Graphing Functions: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

A function $f$ satisfies $f(0) = -5$ . What is the $y$ -intercept of its graph?

Answer choices for question 1 A $(0, -5)$ B $(-5, 0)$ C $(0, 5)$ D There is not enough information, since the rule for $f$ is not given.
2

The point $(5, 2)$ lies on the graph of $y = f(x)$ . Which point must lie on the graph of $y = f(x + 4)$ ?

Answer choices for question 2 A $(9, 2)$ B $(1, 2)$ C $(5, 6)$ D $(9, 6)$
3

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The point $(2, -6)$ lies on the graph of a one-to-one function $f$ . Which point must lie on the graph of $f^{-1}$ ?

Answer choices for question 3 A $(2, 6)$ B $(-2, 6)$ C $(-6, 2)$ D $(6, -2)$
4

The point $(-2, 7)$ lies on the graph of $y = f(x)$ . Which point must lie on the graph of $y = 3f(x)$ ?

Answer choices for question 4 A $(-2, 10)$ B $(-6, 7)$ C $(-6, 21)$ D $(-2, 21)$
5

How is the graph of $y = f\left(\tfrac{x}{2}\right)$ related to the graph of $y = f(x)$ ?

Answer choices for question 5 A It is stretched horizontally away from the vertical axis, to twice its width. B It is compressed horizontally toward the vertical axis, to half its width. C It is compressed vertically, to half its height. D It is shifted right by $2$ .
6

The graph of $f$ is shown, with five of its points marked. What are all the solutions of $f(x) = 3$ ?

Answer choices for question 6 A $x = 1$ and $x = 3$ B $x = 3$ C $x = 0$ and $x = 4$ D $x = 0$
7

The graph of $y = f(x)$ meets the horizontal axis only at $(5, 0)$ . Where does the graph of $y = f(x + 2)$ meet the horizontal axis?

Answer choices for question 7 A $(3, 0)$ B $(7, 0)$ C $(5, 2)$ D It no longer meets the horizontal axis.
8

A function $f$ has range $-3 \le y \le 6$ . What is the range of $y = -f(x)$ ?

Answer choices for question 8 A $-3 \le y \le 6$ B $6 \le y \le -3$ C $-6 \le y \le 3$ D $3 \le y \le 6$
9

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A one-to-one function $g$ satisfies $g(2) = 5$ , $g(5) = -3$ , $g(7) = 2$ , and $g(9) = 0$ . What is $g^{-1}(5)$ ?

Answer choices for question 9 A $-3$ B $2$ C $\tfrac{1}{5}$ D $7$
10

The graph of $y = f(x)$ meets the horizontal axis at $(12, 0)$ . Which point is the matching crossing on the graph of $y = f(3x)$ ?

Answer choices for question 10 A $(36, 0)$ B $(12, 0)$ C $(9, 0)$ D $(4, 0)$
11

The point $(4, 5)$ lies on the graph of $y = f(x)$ . Which point must lie on the graph of $y = 2f(x) + 3$ ?

Answer choices for question 11 A $(4, 16)$ B $(4, 10)$ C $(8, 13)$ D $(4, 13)$
12

Suppose $g(x) = f(x - 2) + 1$ and $g(7) = 10$ . What is $f(5)$ ?

Answer choices for question 12 A $9$ B $10$ C $11$ D It cannot be found, because $7$ and $5$ are different inputs.
13

The whole graph of a function $h$ is drawn for $-5 \le x \le 6$ . It falls from $(-5, 2)$ to a lowest point at $(-1, -4)$ , rises to a highest point at $(3, 6)$ , and then falls to $(6, 1)$ . On which stretch is $h$ increasing, and what is the greatest output it produces?

Answer choices for question 13 A Increasing for $x < -1$ and for $x > 3$ ; greatest output $6$ . B Increasing for $-1 < x < 3$ ; greatest output $6$ . C Increasing for $-1 < x < 3$ ; greatest output $3$ . D Increasing for $-4 < y < 6$ ; greatest output $6$ .
14

The graph of $y = f(x)$ crosses the horizontal axis at $(2, 0)$ and the vertical axis at $(0, 7)$ . Which point is guaranteed to lie on the graph of $y = 4f(x)$ ?

Answer choices for question 14 A $(0, 7)$ B $(8, 0)$ C $(2, 4)$ D $(2, 0)$
15

A function $f$ has domain $-6 \le x \le 8$ . What is the domain of $y = f(2x)$ ?

Answer choices for question 15 A $-3 \le x \le 4$ B $-12 \le x \le 16$ C $-6 \le x \le 8$ D $-4 \le x \le 3$
16

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The graph of $q(x) = x^2 - 1$ is reflected across the line $y = x$ . Which restriction on $q$ makes that reflection the graph of a function?

Answer choices for question 16 A $x \ge -1$ B $y \ge -1$ C $x \ge 0$ D No restriction is needed, since $q$ already passes the vertical line test.
17

Which rule reflects the graph of $y = f(x)$ across the horizontal axis, stretches every height to three times its size, and then slides the result down $2$ ?

Answer choices for question 17 A $y = -3f(x) - 2$ B $y = -3f(x - 2)$ C $y = -3\bigl(f(x) - 2\bigr)$ D $y = 3f(-x) - 2$
18

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Which statement holds for every one-to-one function $f$ and its inverse $f^{-1}$ ?

Answer choices for question 18 A The graphs of $f$ and $f^{-1}$ meet only on the line $y = x$ . B Wherever the graph of $f$ crosses the line $y = x$ , the graph of $f^{-1}$ passes through that same point. C The graphs of $f$ and $f^{-1}$ never meet. D The graph of $f^{-1}$ is the graph of $f$ reflected across the horizontal axis.
19

The graph of $y = f(x)$ meets the horizontal axis exactly at $(-2, 0)$ and $(6, 0)$ . Where does the graph of $y = f(-x)$ meet the horizontal axis?

Answer choices for question 19 A $(-2, 0)$ and $(6, 0)$ B $(2, 0)$ and $(-6, 0)$ C $(-2, 0)$ and $(-6, 0)$ D $(0, -2)$ and $(0, 6)$
20

The graph of $y = f(x)$ crosses the vertical axis at $(0, 3)$ and the horizontal axis at $(4, 0)$ . Where does the graph of $y = -2f(x)$ cross the vertical axis?

Answer choices for question 20 A $(0, 6)$ B $(0, 3)$ C $(0, 1)$ D $(0, -6)$

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Where you stand

Read a value, an intercept, and every solution of $f(x) = b$ off a graph, a table, or a list of points, keeping the $y$ -intercept $(0, f(0))$ apart from the $x$ -intercepts, where the output is $0$ . Review: Graphs of Functions
Read a graph's domain as its horizontal spread and its range as its vertical spread, say where it rises and falls reading left to right, and carry a domain or a range through a scaling or a reflection. Review: Graphs of Functions , Shifting Graphs , Stretching and Reflecting Graphs
Translate a graph: $y = f(x) + k$ moves it up by $k$ and $y = f(x - h)$ moves it right by $h$ , sending $(a, b)$ to $(a + h,\, b + k)$ , and decide from where a number sits, inside or outside the function, which of the two directions it moves. Review: Shifting Graphs
Scale a graph vertically: $y = a\,f(x)$ multiplies every height by $a$ , sending $(x_0, y_0)$ to $(x_0,\, a y_0)$ , the case $a = -1$ reflects it across the horizontal axis, and every crossing of the horizontal axis stays where it is. Review: Stretching and Reflecting Graphs
Scale a graph horizontally: $y = f(bx)$ divides every input by $b$ , sending $(x_0, y_0)$ to $\left(\tfrac{x_0}{b},\, y_0\right)$ , so $b > 1$ compresses toward the vertical axis and $0 < b < 1$ stretches away from it, and $y = f(-x)$ reflects it across the vertical axis. Review: Stretching and Reflecting Graphs
Apply a combined rule $y = a\,f(x - h) + k$ , which sends $(p, q)$ to $(p + h,\; aq + k)$ , scaling a height first and sliding it afterwards, and read the three numbers back out of corresponding points on the original and the transformed graph. Review: Shifting Graphs , Stretching and Reflecting Graphs
Use the reflection across $y = x$ : a point $(a, b)$ on $f$ gives $(b, a)$ on $f^{-1}$ , an inverse value is read backwards from an output to the input that produced it, the domain and range trade places, and a point where $f$ meets the diagonal is shared by both graphs. Review: Graphs of Inverse Functions
Decide whether a reflected graph is still a function: the horizontal line test on $f$ is the vertical line test on the mirror image, so a rule that repeats an output must be cut down to a single branch before it is reflected. Review: Graphs of Inverse Functions

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Which intercept is which

A function $w$ is defined at only six inputs, and nowhere else: $w(-6)=8$ , $w(-3)=0$ , $w(0)=-7$ , $w(4)=0$ , $w(7)=0$ , and $w(10)=6$ . A classmate says $(-3,0)$ is the $y$ -intercept, since it contains a $0$ . Say what is wrong with that claim, name the actual $y$ -intercept, and list every $x$ -intercept.
1. Hint 1
  A $y$ -intercept always has input $0$ ; an $x$ -intercept always has output $0$ . Check which of those two conditions the pair $(-3,0)$ actually meets.
2. Hint 2
  Scan all six pairs for the one whose input is $0$ , then separately for the ones whose output is $0$ .
3. Hint 3
  A function can have several $x$ -intercepts, but only one pair among the six can have input $0$ .
Answer

The pair $(-3,0)$ has output $0$ : it is an $x$ -intercept, not the $y$ -intercept. $y$ -intercept: $(0,-7)$ . $x$ -intercepts: $(-3,0)$ , $(4,0)$ , and $(7,0)$ .

Full solution

The classmate's claim confuses the two intercepts.
Containing a $0$ is not the test: an $x$ -intercept needs output $0$ , and a $y$ -intercept needs input $0$ .
The pair $(-3,0)$ has input $-3$ , not $0$ , so it cannot be the $y$ -intercept.

The $y$ -intercept is the listed pair whose input is $0$ .
Checking all six pairs shows $w(0)=-7$ , so the $y$ -intercept is $(0,-7)$ .

An $x$ -intercept is a listed pair whose output is $0$ .
Scanning the six outputs gives $w(-3)=0$ , $w(4)=0$ , and $w(7)=0$ , so the $x$ -intercepts are $(-3,0)$ , $(4,0)$ , and $(7,0)$ ; the remaining two inputs, $-6$ and $10$ , give the nonzero outputs $8$ and $6$ .

Answer

The pair $(-3,0)$ has output $0$ : it is an $x$ -intercept, not the $y$ -intercept. $y$ -intercept: $(0,-7)$ . $x$ -intercepts: $(-3,0)$ , $(4,0)$ , and $(7,0)$ .

Key idea

A $y$ -intercept is picked out by its input being $0$ , not by its output being $0$ , which is instead the mark of an $x$ -intercept.
Problem 1: Mark complete
Problem 2 The open endpoints

The graph shows the entire function $g$ , with hollow endpoints excluded. Give its domain and range, and state where it increases and decreases.

The complete graph of $g$ .

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 4 to 5 and whose vertical y axis is numbered from negative 1 to 5, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. Two straight segments are joined end to end. The first falls steadily from negative 3, 4 down to 1, 0. The second rises from 1, 0 up to 4, 3. The left end at negative 3, 4 and the right end at 4, 3 are hollow dots, and the joining point at 1, 0 is a filled dot. Nothing else is drawn and no point is labeled with its coordinates.
1. Hint 1
  Endpoint inclusion affects the spans that the graph actually reaches.
2. Hint 2
  Read left to right for direction, and check whether a missing endpoint height occurs anywhere else.
Answer

Domain: $-3<x<4$ . Range: $0\le y<4$ . Decreasing for $-3<x<1$ ; increasing for $1<x<4$ .

Full solution

The graph covers all horizontal positions between $-3$ and $4$ , with both endpoint inputs excluded.
Its lowest point $(1,0)$ is included.
Height four occurs only at an excluded endpoint, so the attained heights satisfy
$0\le y<4.$

Moving left to right, the graph falls until input one and rises afterwards.
Hence it decreases on $-3<x<1$ and increases on $1<x<4$ .
The endpoint choices agree with the hollow and filled marks.

Answer

Domain: $-3<x<4$ . Range: $0\le y<4$ . Decreasing for $-3<x<1$ ; increasing for $1<x<4$ .

Key idea

Graph spans must distinguish a missing endpoint from a height attained elsewhere.
Problem 2: Mark complete
Problem 3 Two matching outlines

The two complete graphs have the same shape. Write $g(x)$ in the form $f(x-h)+k$ , and verify your choice using both endpoints.

The complete graphs of $f$ and $g$ .

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 3 to 6 and whose vertical y axis is numbered from negative 3 to 5, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. Two complete graphs are drawn. The solid one, labeled f, rises from a filled endpoint at negative 2, 1 to a filled corner at 0, 4, then falls to a filled endpoint at 3, 2. The dashed one, labeled g, has the same shape: it rises from a filled endpoint at 0, negative 2 to a filled corner at 2, 1, then falls to a filled endpoint at 5, negative 1. No point is labeled with its coordinates and no arrows join the two graphs.
1. Hint 1
  Match a distinctive corner, then check that each endpoint has the same displacement.
2. Hint 2
  The horizontal displacement determines the number subtracted inside the function, while the vertical displacement is added outside.
Answer

$g(x)=f(x-2)-3$ ; endpoint images are $(0,-2)$ and $(5,-1)$ .

Full solution

The corner of $f$ is $(0,4)$ and the corner of $g$ is $(2,1)$ , a move right two and down three.
This gives
$g(x)=f(x-2)-3.$

The left endpoint $(-2,1)$ moves to $(0,-2)$ , and the right endpoint $(3,2)$ moves to $(5,-1)$ .
Both are the displayed endpoints of $g$ , so the same translation carries the whole outline to the new one.

Answer

$g(x)=f(x-2)-3$ ; endpoint images are $(0,-2)$ and $(5,-1)$ .

Key idea

A graph translation is identified by one displacement that must also match every other corresponding point.
Problem 3: Mark complete
Problem 4 A factor from one height

The graph shows all of $f$ . For a constant $a$ , the graph of $y=af(x)$ contains the point $(-3,-1)$ . Find $a$ , and give the point of $y=af(x)$ that comes from the point of $f$ at input $4$ .

The complete graph of $f$ .

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 4 to 5 and whose vertical y axis is numbered from negative 3 to 3, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. Three straight segments are joined end to end and carry filled dots at all four of their meeting points and ends. The first falls from negative 3, 2 to negative 1, 0. The second lies flat along the horizontal axis from negative 1, 0 to 2, 0. The third falls from 2, 0 to 4, negative 2. Nothing else is drawn and no point is labeled with its coordinates.
1. Hint 1
  Corresponding points share an input, and the second graph's height at that input is built from the first graph's height there.
2. Hint 2
  Read $f(-3)$ from the picture and compare it with $-1$ to find the factor, then apply that same factor to $f(4)$ .
Answer

$a=-\frac12$ , or $-0.5$ ; the point at input $4$ becomes $(4,1)$ .

Full solution

The picture gives $f(-3)=2$ , and the point of $y=af(x)$ with input $-3$ has height $a\times2$ .
Matching that height to $-1$ gives
$2a=-1,$
so
$a=-\frac12.$

The picture also gives $f(4)=-2$ , so the corresponding point of $y=af(x)$ has height
$-\frac12\times(-2)=1.$
That point is $(4,1)$ .

Answer

$a=-\frac12$ , or $-0.5$ ; the point at input $4$ becomes $(4,1)$ .

Key idea

A vertical factor multiplies every stored height by the same number, so a single nonzero height fixes the factor and, with it, every other image point.
Problem 4: Mark complete
Problem 5 A peak and an axis crossing

The complete graph of $f$ has one peak, at $(-8,6)$ , and it meets the vertical axis at $(0,-3)$ . Let $g(x)=f\left(\frac{x}{2}\right)$ . Give the coordinates of the peak of $g$ and of the point where $g$ meets the vertical axis.
1. Hint 1
  At input $x$ the new graph shows the height that $f$ stores at whatever the expression inside the function equals.
2. Hint 2
  Set that inside expression equal to $-8$ for the peak, then put $x=0$ and see which input of $f$ is being read.
Answer

Peak $(-16,6)$ ; $g$ meets the vertical axis at $(0,-3)$ .

Full solution

The height stored at input $-8$ reappears where the inside expression equals $-8$ .
Solving
$\frac{x}{2}=-8$
gives $x=-16$ , so the peak of $g$ is $(-16,6)$ .

At input zero the inside expression is $\frac{0}{2}=0$ , so $g(0)=f(0)=-3$ and $g$ meets the vertical axis at $(0,-3)$ , exactly where $f$ does.

Answer

Peak $(-16,6)$ ; $g$ meets the vertical axis at $(0,-3)$ .

Key idea

A horizontal factor divides the inputs, so the crossing of the vertical axis is a point it can never move.
Problem 5: Mark complete
Problem 6 A value read backward

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The graph shows the complete one-to-one function $f$ . Find $f^{-1}(f(0)+f(2))$ .

The complete graph of the one-to-one function $f$ .

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 4 to 6 and whose vertical y axis is numbered from negative 4 to 6, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. There are extra gridlines and tick marks at every half unit as well. Three straight segments are joined end to end, each one rising from left to right, with filled dots at negative 3, negative 2, then 0, 1, then 2, 2, and finally 5, 4. The first segment rises one unit for each unit right, the second rises half a unit for each unit right, and the third rises two units over three. Nothing else is drawn and no point is labeled with its coordinates.
1. Hint 1
  The expression inside the inverse is an output value that must first be determined.
2. Hint 2
  Read the two forward values, add them, then find the original input at that height.
Answer

$\frac72$ .

Full solution

The graph gives $f(0)=1$ and $f(2)=2$ , so the inverse input is $1+2=3$ .
To find $f^{-1}(3)$ , locate the original point at height three.

That height lies halfway between the endpoint heights two and four on the segment from $(2,2)$ to $(5,4)$ .
Its input is halfway between two and five:
$\frac{2+5}{2}=\frac72.$
Reversing this graph record gives the requested value.

Answer

$\frac72$ .

Key idea

A nested expression can require forward graph readings followed by a backward inverse reading.
Problem 6: Mark complete
Problem 7 The altered input rule

The graph shows the entire function $f$ . For $g(x)=f(-2x)$ , give the domain and range, and state where $g$ increases and decreases.

The complete graph of $f$ .

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 3 to 5 and whose vertical y axis is numbered from negative 2 to 4, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. Two straight segments are joined end to end, with filled dots at both ends and at the corner. The first falls steeply from negative 2, 3 to the corner at 0, negative 1, dropping two units for each unit to the right. The second rises gently from that corner to 4, 1, climbing one unit for every two to the right. Nothing else is drawn and no point is labeled with its coordinates.
1. Hint 1
  The inside negative reverses horizontal order and the factor two changes horizontal distances.
2. Hint 2
  Map the original endpoints and corner by solving for the new input, then read the transformed order.
Answer

Domain: $-2\le x\le1$ . Range: $-1\le y\le3$ . Decreasing for $-2<x<0$ ; increasing for $0<x<1$ .

Full solution

An original input $u$ reappears where $-2x=u$ , so its new input is $-u/2$ .
The endpoints and corner $(-2,3)$ , $(0,-1)$ , and $(4,1)$ therefore become $(1,3)$ , $(0,-1)$ , and $(-2,1)$ .

The transformed horizontal span is $-2\le x\le1$ , while all original heights remain, giving range
$-1\le y\le3.$
Reading the new points left to right, the graph falls from $(-2,1)$ to $(0,-1)$ and then rises to $(1,3)$ .
This gives the stated decreasing and increasing intervals.

Answer

Domain: $-2\le x\le1$ . Range: $-1\le y\le3$ . Decreasing for $-2<x<0$ ; increasing for $0<x<1$ .

Key idea

A negative horizontal factor reverses point order while its size rescales the domain and preserves output heights.
Problem 7: Mark complete
Problem 8 Two pairs of points

A transformation $g(x)=af(x-h)+k$ sends the point $(-1,2)$ to $(3,1)$ and sends $(2,-1)$ to $(6,7)$ . Find $a$ , $h$ , and $k$ , and state the image of $(0,4)$ under the same rule.
1. Hint 1
  Corresponding horizontal coordinates give the same horizontal displacement, while corresponding heights give two equations.
2. Hint 2
  Solve for the scale and the vertical displacement first, then apply the whole recovered rule to the extra point.
Answer

$a=-2$ , $h=4$ , $k=5$ ; $(0,4)$ moves to $(4,-3)$ .

Full solution

The first input moves from $-1$ to $3$ , so $h=4$ .
The other input moves from two to six, confirming that displacement.
The output records give $2a+k=1$ and $-a+k=7$ .
Subtracting gives
$3a=-6,$
hence $a=-2$ and $k=5$ .

So $g(x)=-2f(x-4)+5$ , and this rule sends a point $(p,q)$ of $f$ to $(p+4,\,-2q+5)$ .
Applying it to $(0,4)$ gives
$(4,\;-2\times4+5)=(4,-3).$

Answer

$a=-2$ , $h=4$ , $k=5$ ; $(0,4)$ moves to $(4,-3)$ .

Key idea

Two corresponding points fix the scale and both displacements, and the recovered rule then moves every other point.
Problem 8: Mark complete
Problem 9 A changed height reference

The graph shows all of $f$ . Draw $g(x)=-3f(x-2)+6$ on the same grid. Give all $x$ -intercepts of $g$ , and give the images of the two $x$ -intercepts of $f$ .

The complete graph of $f$ .

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 3 to 5 and whose vertical y axis is numbered from negative 1 to 7, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. One graph is drawn and labeled f. It rises from a filled point at negative 2, 0 to a filled corner at 0, 2, then falls back to a filled point at 2, 0, so it makes a peak two units high. The grid above it is empty. No point is labeled with its coordinates and no second graph is drawn.
1. Hint 1
  Separate the horizontal move from the scaling and final change of height.
2. Hint 2
  Track endpoints and the corner; an old height zero and a new height zero need not correspond once a constant is added.
Answer

$g$ joins $(0,6)$ , $(2,0)$ , and $(4,6)$ . Its sole $x$ -intercept is $(2,0)$ . The old intercepts become $(0,6)$ and $(4,6)$ .

Full solution

Every original point $(p,q)$ moves to $(p+2,-3q+6)$ .
The original points $(-2,0)$ , $(0,2)$ , and $(2,0)$ therefore become $(0,6)$ , $(2,0)$ , and $(4,6)$ .
Draw the two segments connecting these points.

The new graph reaches zero only at its corner $(2,0)$ .
The old zero heights are multiplied by $-3$ to give zero, but the final addition raises each to six.
Their images are consequently $(0,6)$ and $(4,6)$ , both above the horizontal axis.

Answer

$g$ joins $(0,6)$ , $(2,0)$ , and $(4,6)$ . Its sole $x$ -intercept is $(2,0)$ . The old intercepts become $(0,6)$ and $(4,6)$ .

Key idea

A nonzero vertical scale preserves zero heights until an added vertical shift changes their reference level.
Problem 9: Mark complete
Problem 10 Two restrictions compared

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The figure shows the complete function $f$ together with the dashed line $y=x$ . Ben keeps the part of the graph with $-2\le x\le2$ , and Cara keeps the part with $-4\le x\le-2$ ; each kept part is then reflected across the dashed line. Exactly one of the two reflections is the graph of a function: say which, and name an input of the other reflection that receives two outputs. Then give the coordinates of the point where $f$ meets the dashed line.

The complete graph of $f$ , with the line $y=x$ dashed.

Text description of this figure

A coordinate grid whose horizontal x axis is numbered from negative 5 to 4 and whose vertical y axis is numbered from negative 5 to 4, with equal unit lengths, tick marks, gridlines and a number at every whole number, and the origin labeled 0. A dashed diagonal line labeled y equals x runs across the grid from corner to corner through the origin. The graph itself is two joined straight segments with filled dots at both ends and at the turning point: it falls one unit for each unit to the right, from negative 4, 3 down to 0, negative 1, then rises one unit for each unit to the right, from 0, negative 1 up to 2, 1. No point is labeled with its coordinates and no reflected graph is drawn.
1. Hint 1
  A reflection across $y=x$ is the graph of a function exactly when the kept part reaches each height at most once.
2. Hint 2
  Compare the heights at the ends of each kept part with the heights between them. For the last part, use the falling piece's rule and ask where its height equals its input.
Answer

Cara's reflection is a function; Ben's is not: any input with $-1<x\le1$ works, for example input $1$ , giving outputs $-2$ and $2$ . $f$ meets the dashed line at $\left(-\frac12,-\frac12\right)$ .

Full solution

Cara's part falls steadily from $(-4,3)$ to $(-2,1)$ , so each height from one to three occurs once.
Exchanging coordinates turns those ends into $(3,-4)$ and $(1,-2)$ , and the reflected segment gives one output at each of its inputs, so it is the graph of a function.

Ben's part turns at $(0,-1)$ : it falls from $(-2,1)$ to that corner and rises again to $(2,1)$ , so the height one occurs at $x=-2$ and at $x=2$ .
Exchanging coordinates sends those points to $(1,-2)$ and $(1,2)$ , so input $1$ of Ben's reflection receives two outputs and the picture is not a function.
The same doubling happens at every height strictly between $-1$ and $1$ , and at height $1$ itself, since one occurrence sits on each side of the corner; only the corner's own height, $-1$ , is reached once.

The falling piece drops one unit for each unit to the right from $(0,-1)$ , so its heights obey $y=-x-1$ .
Setting the height equal to the input gives
$-x-1=x,$
hence $x=-\frac12$ and the meeting point is $\left(-\frac12,-\frac12\right)$ .
Exchanging its coordinates returns the same point, so wherever a kept part contains it, it stays put under its own reflection (here, Ben's).

The two verdicts differ because Cara kept a piece on one side of the turning point while Ben kept the turn itself, which is what makes a height repeat.

Answer

Cara's reflection is a function; Ben's is not: any input with $-1<x\le1$ works, for example input $1$ , giving outputs $-2$ and $2$ . $f$ meets the dashed line at $\left(-\frac12,-\frac12\right)$ .

Key idea

A kept part reflects to a function exactly when no height repeats, and a point on the line $y=x$ is its own reflection.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 13: Polynomials, Exponentials, and Logarithms →

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