = 4$ . As the workshop makes more items, the average cost falls toward $4$ dollars each, the per-item variable cost, because the fixed $500$ dollars is spread ever more thinly and its share per item shrinks toward 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 2 points

Substitutes $n = 100$ and simplifies to $9$ . . Worth 1 point.

Reports the result as a cost per item in dollars. . Worth 1 point.

Part B 2 points

Compares degrees or rewrites $C(n)$ as $4 + \tfrac{500}{n}$ . . Worth 1 point.

Identifies the horizontal asymptote as $y = 4$ . . Worth 1 point.

Part C 3 points

Reads the asymptote as the average cost the workshop approaches as it makes more items. . Worth 2 points.

Explains that the fixed cost's per-item share shrinks as $n$ grows, leaving the variable cost. . Worth 1 point. needs an explanation, not just an answer
6. Evaluating and graphing a three-piece function . 8 points. Question 6 of 10.

Consider the piecewise function $f(x) = \begin{cases} -1 & \text{if } x < 0, \\ 2x - 1 & \text{if } 0 \le x \le 3, \\ 6 & \text{if } x > 3. \end{cases}$
1. Part A.
  Evaluate $f(-5)$ , $f(0)$ , $f(3)$ , and $f(10)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  State the domain and range of $f$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
3. Part C.
  Determine whether the graph of $f$ connects or jumps at $x = 0$ and at $x = 3$ , and explain each verdict by comparing the heights the neighbouring pieces reach at that boundary.
  
  Carry your own answer forward Use the boundary values you computed in part A, whatever they were, to decide connect or jump at each boundary; the credit is for comparing the two pieces' heights at a boundary, not for a particular pair of numbers.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
The answer

Part A

$f(-5) = -1$ , $f(0) = -1$ , $f(3) = 5$ , $f(10) = 6$ .

Part B

Domain: all real numbers. Range: $-1 \le y \le 5$ together with $y = 6$ .

Part C

At $x = 0$ the left piece heads to $-1$ and the middle piece also gives $-1$ , the same height, so the dots land on one point and the graph connects. At $x = 3$ the middle piece reaches $5$ while the right piece sits at $6$ , different heights, so the graph must leap from $5$ to $6$ and jumps.

Worked solution

Part A

Choose the governing piece for each input.

$f(-5) = -1, \quad f(0) = 2(0) - 1 = -1, \quad f(3) = 2(3) - 1 = 5, \quad f(10) = 6$

Both boundaries $0$ and $3$ satisfy the middle condition, so the middle piece owns them.

Part B

The three conditions cover every real number, so the domain is all real numbers. Restrict each piece to its interval: the constant $-1$ gives $y = -1$ ; the line $2x - 1$ on $[0, 3]$ gives $-1 \le y \le 5$ ; the constant $6$ gives $y = 6$ . Their union is

$-1 \le y \le 5 \ \text{ or } \ y = 6,$

with the values between $5$ and $6$ skipped by the jump.

Part C

Whether a piecewise graph connects or jumps at a boundary is settled by comparing the two pieces' heights there.

$\text{at } x = 0: \ -1 \text{ and } -1 \ (\text{equal}); \qquad \text{at } x = 3: \ 5 \text{ and } 6 \ (\text{differ})$

At $x = 0$ the neighbouring pieces reach the same height $-1$ , so the excluded point of the left piece and the included point of the middle piece coincide and the graph connects. At $x = 3$ the middle piece reaches $5$ (a closed dot) and the right piece $6$ (an open dot), so the graph cannot be drawn without lifting the pen, and it jumps.

In one line

$f(-5) = -1$ , $f(0) = -1$ , $f(3) = 5$ , $f(10) = 6$ . The domain is all real numbers and the range is $-1 \le y \le 5$ together with $y = 6$ . The graph connects at $x = 0$ , where both pieces reach $-1$ , and jumps at $x = 3$ , where the middle piece reaches $5$ but the right piece is $6$ .

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 3 points

Selects the correct piece for each input, including both boundaries. . Worth 2 points.

Notes that the boundaries $0$ and $3$ are governed by the middle piece. . Worth 1 point.

Part B 2 points

States the domain as all real numbers. . Worth 1 point.

Builds the range piece by piece as $[-1, 5]$ together with $\{6\}$ . . Worth 1 point.

Part C 3 points

Decides connect or jump at each boundary by comparing the two pieces' heights there. . Worth 2 points. needs an explanation, not just an answer

Ties the closed and open dots to the including and excluding pieces. . Worth 1 point.
7. A parking garage's charges . 8 points. Question 7 of 10.

A parking garage charges a flat $5$ dollars for the first $2$ hours and $4$ dollars for each hour after that, so the cost for $t$ hours is $C(t) = \begin{cases} 5 & \text{if } 0 < t \le 2, \\ 5 + 4(t - 2) & \text{if } t > 2, \end{cases}$ in dollars.
1. Part A.
  Find $C(1)$ , $C(2)$ , and $C(5)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Does the charge jump or connect at $t = 2$ ? Give the height each tier reaches there.
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
3. Part C.
  A nearby lot charges a flat $4.50$ dollars per hour with no free period, so its cost is $D(t) = 4.5t$ dollars. Compare the two lots for a $1$ -hour stay and for a $4$ -hour stay, and say which is cheaper in each case.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 3 points
The answer

Part A

$C(1) = 5$ dollars, $C(2) = 5$ dollars, $C(5) = 17$ dollars.
Part B

It connects: both tiers reach $5$ dollars at $t = 2$ .
- The two tiers meet at $5$ dollars, so the graph connects at $t = 2$ ; a jump would need them to reach different heights
Part C

For a $1$ -hour stay the garage costs $5$ dollars and the nearby lot $4.50$ dollars, so the nearby lot is cheaper. For a $4$ -hour stay the garage costs $13$ dollars and the nearby lot $18$ dollars, so the garage is cheaper. The garage's flat start costs more for short stays but its lower hourly rate wins for long ones.
Worked solution

Part A

Choose the tier for each time; the boundary $t = 2$ is claimed by the first tier because of its "or equal to."

$C(1) = 5, \quad C(2) = 5, \quad C(5) = 5 + 4(5 - 2) = 17$

Part B

Compare the two tiers at the boundary. The first gives $5$ , and the second gives $5 + 4(2 - 2) = 5$ . The heights match, so the graph connects.

$\text{first tier: } 5, \qquad \text{second tier: } 5 + 4(2 - 2) = 5$

Part C

Evaluate both cost rules at each time.

$C(1) = 5, \ D(1) = 4.5; \qquad C(4) = 5 + 4(2) = 13, \ D(4) = 4.5(4) = 18$

For one hour the nearby lot ( $4.50$ dollars) beats the garage ( $5$ dollars). For four hours the garage ( $13$ dollars) beats the nearby lot ( $18$ dollars). The garage's flat first-two-hours charge is dearer for a short stay, but its cheaper marginal rate of $4$ dollars per hour makes it the better deal for a longer one.

In one line

$C(1) = 5$ , $C(2) = 5$ , and $C(5) = 17$ dollars, and the charge connects at $t = 2$ because both tiers reach $5$ dollars. Against a flat $4.50$ -dollars-per-hour lot, the garage is dearer for a $1$ -hour stay ( $5$ versus $4.50$ dollars) but cheaper for a $4$ -hour stay ( $13$ versus $18$ dollars).

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 3 points

Chooses the correct tier for each time and computes each charge. . Worth 2 points.

Reports each charge in dollars and gives the $2$ -hour charge as $5$ . . Worth 1 point.

Part B 2 points

Evaluates both tiers at $t = 2$ and gets $5$ each. . Worth 1 point.

Concludes the graph connects because the heights match. . Worth 1 point.

Part C 3 points

Computes both lots' costs at $1$ hour and at $4$ hours. . Worth 1 point.

Names the cheaper lot in each case and explains the trade-off between the flat start and the hourly rate. . Worth 2 points.
8. An absolute value written as two pieces . 7 points. Question 8 of 10.

Consider the absolute value function $f(x) = \lvert 2x - 6 \rvert$ .
1. Part A.
  Write $f(x) = \lvert 2x - 6 \rvert$ as a piecewise function with no absolute value bars.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
2. Part B.
  Give the coordinates of the vertex of the graph of $f$ , and the value $f(0)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 2 points
3. Part C.
  Using your two pieces from part A, justify why the graph of $f$ connects with no jump at the boundary between the pieces, and explain why every absolute value function connects rather than jumps at its vertex.
  
  Carry your own answer forward Argue from whichever two pieces you wrote in part A, even if they were not the expected ones; the credit is for showing the pieces reach the same height at the boundary and for the general reason an absolute value connects there, not for a particular formula.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$f(x) = \begin{cases} 2x - 6 & \text{if } x \ge 3 \\ 6 - 2x & \text{if } x < 3 \end{cases}$

Part B

Vertex $(3, 0)$ ; $f(0) = 6$ .

Part C

At $x = 3$ the top piece gives $2(3) - 6 = 0$ and the bottom gives $6 - 2(3) = 0$ , the same height, so the pieces meet with no jump. This always happens at a vertex: the boundary is where the inside is zero, and both pieces equal that inside there (one as $+0$ , one as $-0$ ), so they must agree and the graph connects.

Worked solution

Part A

The bars keep the inside when it is nonnegative and negate it when negative. The inside $2x - 6$ is nonnegative for $x \ge 3$ , and $-(2x - 6) = 6 - 2x$ .

$f(x) = \begin{cases} 2x - 6 & \text{if } x \ge 3, \\ 6 - 2x & \text{if } x < 3 \end{cases}$

Part B

The vertex sits where the inside is zero.

$2x - 6 = 0 \Longrightarrow x = 3, \quad f(3) = 0$

So the vertex is $(3, 0)$ , and $f(0) = \lvert 2(0) - 6 \rvert = 6$ .

Part C

At the boundary the two pieces from part A agree.

$2(3) - 6 = 0 \quad\text{and}\quad 6 - 2(3) = 0$

Both reach height $0$ , so the closed dot of one piece lands on the value the other heads toward, and the graph connects. This is general: an absolute value splits at the input where the inside is zero, and there both pieces equal that inside, one as $+(\text{inside})$ and the other as $-(\text{inside})$ , both $0$ . Equal heights at the boundary mean the graph always connects at the vertex, forming the sharp corner of a V rather than a jump.

In one line

$\lvert 2x - 6 \rvert$ equals $2x - 6$ for $x \ge 3$ and $6 - 2x$ for $x < 3$ , with vertex $(3, 0)$ and $f(0) = 6$ . At $x = 3$ both pieces give $0$ , so the graph connects; this holds at every absolute value's vertex, because the split is where the inside is zero and both pieces equal that zero, giving a corner, not a jump.

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 2 points

Splits on the sign of the inside $2x - 6$ , with the boundary at $x = 3$ . . Worth 1 point.

Writes the lower piece as $-(2x - 6) = 6 - 2x$ . . Worth 1 point.

Part B 2 points

Finds the vertex at $(3, 0)$ , where the inside is zero. . Worth 1 point.

Evaluates $f(0) = 6$ . . Worth 1 point.

Part C 3 points

Shows both pieces reach the same height at the boundary, so the graph connects there. . Worth 2 points. needs an explanation, not just an answer

Generalizes: at a vertex both pieces equal the inside, which is zero, so an absolute value always connects. . Worth 1 point.
9. A piecewise rule with a fraction inside . 8 points. Question 9 of 10.

Consider the piecewise function $g(x) = \begin{cases} \dfrac{12}{x + 2} & \text{if } x \le 0, \\ x - 1 & \text{if } x > 0. \end{cases}$
1. Part A.
  Evaluate $g(-4)$ , $g(0)$ , and $g(3)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  State the domain of $g$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
3. Part C.
  Part B restricted the domain of $g$ . Justify that restriction: name which of the two pieces governs each excluded input, say what goes wrong there, and explain why the other piece cannot rescue it.
  
  Carry your own answer forward Argue about whichever input you excluded in part B, even if it was not the expected one; the credit is for identifying the governing piece, showing what fails there, and noting the other piece does not cover it, not for a particular number.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$g(-4) = -6$ , $g(0) = 6$ , $g(3) = 2$ .

Part B

All real numbers except $x = -2$ .

Part C

The excluded input satisfies $x \le 0$ , so the rational piece $\frac{12}{x + 2}$ governs it. There the denominator $x + 2$ is zero, so the piece divides by zero and has no value. The other piece, $x - 1$ , governs only $x > 0$ , so it cannot cover that input. With no defined piece there, $g$ is undefined.

Worked solution

Part A

Pick the governing piece first.

$g(-4) = \frac{12}{-4 + 2} = -6, \quad g(0) = \frac{12}{0 + 2} = 6, \quad g(3) = 3 - 1 = 2$

The boundary $x = 0$ satisfies $x \le 0$ , so the rational piece owns it.

Part B

The piece $x > 0$ is defined for every input in its interval. The piece $\frac{12}{x + 2}$ is undefined at $x = -2$ , and $-2 \le 0$ , so that input falls in the interval where the rational piece governs. Hence $x = -2$ is excluded, and it is the only exclusion.

$\text{domain: all real } x \text{ except } x = -2$

Part C

The excluded input is the one where the rational piece's denominator vanishes, and it lies in that piece's interval $x \le 0$ . There

$x + 2 = 0,$

so $\frac{12}{x + 2}$ divides by zero and returns no value. The piece $x - 1$ governs only $x > 0$ , so it never gets a say at that input. With the governing piece undefined and no other piece covering the input, $g$ is undefined there, which is why it is the one exclusion from the domain.

In one line

$g(-4) = -6$ , $g(0) = 6$ , $g(3) = 2$ . The domain is all real numbers except $x = -2$ : at $x = -2$ the condition $x \le 0$ hands the input to the piece $\frac{12}{x + 2}$ , whose denominator is zero there, so the function divides by zero and is undefined. Every other input has a well-defined governing piece.

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 3 points

Selects the correct piece for each input and computes it, including the boundary $x = 0$ . . Worth 2 points.

Assigns the boundary $x = 0$ to the piece governing $x \le 0$ . . Worth 1 point.

Part B 2 points

Checks where each piece is undefined within its own interval. . Worth 1 point.

Excludes $x = -2$ and nothing else. . Worth 1 point.

Part C 3 points

Names the rational piece as the one governing the excluded input, and shows its denominator is zero there. . Worth 2 points. needs an explanation, not just an answer

Notes the other piece governs a different interval and cannot supply a value, so $g$ is undefined there. . Worth 1 point.
10. The floor function up close . 9 points. Question 10 of 10.

This question looks at the floor function $\lfloor x \rfloor$ on the interval $-1 \le x < 2$ and asks what kind of function it is.
1. Part A.
  Evaluate $\lfloor -1.5 \rfloor$ , $\lfloor 0.9 \rfloor$ , and $\lfloor 1 \rfloor$ , and find every $x$ with $\lfloor x \rfloor = 3$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Write $\lfloor x \rfloor$ as a piecewise function on the interval $-1 \le x < 2$ , using one constant piece per integer step, and say for each step which end is closed and which is open.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  A step function is a piecewise function whose pieces are all constants, with a jump at each boundary. Using your pieces from part B, justify why the floor is a step function that jumps by exactly $1$ at every integer and never connects, and contrast this with the absolute value, which connects at its vertex.
  
  Carry your own answer forward Argue from whichever pieces you wrote in part B, even if they were not the expected ones; the credit is for showing constant pieces one unit apart force a jump at each integer, and for contrasting that with an absolute value's equal boundary heights, not for a particular list of steps.
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$\lfloor -1.5 \rfloor = -2$ , $\lfloor 0.9 \rfloor = 0$ , $\lfloor 1 \rfloor = 1$ ; and $\lfloor x \rfloor = 3$ for $3 \le x < 4$ .

Part B

$\lfloor x \rfloor = \begin{cases} -1 & -1 \le x < 0 \\ 0 & 0 \le x < 1 \\ 1 & 1 \le x < 2 \end{cases}$ , each step closed on the left and open on the right.

Part C

Each floor piece is a constant, and neighbouring pieces differ by $1$ , so at every integer boundary the two pieces reach heights one apart and the graph jumps by $1$ , never meeting. That is what makes it a step function. The absolute value's two pieces instead reach the same height at the vertex, so it connects there.

Worked solution

Part A

Round each input down to the integer at or below it.

$\lfloor -1.5 \rfloor = -2, \quad \lfloor 0.9 \rfloor = 0, \quad \lfloor 1 \rfloor = 1$

The floor equals $3$ on the step it begins, $3 \le x < 4$ .

Part B

On each unit interval the floor holds the value of the left integer, since $\lfloor x \rfloor = n$ for $n \le x < n + 1$ .

$\lfloor x \rfloor = \begin{cases} -1 & \text{if } -1 \le x < 0, \\ 0 & \text{if } 0 \le x < 1, \\ 1 & \text{if } 1 \le x < 2 \end{cases}$

Each step includes its left endpoint (a closed dot) and excludes its right endpoint (an open dot), because the value jumps up at the next integer.

Part C

Every piece of $\lfloor x \rfloor$ is a constant, so it is a piecewise function of the constant kind, a step function. At each integer boundary the piece to the left holds $n - 1$ and the piece to the right holds $n$ , heights exactly $1$ apart:

$\text{left of } n: \ n - 1; \qquad \text{at and right of } n: \ n$

Different heights at the boundary force a jump, and the gap is always $1$ , so the floor jumps by $1$ at every integer and never connects, giving a staircase. The absolute value is the opposite case: its two pieces meet at the same height at the vertex, so equal boundary heights make it connect into a V. The same test, comparing the two pieces' heights, gives opposite verdicts.

In one line

$\lfloor -1.5 \rfloor = -2$ , $\lfloor 0.9 \rfloor = 0$ , $\lfloor 1 \rfloor = 1$ , and $\lfloor x \rfloor = 3$ on $3 \le x < 4$ . On $-1 \le x < 2$ the floor is the three constant pieces $-1$ , $0$ , $1$ , each closed on the left and open on the right. Being constant pieces one unit apart, the floor jumps by $1$ at every integer and never connects, a step function, unlike the absolute value, whose pieces meet at the vertex and connect.

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 3 points

Evaluates each floor correctly, rounding $-1.5$ down to $-2$ . . Worth 2 points.

Solves $\lfloor x \rfloor = 3$ as the half-open interval $3 \le x < 4$ . . Worth 1 point.

Part B 3 points

Gives the three constant pieces with the correct half-open intervals. . Worth 2 points.

States each step is closed on the left and open on the right. . Worth 1 point.

Part C 3 points

Argues that constant pieces one unit apart give a jump of $1$ at every integer, so the floor is a step function that never connects. . Worth 2 points. needs an explanation, not just an answer

Contrasts this with the absolute value, whose pieces meet at equal height and so connect. . Worth 1 point.

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. A square root, its domain, and an equation to solve 10 points

Part A 3

Sets the radicand $\ge 0$ and solves for the domain, and evaluates $h(11)$ as the principal root. 2
States the domain as an inequality and reports $h(11)$ as a single nonnegative value. 1

Part B 4

Squares both sides and reaches the quadratic $x^2 - 2x - 3 = 0$ . 1
Factors and finds both candidates $x = 3$ and $x = -1$ . 2
Tests each candidate in the original equation and rejects the one that fails. 1

Part C 3

Argues that $A = B$ forces $A^2 = B^2$ but not conversely, so squaring can admit a root of $A = -B$ . 2
Concludes that only substitution into the original equation can reject the extraneous candidate. 1

2. An absolute-value equation and an inequality 9 points

Part A 3

Recognizes a positive right side and sets the inside equal to both $9$ and $-9$ . 2
Solves both linear equations and reports both values. 1

Part B 3

Unwraps the "less than" absolute value into a single double inequality. 1
Adds $6$ and divides by $3$ across all three parts correctly. 1
Reports one interval rather than two rays. 1

Part C 3

Grounds the three cases in an absolute value being a nonnegative distance, not in a rule about signs. 2
States the correct count (two, one, none) for $c$ positive, zero, and negative. 1

3. Evaluating floors, ceilings, and fractional parts 9 points

Part A 3

Rounds the floor down to $-4$ and the ceiling to $-3$ , not toward zero. 2
Computes the fractional part from $x - \lfloor x \rfloor$ , landing in $[0, 1)$ . 1

Part B 3

Gives each solution as the correct half-open interval. 2
Places the closed end on the correct side: the right for the ceiling, the left for the floor. 1

Part C 3

Explains the fractional part as the distance from the input up to its floor, the leftover above the step. 2
Argues from $\lfloor x \rfloor \le x$ that a fractional part is never negative. 1

4. Reading the features of a rational function 10 points

Part A 2

Factors the denominator to find its zeros. 1
Excludes both $x = 2$ and $x = 3$ from the domain. 1

Part B 3

Factors and cancels the common factor $x - 2$ . 1
Locates the vertical asymptote at the non-cancelling zero $x = 3$ . 1
Evaluates the reduced form at $x = 2$ to place the hole at $(2, -4)$ . 1

Part C 2

Compares the degrees of top and bottom and finds them equal. 1
Reports the asymptote as the ratio of leading coefficients, $y = 1$ . 1

Part D 3

Explains that a denominator zero is an asymptote only when its factor does not cancel with the numerator. 2
Names the feature produced when the factor cancels: a hole. 1

5. Average cost per item 7 points

Part A 2

Substitutes $n = 100$ and simplifies to $9$ . 1
Reports the result as a cost per item in dollars. 1

Part B 2

Compares degrees or rewrites $C(n)$ as $4 + \tfrac{500}{n}$ . 1
Identifies the horizontal asymptote as $y = 4$ . 1

Part C 3

Reads the asymptote as the average cost the workshop approaches as it makes more items. 2
Explains that the fixed cost's per-item share shrinks as $n$ grows, leaving the variable cost. 1

6. Evaluating and graphing a three-piece function 8 points

Part A 3

Selects the correct piece for each input, including both boundaries. 2
Notes that the boundaries $0$ and $3$ are governed by the middle piece. 1

Part B 2

States the domain as all real numbers. 1
Builds the range piece by piece as $[-1, 5]$ together with $\{6\}$ . 1

Part C 3

Decides connect or jump at each boundary by comparing the two pieces' heights there. 2
Ties the closed and open dots to the including and excluding pieces. 1

7. A parking garage's charges 8 points

Part A 3

Chooses the correct tier for each time and computes each charge. 2
Reports each charge in dollars and gives the $2$ -hour charge as $5$ . 1

Part B 2

Evaluates both tiers at $t = 2$ and gets $5$ each. 1
Concludes the graph connects because the heights match. 1

Part C 3

Computes both lots' costs at $1$ hour and at $4$ hours. 1
Names the cheaper lot in each case and explains the trade-off between the flat start and the hourly rate. 2

8. An absolute value written as two pieces 7 points

Part A 2

Splits on the sign of the inside $2x - 6$ , with the boundary at $x = 3$ . 1
Writes the lower piece as $-(2x - 6) = 6 - 2x$ . 1

Part B 2

Finds the vertex at $(3, 0)$ , where the inside is zero. 1
Evaluates $f(0) = 6$ . 1

Part C 3

Shows both pieces reach the same height at the boundary, so the graph connects there. 2
Generalizes: at a vertex both pieces equal the inside, which is zero, so an absolute value always connects. 1

9. A piecewise rule with a fraction inside 8 points

Part A 3

Selects the correct piece for each input and computes it, including the boundary $x = 0$ . 2
Assigns the boundary $x = 0$ to the piece governing $x \le 0$ . 1

Part B 2

Checks where each piece is undefined within its own interval. 1
Excludes $x = -2$ and nothing else. 1

Part C 3

Names the rational piece as the one governing the excluded input, and shows its denominator is zero there. 2
Notes the other piece governs a different interval and cannot supply a value, so $g$ is undefined there. 1

10. The floor function up close 9 points

Part A 3

Evaluates each floor correctly, rounding $-1.5$ down to $-2$ . 2
Solves $\lfloor x \rfloor = 3$ as the half-open interval $3 \le x < 4$ . 1

Part B 3

Gives the three constant pieces with the correct half-open intervals. 2
States each step is closed on the left and open on the right. 1

Part C 3

Argues that constant pieces one unit apart give a jump of $1$ at every integer, so the floor is a step function that never connects. 2
Contrasts this with the absolute value, whose pieces meet at equal height and so connect. 1

Multiple choice

Free response