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Graphing 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, 127 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. One short list, several different questions . 11 points. Question 1 of 10.

A function $f$ is known at exactly five inputs: $f(-3) = 4$ , $f(-1) = 0$ , $f(0) = -2$ , $f(2) = 4$ , and $f(5) = 1$ . Every part below is answerable from that list alone.
1. Part A.
  Give the value of $f(2)$ , and give the coordinates of the point where the graph meets the vertical axis.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  List every input in the list for which $f(x) = 4$ , and give the coordinates of the one crossing of the horizontal axis that the list contains.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Part A asked you to evaluate $f$ at an input, and part B asked you for the inputs that produce a given output. One of those two tasks is guaranteed to have exactly one answer for any function whatever, and the other could have come out with none, one, or several. Say which is which, and explain what it is about a function that forces the difference.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$f(2) = 4$ , and the graph meets the vertical axis at $(0, -2)$ .

Part B

$f(x) = 4$ at $x = -3$ and at $x = 2$ ; the crossing of the horizontal axis is $(-1, 0)$ .

Part C

Evaluating, as in $f(2)$ , always returns exactly one number, because a function pairs each input with a single output. Solving $f(x) = 4$ runs the other way and can return none, one, or several inputs, since nothing forbids two different inputs from sharing an output.

Worked solution

Part A

The value $f(2)$ is the output paired with the input $2$ , read straight off the list. The graph meets the vertical axis where the input is $0$ , so that point is $(0, f(0))$ .

$f(2) = 4, \qquad \bigl(0,\, f(0)\bigr) = (0, -2).$

Part B

Solving $f(x) = 4$ means searching the outputs for $4$ and reporting every input that produced it.

$f(-3) = 4 \quad \text{and} \quad f(2) = 4 \;\Rightarrow\; x = -3 \text{ or } x = 2.$

A crossing of the horizontal axis is a point whose height is $0$ , and the list has exactly one, at the input $-1$ , giving the point $(-1, 0)$ .

Part C

A function is defined by a one-way promise: each input in the domain is paired with exactly one output. Evaluating uses that promise directly, so $f(2)$ names one number and can never name two.

$\text{input} \longmapsto \text{exactly one output}.$

Solving $f(x) = 4$ travels backwards along the pairing, and the definition says nothing about that direction. Two different inputs are free to share an output, which is exactly what this list does at the height $4$ , so the equation can have several solutions, or one, or none at all if the height is never reached.

In one line

$f(2) = 4$ and the graph meets the vertical axis at $(0, -2)$ ; $f(x) = 4$ at both $x = -3$ and $x = 2$ , and the listed crossing of the horizontal axis is $(-1, 0)$ . Evaluating always returns exactly one output, while solving $f(x) = b$ can return none, one, or several inputs, because the definition of a function controls only the input-to-output direction.

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

Reads the output paired with the input $2$ , rather than an input paired with the output $2$ . . Worth 2 points.

Reports the crossing of the vertical axis as an ordered pair built from the input $0$ , not as a bare number. . Worth 1 point.

Part B 4 points

Collects BOTH inputs whose output is $4$ , rather than stopping at the first one found. . Worth 2 points.

Identifies the crossing of the horizontal axis as the point whose OUTPUT is zero, and reports it as an ordered pair. . Worth 2 points.

Part C 4 points

Names which of the two tasks is the guaranteed single-answer one and which is the possibly-several one. . Worth 2 points.

Ties the difference to the definition of a function, which constrains the input-to-output direction only. . Worth 2 points. needs an explanation, not just an answer
2. One point, two slides . 12 points. Question 2 of 10.

The point $(2, -5)$ lies on the graph of $y = f(x)$ , and nothing else about $f$ is given.
1. Part A.
  Give the point that must lie on the graph of $y = f(x) + 6$ , and the point that must lie on the graph of $y = f(x - 6)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  The graph of $y = f(x)$ is slid $4$ to the left and $3$ down. Write the rule of the resulting graph in terms of $f$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  In part A, one of the two rules moved the point sideways. Say which one, and explain why that move goes in the direction it does, arguing from the single point you tracked rather than from a remembered rule.
  
  Carry your own answer forward Argue from the two images you reported in part A, whatever they were, and from which coordinate each rule changed.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$(2, 1)$ lies on $y = f(x) + 6$ , and $(8, -5)$ lies on $y = f(x - 6)$ .
Part B

$y = f(x + 4) - 3$ .
- $f(x + 4) - 3$
- $y = f\bigl(x - (-4)\bigr) + (-3)$
Part C

The rule with the $6$ inside the function is the sideways one. To hand $f$ the old input $2$ again, the inside $x - 6$ must equal $2$ , which forces $x = 8$ , a larger input, so the point moves right even though the sign written inside is a minus.
Worked solution

Part A

The $+6$ sits outside the function, so it acts on the height and leaves the input alone.

$(2,\; -5 + 6) = (2, 1).$

The $-6$ sits inside, so it decides which input reports the stored height: the inside expression has to equal the old input $2$ .

$x - 6 = 2 \;\Rightarrow\; x = 8, \qquad (2, -5) \longrightarrow (8, -5).$

Part B

Match the description against $y = f(x - h) + k$ . A slide to the left by $4$ is a slide right by $-4$ , so $h = -4$ and the inside becomes $x - (-4)$ . A slide down by $3$ is $k = -3$ , added outside.

$y = f\bigl(x - (-4)\bigr) + (-3) = f(x + 4) - 3.$

Part C

A number inside the function does not change the height it stores; it changes which input reaches that stored height. The height $-5$ lives at the input $2$ of $f$ , so on $y = f(x - 6)$ it appears where

$x - 6 = 2 \;\Rightarrow\; x = 8.$

Subtracting $6$ inside means the input has to be $6$ larger before the inside catches up to the old value, so every feature of the graph shows up $6$ units later, that is, further right. The minus sign describes what is done to $x$ , not the direction the picture travels, which is why the two look opposite.

In one line

$(2, -5)$ goes to $(2, 1)$ on $y = f(x) + 6$ and to $(8, -5)$ on $y = f(x - 6)$ ; sliding left $4$ and down $3$ gives $y = f(x + 4) - 3$ . The inside change is the sideways one, and it runs right because $x - 6$ only reaches the old input $2$ once $x$ has grown to $8$ .

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

Changes the height only, for the number that sits outside the function. . Worth 2 points.

Finds the new input by asking which value of $x$ makes the inside expression equal the old input. . Worth 2 points.

Reports both images as ordered pairs, each with exactly one coordinate changed. . Worth 1 point.

Part B 3 points

Writes the sideways slide inside the function with the sign the direction forces, rather than the sign the description names. . Worth 2 points.

Puts the up-and-down slide outside the function, where it acts on the height. . Worth 1 point.

Part C 4 points

Identifies which of the two rules moved the point sideways, from where its number sits. . Worth 1 point.

Argues from the equation that makes the inside expression equal the old input, and reads the direction off the size of the new input. . Worth 3 points. needs an explanation, not just an answer
3. Three points under two factors . 11 points. Question 3 of 10.

The graph of $y = f(x)$ passes through $(-3, 2)$ and $(0, -4)$ , and it meets the horizontal axis at $(5, 0)$ .
1. Part A.
  Give the images of all three points on the graph of $y = 3f(x)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Give the images of the same three points on the graph of $y = -f(x)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain what any rule of the form $y = a\,f(x)$ does to a point whose height is $0$ , and say what that means for where the graph meets the horizontal axis. Then say which of the two rules above leaves the graph the same shape it had, and why.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$(-3, 6)$ , $(0, -12)$ , and $(5, 0)$ .

Part B

$(-3, -2)$ , $(0, 4)$ , and $(5, 0)$ .

Part C

A height of $0$ becomes $a \times 0 = 0$ , so every point on the horizontal axis stays exactly where it is and the crossings never move. The reflection $y = -f(x)$ keeps the shape, since it only reverses signs, while the factor $3$ makes every height three times as large and so does change the shape.

Worked solution

Part A

The factor sits outside the function, so it multiplies each height and leaves each input alone.

$(-3,\; 3 \times 2) = (-3, 6), \qquad (0,\; 3 \times (-4)) = (0, -12), \qquad (5,\; 3 \times 0) = (5, 0).$

Part B

The minus sign is the factor $-1$ outside the function, so every height changes sign while every input stays.

$(-3,\; -2), \qquad (0,\; 4), \qquad (5,\; 0).$

The point that was $2$ above the horizontal axis is now $2$ below it, and the point that was $4$ below is now $4$ above.

Part C

Whatever the factor $a$ is, it multiplies heights, and multiplying $0$ by anything returns $0$ .

$(x_0,\, 0) \longrightarrow (x_0,\; a \times 0) = (x_0,\, 0).$

So every point sitting on the horizontal axis is its own image, and the crossings of that axis are exactly where they were before, which the point $(5, 0)$ shows in both parts above.

Shape is a separate question, decided by the SIZE of the factor. The size of $-1$ is $1$ , so $y = -f(x)$ flips the graph without making it taller or shorter: same shape, upside down. The size of $3$ is greater than $1$ , so $y = 3f(x)$ pulls every nonzero height three times further from the axis and genuinely changes the shape.

In one line

Under $y = 3f(x)$ the three points become $(-3, 6)$ , $(0, -12)$ and $(5, 0)$ ; under $y = -f(x)$ they become $(-3, -2)$ , $(0, 4)$ and $(5, 0)$ . Any factor leaves a height of $0$ at $0$ , so the crossing $(5, 0)$ never moves, and only the factor whose size differs from $1$ , the $3$ , changes the shape; the reflection flips the graph without reshaping 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 3 points

Multiplies every height by the factor and leaves every input unchanged. . Worth 2 points.

Reports all three images as ordered pairs, the point whose height is zero included. . Worth 1 point.

Part B 3 points

Negates every height, sending a point above the horizontal axis to the matching place below it and the reverse. . Worth 2 points.

Keeps every input unchanged and reports each image as an ordered pair. . Worth 1 point.

Part C 5 points

Argues from the fact that any factor times zero is zero that points at height zero are fixed, so the crossings of the horizontal axis do not move. . Worth 3 points. needs an explanation, not just an answer

States which of the two rules preserves the shape and which does not, deciding it by the size of the factor rather than by its sign. . Worth 2 points.
4. A conversion read the other way . 12 points. Question 4 of 10.

A conversion graph turns a distance in kilometres into the same distance in miles: an input of $k$ kilometres returns $m(k)$ miles. The graph is a straight line through $(0, 0)$ , $(8, 5)$ and $(16, 10)$ , drawn for distances from $0$ up to $16$ kilometres, so $m$ has domain $0 \le k \le 16$ and range $0 \le y \le 10$ .
1. Part A.
  Find $m^{-1}(10)$ and $m^{-1}(5)$ , and say what each of the two answers means in the situation.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  State the domain and the range of $m^{-1}$ , and give the coordinates of two points on its graph.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  The origin lies on the graph of $m$ and also on the graph of $m^{-1}$ . Explain what makes it a point these two graphs are bound to share. Then determine whether these two particular graphs share any other point, and justify your answer.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$m^{-1}(10) = 16$ and $m^{-1}(5) = 8$ : a distance of $10$ miles is $16$ kilometres, and $5$ miles is $8$ kilometres.

Part B

$m^{-1}$ has domain $0 \le x \le 10$ and range $0 \le y \le 16$ , and its graph passes through $(5, 8)$ and $(10, 16)$ .

Part C

The origin has equal coordinates, so it sits on the line $y = x$ and is its own mirror image: a point of $m$ on that diagonal has to lie on $m^{-1}$ as well. No other point is shared here because the two graphs are different straight lines through the origin, and two different lines cross only once.

Worked solution

Part A

An inverse value runs the conversion backwards, so $m^{-1}(10)$ asks which input produced the output $10$ . Search the outputs, not the inputs.

$m(16) = 10 \;\Rightarrow\; m^{-1}(10) = 16, \qquad m(8) = 5 \;\Rightarrow\; m^{-1}(5) = 8.$

The two given points hold both answers, so no formula is needed anywhere.

Part B

The graph of $m^{-1}$ is the graph of $m$ reflected across $y = x$ , and that reflection trades the two axes, so a horizontal spread becomes a vertical one.

$\text{domain of } m^{-1} = \text{range of } m = 0 \le x \le 10, \qquad \text{range of } m^{-1} = \text{domain of } m = 0 \le y \le 16.$

Each point of $m$ gives a point of $m^{-1}$ with its coordinates exchanged, so $(8, 5)$ gives $(5, 8)$ and $(16, 10)$ gives $(10, 16)$ .

Part C

Reflecting across $y = x$ exchanges a point's coordinates, and the origin's two coordinates are already equal, so the reflection leaves it exactly where it is. A point that is its own image belongs to both graphs at once.

$(0, 0) \longrightarrow (0, 0).$

For the second question, compare the two lines. The line $m$ climbs $5$ miles for every $8$ kilometres, a slope of $\tfrac{5}{8}$ , and its mirror image climbs $8$ for every $5$ , a slope of $\tfrac{8}{5}$ . Two straight lines with different slopes meet in exactly one point, and the origin is already that point, so there is no other. This argument is about these two lines; a shared point off the diagonal is not impossible in general, and a decreasing function can manage it.

In one line

$m^{-1}(10) = 16$ kilometres and $m^{-1}(5) = 8$ kilometres. The inverse has domain $0 \le x \le 10$ and range $0 \le y \le 16$ , and its graph passes through $(5, 8)$ and $(10, 16)$ . The origin is shared because a point on the line $y = x$ is its own reflection, and nothing else is shared because these two graphs are straight lines of different slopes, which can meet only once.

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 4 points

Searches the OUTPUTS for the given value and reports the input that produced it, rather than evaluating the conversion forwards. . Worth 2 points.

Attaches the right unit to each answer, kilometres rather than miles, and says what the number means for a traveller. . Worth 2 points.

Part B 4 points

Takes the inverse's domain from the original's range and its range from the original's domain, rather than carrying either interval across unchanged. . Worth 2 points.

Produces points of the inverse by exchanging both coordinates of points of the original, and reports them as pairs. . Worth 2 points.

Part C 4 points

Explains that a point with equal coordinates is unmoved by the reflection, so it belongs to both graphs. . Worth 2 points. needs an explanation, not just an answer

Rules out further shared points with an argument about these two particular graphs, such as their slopes, rather than by a general claim about every function and its inverse. . Worth 2 points.
5. One factor inside, twice over . 14 points. Question 5 of 10.

A function $f$ has domain $-4 \le x \le 10$ and range $1 \le y \le 7$ , and its graph passes through $(6, 7)$ .
1. Part A.
  Give the point on the graph of $y = f(2x)$ that carries the same height as $(6, 7)$ , and state the domain of $y = f(2x)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  State the range of $y = f(2x)$ , and state the domain of $y = f\left(\tfrac{x}{2}\right)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Compare the two rules. Which one squeezes the graph toward the vertical axis and which spreads it away, what does each do to the domain and to the range, and how does the direction each one moves the graph follow from the number written inside?
  
  Carry your own answer forward Compare the two rules using the domains you reported in parts A and B, whatever they turned out to be.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer

Part A

The point is $(3, 7)$ , and the domain is $-2 \le x \le 5$ .

Part B

The range of $y = f(2x)$ is $1 \le y \le 7$ , exactly as before; the domain of $y = f\left(\tfrac{x}{2}\right)$ is $-8 \le x \le 20$ .

Part C

$y = f(2x)$ squeezes the graph toward the vertical axis and halves the domain; $y = f\left(\tfrac{x}{2}\right)$ spreads it away and doubles the domain. Neither touches the range. Each direction looks backwards because the inside has to reach an old input, so a larger factor is satisfied at a smaller $x$ .

Worked solution

Part A

A stored height comes back where the inside expression equals the old input.

$2x = 6 \;\Rightarrow\; x = 3, \qquad (6, 7) \longrightarrow (3, 7).$

The new rule accepts exactly those $x$ for which $2x$ is an input $f$ accepts, so solve at both ends.

$2x = -4 \;\Rightarrow\; x = -2, \qquad 2x = 10 \;\Rightarrow\; x = 5.$

Part B

A number combined with the input never reaches a height, so every output $f$ can produce is still produced and the range is unchanged at $1 \le y \le 7$ .

For the second rule, ask again which $x$ make the inside an input $f$ accepts.

$\frac{x}{2} = -4 \;\Rightarrow\; x = -8, \qquad \frac{x}{2} = 10 \;\Rightarrow\; x = 20.$

This time each endpoint doubles, so the graph needs more horizontal room, not less.

Part C

Both rules leave heights alone, so both leave the range at $1 \le y \le 7$ ; only the domain moves, from $-4 \le x \le 10$ to $-2 \le x \le 5$ in the first case and to $-8 \le x \le 20$ in the second. Half the room means the picture is squeezed toward the vertical axis; twice the room means it is spread away from it.

$bx = a \;\Rightarrow\; x = \frac{a}{b}.$

That one line explains the reversal. The new input is the old one DIVIDED by the factor, so a factor larger than $1$ pulls every point closer to the vertical axis and a factor smaller than $1$ pushes it further out. The point at input $0$ is the exception that proves it: dividing $0$ by anything leaves $0$ , so the crossing of the vertical axis never moves.

In one line

$(6, 7)$ moves to $(3, 7)$ under $y = f(2x)$ , whose domain is $-2 \le x \le 5$ and whose range is still $1 \le y \le 7$ ; the domain of $y = f\left(\tfrac{x}{2}\right)$ is $-8 \le x \le 20$ . The factor $2$ compresses the graph toward the vertical axis and the factor $\tfrac{1}{2}$ stretches it away, because each input is divided by the factor rather than multiplied by 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 5 points

Finds the new input by setting the inside expression equal to the old input, rather than multiplying the old input by the factor. . Worth 2 points.

Applies the same reasoning to both ends of the domain and reports an interval, not a single number. . Worth 2 points.

Leaves the height untouched and reports the image as an ordered pair. . Worth 1 point.

Part B 4 points

Reaches a verdict on the range and grounds it in which of the two coordinates a factor inside the function can reach. . Worth 2 points.

Solves the boundary conditions for the second rule and reports the resulting interval. . Worth 2 points.

Part C 5 points

Says which rule compresses and which stretches, matching each to what it did to the domain. . Worth 2 points.

States that neither rule changes the range, and explains the reversal by way of the input being divided by the factor. . Worth 3 points. needs an explanation, not just an answer
6. Two rules built from the same three pieces . 13 points. Question 6 of 10.

The point $(4, 6)$ lies on the graph of $y = f(x)$ .
1. Part A.
  Find the point that must lie on the graph of $y = 3f(x - 1) + 2$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Find the point that the same original point produces on the graph of $y = 3\bigl(f(x - 1) + 2\bigr)$ , and state how far apart the two heights are.
  
  Carry your own answer forward Compare with the height you reported in part A, whatever it was, and give the gap between the two.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  A description of $y = 3f(x - 1) + 2$ reads: 'raise every height by $2$ , then triple the result.' Decide whether that description matches the rule. If it does not, name the one thing it gets wrong and write a description that does match.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
The answer

Part A

$(5, 20)$ .

Part B

$(5, 24)$ , a height $4$ above the one in part A.

Part C

It runs the two output steps in the wrong order: the rule triples a height first and adds $2$ to what comes out, while the description adds first. Correctly: slide the graph right $1$ , triple every height, then raise the result by $2$ .

Worked solution

Part A

The inside expression decides the input: it reaches the old input $4$ when $x = 5$ . Everything outside acts on the height, tripling it and then adding $2$ .

$x - 1 = 4 \;\Rightarrow\; x = 5, \qquad 3 \times 6 + 2 = 20.$

Part B

The inside expression is untouched, so the input is the same as before. The parentheses now demand the addition first and the tripling second.

$3(6 + 2) = 24, \qquad 24 - 20 = 4.$

The two rules use the same three pieces and the same starting point, yet they land $4$ apart, because the $2$ has been tripled along with the height.

Part C

The sideways part of the description is not the problem; the order of the two things done to the height is. In $y = 3f(x - 1) + 2$ the factor is applied to $f(x - 1)$ and the $2$ is added to the number that comes out of that multiplication.

$(p, q) \longrightarrow (p + 1,\; 3q + 2), \qquad \text{not} \qquad (p + 1,\; 3(q + 2)).$

The quoted description is a description of the other rule, the one with the parentheses, which is why the two heights in parts A and B disagree. A correct reading of the written rule is: slide the graph right $1$ , then stretch every height to three times its size, then raise the whole thing by $2$ .

In one line

$(4, 6)$ goes to $(5, 20)$ on $y = 3f(x - 1) + 2$ and to $(5, 24)$ on $y = 3\bigl(f(x - 1) + 2\bigr)$ , a gap of $4$ . The quoted description adds before it multiplies; the written rule sends $(p, q)$ to $(p + 1,\ 3q + 2)$ , so it slides right $1$ , triples every height, and only then raises the result by $2$ .

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

Finds the new input from the inside expression alone, leaving the factor and the added number out of it. . Worth 2 points.

Scales the height first and adds afterwards, rather than the other way round. . Worth 2 points.

Reports the image as an ordered pair. . Worth 1 point.

Part B 4 points

Adds the $2$ to the height before multiplying, following the parentheses written in this second rule. . Worth 2 points.

Compares the two heights and states the gap, keeping the shared input separate from the heights being compared. . Worth 2 points.

Part C 4 points

Reaches a verdict on the description and names the single fault precisely, rather than blaming the sideways slide or the factor itself. . Worth 2 points. needs an explanation, not just an answer

Gives a corrected description that scales before adding and still accounts for the change made inside the function. . Worth 2 points.
7. Everything one drawn curve is willing to say . 13 points. Question 7 of 10.

The curve below is the whole graph of a function $g$ . It is drawn only between its two endpoints, and both endpoints are filled. No formula for $g$ is given, and none is needed.

The whole graph of $g$ , with seven of its points marked.

Text description of this figure

A grid running from negative 3 to 5 across and from negative 5 to 4 upward carries one smooth curve, drawn only between two filled endpoints, the left one at negative 2 comma negative 5 and the right one at 4 comma negative 5. Dots also mark the points negative 1 comma 0, 0 comma 3, 1 comma 4, 2 comma 3 and 3 comma 0. The curve climbs from the left endpoint to the dot at 1 comma 4 and comes back down to the right endpoint.
1. Part A.
  State the domain and the range of $g$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Give $g(0)$ and every solution of $g(x) = 0$ , and say which of your answers is the crossing of the vertical axis and which are crossings of the horizontal axis.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  State the stretch of inputs on which $g$ is increasing and the stretch on which it is decreasing, and decide, from the direction the curve travels alone, whether the height at the turning point is the largest output $g$ produces, justifying your answer.
  
  Carry your own answer forward When you name the largest output, use the upper end of the range you reported in part A, whatever you reported.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

Domain $-2 \le x \le 4$ ; range $-5 \le y \le 4$ .

Part B

$g(0) = 3$ , the crossing of the vertical axis at $(0, 3)$ ; $g(x) = 0$ at $x = -1$ and $x = 3$ , the two crossings of the horizontal axis.

Part C

$g$ increases from $x = -2$ to $x = 1$ and decreases from $x = 1$ to $x = 4$ . Every input left of $1$ gives a smaller output, because the curve is still climbing toward the turning point, and every input right of $1$ gives a smaller output, because the curve is falling away from it, so nothing in the domain beats the height there.

Worked solution

Part A

The domain is the horizontal spread of the drawing. The curve occupies every horizontal position between its two endpoints and nothing outside them, and both ends are filled, so both are included.

$\text{domain} \colon \quad -2 \le x \le 4.$

The range is the vertical spread. The lowest the curve reaches is the height $-5$ , shared by the two endpoints, and the highest is the height $4$ at the top of the climb.

$\text{range} \colon \quad -5 \le y \le 4.$

Part B

Finding $g(0)$ starts at the input $0$ and reads the height there, which is the marked dot at $(0, 3)$ . That point is on the vertical axis, so it is the crossing of that axis.

$g(0) = 3.$

Solving $g(x) = 0$ starts at the height $0$ and reads across, collecting every input beneath a meeting point. The curve sits on the horizontal axis at two marked dots.

$g(-1) = 0 \quad \text{and} \quad g(3) = 0 \;\Rightarrow\; x = -1 \text{ or } x = 3.$

The zero output is what makes those two the crossings of the horizontal axis, while the zero input is what makes $(0, 3)$ the crossing of the vertical one.

Part C

Direction is read left to right, in the direction of increasing input. From the left endpoint the curve climbs until it turns, and after the turn it falls to the right endpoint.

$\text{increasing on } -2 \le x \le 1, \qquad \text{decreasing on } 1 \le x \le 4.$

That single description settles which output is largest, with no measuring. Take any input to the left of the turning point: the curve is increasing there, so its height is below the height it will reach at the turn. Take any input to the right: the curve has been decreasing since the turn, so its height is below the height it started that descent from. Since the domain is only those two stretches and the turning point itself, no input produces a greater output than the turning point does.

In one line

$g$ has domain $-2 \le x \le 4$ and range $-5 \le y \le 4$ ; $g(0) = 3$ gives the crossing $(0, 3)$ of the vertical axis, while $g(x) = 0$ at $x = -1$ and $x = 3$ gives the two crossings of the horizontal axis. Reading left to right, $g$ increases up to $x = 1$ and decreases after it, so every other input is either still climbing toward the turning point or already falling away from it, which is why the height $4$ there is the largest output.

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 4 points

Reads the domain along the horizontal axis, from one endpoint's input to the other's. . Worth 2 points.

Reads the range along the vertical axis, from the lowest height reached to the highest, rather than from the endpoints' inputs. . Worth 2 points.

Part B 4 points

Reads $g(0)$ upward from the input $0$ and reports the height found there. . Worth 2 points.

Collects both inputs at height zero, not one of them, and labels which answer is which kind of crossing. . Worth 2 points.

Part C 5 points

Names both stretches as sets of inputs, read in the direction of increasing input. . Worth 2 points.

Argues that inputs on either side give smaller outputs because of the direction the curve travels on each side, rather than asserting the largest output from the picture alone. . Worth 3 points. needs an explanation, not just an answer
8. Which half survives the mirror . 13 points. Question 8 of 10.

The graph of $q(x) = (x - 3)^2$ is a parabola whose lowest point is $(3, 0)$ .
1. Part A.
  Name two different inputs of $q$ that produce the same output, and say what that fact means for the horizontal line test.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
2. Part B.
  Restrict the domain of $q$ so that the branch that remains is one-to-one, state your restriction, and give the coordinates of two points on the reflection of that branch across the line $y = x$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Show that reflecting the WHOLE parabola across $y = x$ cannot produce the graph of a function: name one vertical line that meets the reflection twice, give the two points where it does, and say what your inputs from part A have to do with it.
  
  Carry your own answer forward Reflect the two points you produced in part A, whatever they were, and build your vertical line from their images.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 5 points
The answer

Part A

$q(1) = 4$ and $q(5) = 4$ , so the horizontal line at the height $4$ meets the graph twice and $q$ fails the horizontal line test.
Part B

Restrict to $x \ge 3$ ; then the reflection passes through $(0, 3)$ and $(4, 5)$ .
- Restrict to $x \le 3$ ; the reflection then passes through $(0, 3)$ and $(4, 1)$ .
Part C

The two points of $q$ at the height $4$ reflect to $(4, 1)$ and $(4, 5)$ , so the vertical line $x = 4$ meets the reflection twice. A curve that some vertical line meets twice is not the graph of a function.
Worked solution

Part A

Squaring destroys the difference between a number and its opposite, so inputs the same distance either side of $3$ share an output.

$q(1) = (1 - 3)^2 = 4, \qquad q(5) = (5 - 3)^2 = 4.$

The two points $(1, 4)$ and $(5, 4)$ sit at the same height, so the horizontal line $y = 4$ crosses the graph twice. The horizontal line test asks whether any horizontal line meets the graph more than once, and this one does, so $q$ is not one-to-one.

Part B

The turning point sits above the input $3$ , and outputs repeat only across it, so either side on its own is one-to-one. Keep the right-hand branch, $x \ge 3$ .

On that branch, $q(3) = 0$ and $q(5) = 4$ , so the branch holds $(3, 0)$ and $(5, 4)$ . Reflecting across $y = x$ exchanges the coordinates of each point.

$(3, 0) \longrightarrow (0, 3), \qquad (5, 4) \longrightarrow (4, 5).$

The left-hand branch $x \le 3$ is an equally good choice; it reflects to a different curve, through $(0, 3)$ and $(4, 1)$ .

Part C

Reflect the two points that shared an output.

$(1, 4) \longrightarrow (4, 1), \qquad (5, 4) \longrightarrow (4, 5).$

Both images have the first coordinate $4$ , so the vertical line $x = 4$ passes through both of them, meeting the reflected curve twice. The vertical line test says a curve meeting some vertical line twice is not the graph of a function, so the reflection of the whole parabola is not one.

That is the horizontal line test from part A seen in the mirror: the reflection turns the horizontal line $y = 4$ , which met $q$ twice, into the vertical line $x = 4$ , which meets the image twice. Restricting to one branch first is what removes the second point.

In one line

$q(1) = q(5) = 4$ , so a horizontal line meets $q$ twice and $q$ is not one-to-one. Restricting to $x \ge 3$ leaves a one-to-one branch whose reflection passes through $(0, 3)$ and $(4, 5)$ . Reflecting the whole parabola instead sends $(1, 4)$ and $(5, 4)$ to $(4, 1)$ and $(4, 5)$ , so the vertical line $x = 4$ meets the image twice and the image is not the graph of a function.

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 4 points

Produces two genuinely different inputs and shows that each gives the same output. . Worth 2 points.

Connects the repeated output to a horizontal line meeting the graph more than once, and states the verdict of the test. . Worth 2 points.

Part B 4 points

Cuts the domain at the INPUT of the turning point, keeping exactly one branch, rather than at its height. . Worth 2 points.

Produces points of the mirror image by exchanging both coordinates of points on the kept branch. . Worth 2 points.

Part C 5 points

Reflects the two points from part A and reads off a single vertical line passing through both images. . Worth 2 points.

States that meeting a vertical line twice disqualifies the reflection as the graph of a function, and ties that back to the repeated output found earlier. . Worth 3 points. needs an explanation, not just an answer
9. One factor outside, one factor inside . 14 points. Question 9 of 10.

The graph of $y = f(x)$ has a single turning point, a maximum at $(6, 2)$ , and it meets the horizontal axis exactly at $(0, 0)$ and $(12, 0)$ .
1. Part A.
  Give the maximum point of $y = 3f(x)$ and the maximum point of $y = f(3x)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Give the points where each of those two graphs meets the horizontal axis.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Compare the two rules using your four answers: for each rule, say which features of the graph came through unmoved and which did not, and say what that shows about where each factor acts.
  
  Carry your own answer forward Compare the points you reported in parts A and B, whatever they turned out to be.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer

Part A

$y = 3f(x)$ has its maximum at $(6, 6)$ ; $y = f(3x)$ has its maximum at $(2, 2)$ .

Part B

$y = 3f(x)$ meets it at $(0, 0)$ and $(12, 0)$ , exactly where $f$ does; $y = f(3x)$ meets it at $(0, 0)$ and $(4, 0)$ .

Part C

The factor outside leaves the crossings alone, since three times a height of $0$ is still $0$ , though it triples the maximum's height. The factor inside leaves every height alone, so the maximum keeps its height and only slides in. A factor outside acts on heights, a factor inside on inputs.

Worked solution

Part A

The factor outside multiplies the height and leaves the input alone.

$\bigl(6,\; 3 \times 2\bigr) = (6, 6).$

The factor inside leaves the height alone and decides which input reports it: the inside expression has to equal $6$ .

$3x = 6 \;\Rightarrow\; x = 2, \qquad (6, 2) \longrightarrow (2, 2).$

Part B

Multiplying a height of $0$ by $3$ returns $0$ , so both crossings survive untouched.

$\bigl(0,\; 3 \times 0\bigr) = (0, 0), \qquad \bigl(12,\; 3 \times 0\bigr) = (12, 0).$

For the inside factor, solve for the inputs that reach the old crossings.

$3x = 0 \;\Rightarrow\; x = 0, \qquad 3x = 12 \;\Rightarrow\; x = 4.$

Part C

Line the four answers up. The rule $y = 3f(x)$ kept both crossings, $(0, 0)$ and $(12, 0)$ , and changed the maximum's height from $2$ to $6$ . The rule $y = f(3x)$ kept the height $2$ and moved the maximum's input from $6$ to $2$ , and moved the crossing at $12$ in to $4$ while leaving the one at $0$ alone.

$(x_0, y_0) \longrightarrow (x_0,\; 3y_0) \qquad \text{against} \qquad (x_0, y_0) \longrightarrow \left(\frac{x_0}{3},\; y_0\right).$

Each mapping changes one coordinate and leaves the other alone, which is the whole explanation. A factor outside multiplies heights, so a height of $0$ is fixed and every crossing of the horizontal axis survives. A factor inside divides inputs, so an input of $0$ is fixed and heights survive, which is why this maximum keeps its height and only the crossing away from the vertical axis moves.

In one line

$y = 3f(x)$ has its maximum at $(6, 6)$ and still meets the horizontal axis at $(0, 0)$ and $(12, 0)$ ; $y = f(3x)$ has its maximum at $(2, 2)$ and meets the horizontal axis at $(0, 0)$ and $(4, 0)$ . The outside factor multiplies heights, so it pins every crossing of the horizontal axis, and the inside factor divides inputs, so it pins every height.

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

Applies the outside factor to the height and leaves that point's input where it was. . Worth 2 points.

Applies the inside factor by solving for the input that makes the inside expression equal the old one, and leaves the height where it was. . Worth 2 points.

Reports both maxima as ordered pairs. . Worth 1 point.

Part B 4 points

Works out what the outside factor does to a height of zero, and uses it to place the crossings. . Worth 2 points.

Divides each crossing's input by the inside factor, rather than multiplying by it. . Worth 2 points.

Part C 5 points

Matches each rule to the feature it leaves fixed, arguing from the computed points rather than from a remembered rule. . Worth 2 points.

States that a factor outside acts on heights while a factor inside acts on inputs, and explains why the fixed feature follows from that. . Worth 3 points. needs an explanation, not just an answer
10. Three numbers read back out of a picture . 14 points. Question 10 of 10.

The graph of $y = f(x)$ has its only turning point, a minimum, at $(2, -1)$ , and it passes through $(0, 3)$ and $(4, 3)$ . A second graph is built from $f$ by a rule of the form $y = a\,f(x - h) + k$ . On that second graph the minimum sits at $(5, 7)$ , and the point that came from $(0, 3)$ sits at $(3, 15)$ .
1. Part A.
  Find $h$ , and then find $a$ and $k$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Write the rule of the second graph in terms of $f$ , and give the point that $(6, 5)$ of $f$ becomes under it.
  
  Carry your own answer forward Use the three numbers you found in part A, whatever they were.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Test both of the mappings $(p, q) \to (p + h,\; aq + k)$ and $(p, q) \to \bigl(p + h,\; a(q + k)\bigr)$ on the turning point of $f$ . Use the two results to say which mapping this second graph performs, and explain why the other one is not simply a different way of writing the same thing.
  
  Carry your own answer forward Use your own $a$ , $h$ and $k$ from part A, applied to the turning point given in the stem.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$h = 3$ , $a = 2$ , and $k = 9$ .

Part B

$y = 2f(x - 3) + 9$ , and $(6, 5)$ becomes $(9, 19)$ .

Part C

The turning point $(2, -1)$ goes to $(5, 7)$ under the first mapping and to $(5, 16)$ under the second, and only the first matches the minimum the picture shows. The second is a different rule, not a rewriting: it multiplies the vertical slide by the factor as well, adding $ak$ instead of $k$ .

Worked solution

Part A

Only $h$ can move an input, so read it off the turning point's first coordinate.

$2 + h = 5 \;\Rightarrow\; h = 3.$

The heights give two equations, because each height is scaled by $a$ and then raised by $k$ . The minimum's height $-1$ became $7$ , and the height $3$ became $15$ .

$-a + k = 7, \qquad 3a + k = 15.$

Subtracting the first equation from the second removes $k$ and leaves $4a = 8$ , so $a = 2$ , and then $k = 7 + a = 9$ .

Part B

Put the three numbers into the given shape, the slide inside and the factor and the raise outside.

$y = 2f(x - 3) + 9.$

Then apply the mapping to the new point: the input picks up the slide, and the height is scaled and then raised.

$(6, 5) \longrightarrow \bigl(6 + 3,\; 2 \times 5 + 9\bigr) = (9, 19).$

Part C

Run both on the same point, with the numbers from part A.

$\bigl(2 + 3,\; 2(-1) + 9\bigr) = (5, 7), \qquad \bigl(2 + 3,\; 2(-1 + 9)\bigr) = (5, 16).$

The picture puts the minimum at $(5, 7)$ , so the first mapping is the one this graph performs: the height is scaled first and raised afterwards.

The second is not the same rule written differently. Expanding it gives $aq + ak$ , which raises every height by $ak$ rather than by $k$ , so the two agree only when $ak = k$ , that is, when the factor is $1$ or the slide is $0$ . Here $ak = 18$ against $k = 9$ , and the $9$ unit gap between the two answers is exactly that difference.

In one line

The rule is $y = 2f(x - 3) + 9$ , so $h = 3$ , $a = 2$ and $k = 9$ , and the point $(6, 5)$ of $f$ becomes $(9, 19)$ . Testing the turning point shows why the order matters: $(2, -1)$ goes to $(5, 7)$ when the height is scaled before the slide is added, but to $(5, 16)$ when the slide is added first, and only $(5, 7)$ is the minimum the picture shows.

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

Gets the sideways slide from the change in an input alone, without involving the two other numbers. . Worth 2 points.

Turns the two changes in height into two equations in the scale factor and the vertical slide, with the scaling applied before the slide. . Worth 2 points.

Solves the pair of equations and reports all three numbers, saying which is which. . Worth 1 point.

Part B 4 points

Writes the rule with the sideways slide inside the function and both the factor and the vertical slide outside it. . Worth 2 points.

Applies the mapping to the new point, scaling the height before adding the vertical slide. . Worth 2 points.

Part C 5 points

Tests both mappings on the same point with its own three numbers and reports the two different heights. . Worth 2 points.

Names the mapping that matches the picture and explains that the other multiplies the vertical slide by the factor as well, making it a different rule rather than a rewriting. . Worth 3 points. needs an explanation, not just an answer

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. One short list, several different questions 11 points

Part A 3

Reads the output paired with the input $2$ , rather than an input paired with the output $2$ . 2
Reports the crossing of the vertical axis as an ordered pair built from the input $0$ , not as a bare number. 1

Part B 4

Collects BOTH inputs whose output is $4$ , rather than stopping at the first one found. 2
Identifies the crossing of the horizontal axis as the point whose OUTPUT is zero, and reports it as an ordered pair. 2

Part C 4

Names which of the two tasks is the guaranteed single-answer one and which is the possibly-several one. 2
Ties the difference to the definition of a function, which constrains the input-to-output direction only. 2

2. One point, two slides 12 points

Part A 5

Changes the height only, for the number that sits outside the function. 2
Finds the new input by asking which value of $x$ makes the inside expression equal the old input. 2
Reports both images as ordered pairs, each with exactly one coordinate changed. 1

Part B 3

Writes the sideways slide inside the function with the sign the direction forces, rather than the sign the description names. 2
Puts the up-and-down slide outside the function, where it acts on the height. 1

Part C 4

Identifies which of the two rules moved the point sideways, from where its number sits. 1
Argues from the equation that makes the inside expression equal the old input, and reads the direction off the size of the new input. 3

3. Three points under two factors 11 points

Part A 3

Multiplies every height by the factor and leaves every input unchanged. 2
Reports all three images as ordered pairs, the point whose height is zero included. 1

Part B 3

Negates every height, sending a point above the horizontal axis to the matching place below it and the reverse. 2
Keeps every input unchanged and reports each image as an ordered pair. 1

Part C 5

Argues from the fact that any factor times zero is zero that points at height zero are fixed, so the crossings of the horizontal axis do not move. 3
States which of the two rules preserves the shape and which does not, deciding it by the size of the factor rather than by its sign. 2

4. A conversion read the other way 12 points

Part A 4

Searches the OUTPUTS for the given value and reports the input that produced it, rather than evaluating the conversion forwards. 2
Attaches the right unit to each answer, kilometres rather than miles, and says what the number means for a traveller. 2

Part B 4

Takes the inverse's domain from the original's range and its range from the original's domain, rather than carrying either interval across unchanged. 2
Produces points of the inverse by exchanging both coordinates of points of the original, and reports them as pairs. 2

Part C 4

Explains that a point with equal coordinates is unmoved by the reflection, so it belongs to both graphs. 2
Rules out further shared points with an argument about these two particular graphs, such as their slopes, rather than by a general claim about every function and its inverse. 2

5. One factor inside, twice over 14 points

Part A 5

Finds the new input by setting the inside expression equal to the old input, rather than multiplying the old input by the factor. 2
Applies the same reasoning to both ends of the domain and reports an interval, not a single number. 2
Leaves the height untouched and reports the image as an ordered pair. 1

Part B 4

Reaches a verdict on the range and grounds it in which of the two coordinates a factor inside the function can reach. 2
Solves the boundary conditions for the second rule and reports the resulting interval. 2

Part C 5

Says which rule compresses and which stretches, matching each to what it did to the domain. 2
States that neither rule changes the range, and explains the reversal by way of the input being divided by the factor. 3

6. Two rules built from the same three pieces 13 points

Part A 5

Finds the new input from the inside expression alone, leaving the factor and the added number out of it. 2
Scales the height first and adds afterwards, rather than the other way round. 2
Reports the image as an ordered pair. 1

Part B 4

Adds the $2$ to the height before multiplying, following the parentheses written in this second rule. 2
Compares the two heights and states the gap, keeping the shared input separate from the heights being compared. 2

Part C 4

Reaches a verdict on the description and names the single fault precisely, rather than blaming the sideways slide or the factor itself. 2
Gives a corrected description that scales before adding and still accounts for the change made inside the function. 2

7. Everything one drawn curve is willing to say 13 points

Part A 4

Reads the domain along the horizontal axis, from one endpoint's input to the other's. 2
Reads the range along the vertical axis, from the lowest height reached to the highest, rather than from the endpoints' inputs. 2

Part B 4

Reads $g(0)$ upward from the input $0$ and reports the height found there. 2
Collects both inputs at height zero, not one of them, and labels which answer is which kind of crossing. 2

Part C 5

Names both stretches as sets of inputs, read in the direction of increasing input. 2
Argues that inputs on either side give smaller outputs because of the direction the curve travels on each side, rather than asserting the largest output from the picture alone. 3

8. Which half survives the mirror 13 points

Part A 4

Produces two genuinely different inputs and shows that each gives the same output. 2
Connects the repeated output to a horizontal line meeting the graph more than once, and states the verdict of the test. 2

Part B 4

Cuts the domain at the INPUT of the turning point, keeping exactly one branch, rather than at its height. 2
Produces points of the mirror image by exchanging both coordinates of points on the kept branch. 2

Part C 5

Reflects the two points from part A and reads off a single vertical line passing through both images. 2
States that meeting a vertical line twice disqualifies the reflection as the graph of a function, and ties that back to the repeated output found earlier. 3

9. One factor outside, one factor inside 14 points

Part A 5

Applies the outside factor to the height and leaves that point's input where it was. 2
Applies the inside factor by solving for the input that makes the inside expression equal the old one, and leaves the height where it was. 2
Reports both maxima as ordered pairs. 1

Part B 4

Works out what the outside factor does to a height of zero, and uses it to place the crossings. 2
Divides each crossing's input by the inside factor, rather than multiplying by it. 2

Part C 5

Matches each rule to the feature it leaves fixed, arguing from the computed points rather than from a remembered rule. 2
States that a factor outside acts on heights while a factor inside acts on inputs, and explains why the fixed feature follows from that. 3

10. Three numbers read back out of a picture 14 points

Part A 5

Gets the sideways slide from the change in an input alone, without involving the two other numbers. 2
Turns the two changes in height into two equations in the scale factor and the vertical slide, with the scaling applied before the slide. 2
Solves the pair of equations and reports all three numbers, saying which is which. 1

Part B 4

Writes the rule with the sideways slide inside the function and both the factor and the vertical slide outside it. 2
Applies the mapping to the new point, scaling the height before adding the vertical slide. 2

Part C 5

Tests both mappings on the same point with its own three numbers and reports the two different heights. 2
Names the mapping that matches the picture and explains that the other multiplies the vertical slide by the factor as well, making it a different rule rather than a rewriting. 3

Multiple choice

Free response