Quarry School

Follow a composition across two graphs

Explain it like I am five

A graph draws a function's input-output pairs as points. Think of it as a map: the horizontal address is the input and the vertical address is the output. For one function, start at the input on the x-axis, find the graph directly above or below it, and read the height on the y-axis. A composition uses two maps. The height you read on the first map becomes a horizontal address on the second map. You do not carry the first point across unchanged. You carry its output number, then use that number as the next input.

x = 2 on gread g height4use x = 4 on fheight 16firstsecond
A vertical output becomes the next horizontal input.
Reminder
  • Ordered pair. (2, 4) means input 2 and output 4; the first coordinate is horizontal.
  • Axes. The x-axis runs horizontally; the y-axis runs vertically.
  • Negative squares. (−1)2 = (−1) × (−1) = 1.
  • Composition. f(g(2)) uses g's output as f's input.
Why it works. An ordered pair (a, b) on a function graph means that input a produces output b. The first graph therefore turns the starting x-coordinate into an intermediate y-coordinate. The definition of composition makes this intermediate output the outer input, so it must be located along the second graph's horizontal axis. Reading the height there finishes the second function evaluation. The graphs can have different scales, so read their numbered axes instead of comparing the physical heights on the page.
RuleFor f(g(a)): read the y-value of g at x = a; use that number as the x-value on f; read f's y-value.
A point (a, b) on g means g(a) = b.
The same idea, five ways
Say it

f of g of two

Write it

Read g’s height at horizontal input 2, then use that number as f’s horizontal input.

In math
  • g(2) = 4
  • f(4) = 16
  • f(g(2)) = 16
  • On g: (2, 4); on f: (4, 16)
Like

Use the destination on one map as the starting address on a second map.

See it
x = 2 on gread g height4use x = 4 on fheight 16firstsecond
A vertical output becomes the next horizontal input.
The same idea, other ways
As two address maps

The first map sends horizontal address 2 to vertical address 4. The second map starts at horizontal address 4 and sends it to vertical address 16.

2g graph4f graph16firstsecond
The maps connect through the number 4.
As two ordered pairs

The pair (2, 4) says g(2) = 4. The pair (4, 16) says f(4) = 16. The output coordinate of the first pair matches the input coordinate of the second.

On g: (2, 4)
On f: (4, 16)
Match the middle 4
Matching coordinates connects the two graph readings.
.1Read the inner graph

The inner graph handles the original input. An output can be below zero, and it still becomes a valid next input if the outer function accepts it. In this part, g(x) = x + 2 and f(x) = x2 are the two rules you will connect.

  • Read input horizontally and output vertically.
  • A point below the x-axis gives a negative output.
−4−22−224g(−3) = −1
Input −3 gives height −1 on the line.
Worked exampleA negative intermediate output

Using g(x) = x + 2, find the first graph reading for f(g(−3)). You only need the output of g at input −3. Plan: read the inner curve at the starting horizontal input, then carry its height to the outer x-axis.

−4−22−224(−3, −1)
The first output lies below the horizontal axis.
  1. Find (−3, −1) on g, so g(−3) = −1.The point records input −3 and output −1.
Answer
The next input is −1.
Check −3 + 2 = −1, agreeing with the point's height.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: f(g(−3)) = f(−3) = 9.
This skipped the inner graph, which gives g(−3) = −1.
✓ Instead: Read g at −3 first, then read f at −1 to obtain 1.
Tips and tricks
  • A point below the x-axis has a negative output. Keep its minus sign at the handoff.
.2Move to the outer graph

The outer graph starts from the intermediate number on its own horizontal axis. Its height is the final answer. A negative input is a location to the left of zero, not a negative height you must keep.

  • f(−1) is read at horizontal input −1.
  • For the square graph, input −1 gives output 1.
−22246810f(−1) = 1
A negative horizontal input can produce a positive height.
Worked exampleFinish the negative-input path

The inner graph gave −1, and f(x) = x2. Find the final output in f(g(−3)). You now read f at input −1. Plan: the first reading is already −1. Put −1 on the outer x-axis and read the curve’s height there.

−22246810(−1, 1)
Read the new output above horizontal input −1.
  1. Locate −1 on f's x-axis and read the point (−1, 1).The intermediate value is now the outer input.
  2. f(−1) = 1.The point's vertical coordinate is the final output.
Answer
f(g(−3)) = 1.
Check Using formulas, (−3 + 2)2 = (−1)2 = 1.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The in-between −1 must stay a height of −1 on f.
The in-between output becomes the next horizontal input. It does not force the outer output to match.
✓ Instead: Use x = −1 on f and read its height 1.
Tips and tricks
  • Move the first output to the second x-axis before reading the next height.
.3Read the unlabeled curve from the grid

An exam graph may give a grid with no printed answer beside its points. Count along the numbered x-axis to your input. Travel straight up or down to the curve. Then read across to the y-axis. A height of 0 means the curve is on the x-axis. It is an answer, rather than a failed lookup.

  • A point’s horizontal coordinate is its input and its vertical coordinate is its output.
  • Two graphs may use different scales. Read their numbers separately.
246246
Read g from its axes and grid: its line slopes down from height 4.
Worked exampleRead both orders from the grid

Use these graphs of g(x) = 4 − x and f(x) = x2 without labeled answer points. Find f(g(2)) and g(f(2)). You start at 2 in each order. Plan: find the first curve’s height from the grid, then carry that number to the other x-axis.

246246
Read g from its axes and grid: its line slopes down from height 4.
−22424681012141618
Read f from its grid: both inputs 2 and −2 have height 4.
2g graph2f graph4firstsecond
The path carries g’s height 2 onto f’s x-axis, then reads final height 4.
  1. At horizontal input 2 on g, the line meets height 2, so g(2) = 2.The grid crossing pairs input 2 with output 2.
  2. At input 2 on f, the curve meets height 4, so f(g(2)) = 4.The first output 2 becomes the second input.
  3. For the reverse order, f(2) = 4 from the curve.The innermost f receives 2 first.
  4. At input 4 on g, the line meets the x-axis at height 0, so g(f(2)) = 0.A point on the x-axis has output 0, which is a valid numerical answer.
Answer
  • f(g(2)) = 4.
  • g(f(2)) = 0.
Check Check with formulas: (4 − 2)2 = 22 = 4, and 4 − 22 = 4 − 4 = 0. The two formula calculations match the readings.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: g(f(2)) has no answer because the last point has height 0.
Zero is a valid output. Undefined means there is no permitted answer, rather than an answer equal to zero.
✓ Instead: f(2) = 4 and g(4) = 0, so g(f(2)) = 0.
Tips and tricks
  • Pencil on the x-axis, vertically to the curve, then horizontally to the y-axis.
Strategy: step by step
  1. 1. Choose the inner graph from the function nearest the starting input.
  2. 2. Locate that input on the inner graph's x-axis and move vertically to its curve.
  3. 3. Read the curve's y-coordinate as the intermediate output.
  4. 4. Locate that number on the outer graph's x-axis and move vertically to its curve.
  5. 5. Read the outer y-coordinate as the final answer. Check each graph's own scale.
Strategy
Read a two-graph composition
1
Can the necessary intersection be read in the drawn portion of the graph?
YesRead the numbered axes and carry that output to the next graph.
NoUse a supplied formula or other data if available; otherwise the picture does not determine that value.
↓
2
Does the question give the starting input?
YesFollow each curve forward to its output.
NoIf the final output is given instead, read backward to every matching input, then verify each forward path.
  1. 1. Rewrite the request with nested parentheses and choose the graph nearest the starting input.
  2. 2. Put your pencil at the starting input on that graph’s x-axis. Move vertically to the curve, then horizontally to read its y-axis value.
  3. 3. Write the in-between answer on a separate line.
  4. 4. Place that number on the outer graph’s x-axis. Move vertically to its curve and horizontally to read its output.
  5. 5. Check each numbered scale. Give exact values at clear grid intersections and describe uncertain readings as estimates.
Worked exampleA line feeds a square graph

The first picture is g(x) = x + 2 and the second is f(x) = x2. Use the labeled points to find f(g(2)). The input is 2 on g; the resulting height selects the input on f. Plan: read the inner curve at the starting horizontal input, then carry its height to the outer x-axis.

242468g(2) = 4
On g, input 2 meets the line at height 4.
2424681012141618f(4) = 16
On f, the new input 4 meets the curve at height 16.
  1. On g, find the labeled point (2, 4), so g(2) = 4.The point's horizontal coordinate is the input and its vertical coordinate is the output.
  2. Move to input 4 on f's horizontal axis.g's output becomes f's input, rather than staying a height.
  3. Read the labeled point (4, 16), so f(4) = 16.The outer graph gives the final output at the intermediate input.
Answer
f(g(2)) = 16.
Check The formulas give g(2) = 2 + 2 = 4, then f(4) = 42 = 16. This separate formula calculation agrees with the graph reading.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: two clear labeled points

The graphs show g(x) = x + 2 and f(x) = x2. Find f(g(0)). The starting input is 0; you want the final height after visiting both graphs. Plan: read the height on g first, then use that number as the horizontal input on f.

−4−224−4−22468(0, 2)
On g(x) = x + 2, read a horizontal input and its matching height.
−4−2242468101214161820(2, 4)
On f(x) = x2, the incoming number is a horizontal input, and its square is the height.
  1. At horizontal input 0, g has point (0, 2), so g(0) = 2.The first coordinate is input, and the second coordinate is output.
  2. Move the output 2 to f’s horizontal axis. The point (2, 4) gives f(2) = 4.Composition makes the first height the second input.
  3. f(g(0)) = 4.The second height is the final output.
Answer
f(g(0)) = 4.
Check The formula route gives (0 + 2)2 = 22 = 4. This agrees with the two graph readings.
Rung 2Rung 2: carry a height to a new horizontal position

Using g(x) = x + 2 and f(x) = x2, find f(g(2)) from the graphs. The input is 2; the first height tells you where to start on the second graph. Plan: read g at 2, carry the height across, then read f at that new input.

−4−224−4−22468(2, 4)
On g(x) = x + 2, read a horizontal input and its matching height.
−4−2242468101214161820(4, 16)
On f(x) = x2, the incoming number is a horizontal input, and its square is the height.
  1. On g, (2, 4) means g(2) = 4.The graph’s height over input 2 is 4.
  2. On f, start at horizontal input 4. Its point (4, 16) gives f(4) = 16.The handoff 4 becomes a horizontal coordinate, not a requested height.
  3. f(g(2)) = 16.The completed two-stage path ends at height 16.
Answer
f(g(2)) = 16.
Check The combined formula gives (2 + 2)2 = 42 = 16. Reading both original graphs at horizontal input 2 would instead give two separate outputs and would miss this path.
Rung 3Rung 3: a negative handoff can work

Using g(x) = x + 2 and f(x) = x2, find f(g(−3)). The input is negative, and the first output is also negative. You want the second graph’s output at that handoff. Plan: read g at −3, then start on f to the left of zero.

−4−224−4−22468(−3, −1)
On g(x) = x + 2, read a horizontal input and its matching height.
−4−2242468101214161820(−1, 1)
On f(x) = x2, the incoming number is a horizontal input, and its square is the height.
  1. On g, the point (−3, −1) gives g(−3) = −1.A point below the horizontal axis has a negative output.
  2. Put −1 on f’s horizontal axis. The point (−1, 1) gives f(−1) = 1.Negative inputs are positions to the left of zero; their square is nonnegative.
  3. f(g(−3)) = 1.The final output is the second graph’s height, not the first output’s sign.
Answer
f(g(−3)) = 1.
Check The original formulas give g(−3) = −3 + 2 = −1 and f(−1) = (−1)(−1) = 1. The combined formula gives (−3 + 2)2 = 1 as well.
Rung 4Rung 4: reverse which graph receives the start

Using g(x) = x + 2 and f(x) = x2, find g(f(2)) and compare it with the earlier f(g(2)). The same input now enters f first. Plan: restart at 2 on f, carry its height to g, and compare the final heights.

−4−224−4−22468(4, 6)
On g(x) = x + 2, read a horizontal input and its matching height.
−4−2242468101214161820(2, 4)
On f(x) = x2, the incoming number is a horizontal input, and its square is the height.
  1. On f, (2, 4) gives f(2) = 4.f is now inside the parentheses and receives the starting input.
  2. On g, horizontal input 4 gives point (4, 6), so g(4) = 6.The first output becomes g’s input.
  3. g(f(2)) = 6, while the earlier f(g(2)) = 16.Changing the first graph changes the route even when the first handoff happens to be 4 in both cases.
Answer
  • g(f(2)) = 6.
  • f(g(2)) = 16.
  • The two orders differ at input 2.
Check The reverse combined formula is x2 + 2, which gives 4 + 2 = 6. The original order gives (x + 2)2, which gives 16 at the same input.
Rung 5Rung 5: solve backward and keep every matching path

The graphs show g(x) = x + 2 and f(x) = x2, both accepting every real input. Solve f(g(x)) = 4. The final height is supplied; you want every starting input that leads to it. Plan: find every input where the outer f graph has height 4, then find the g inputs that produce those numbers.

−4−22−4−2246(−4, −2)(0, 2)
On g(x) = x + 2, read a horizontal input and its matching height.
−22246810(−2, 4)(2, 4)
On f(x) = x2, the incoming number is a horizontal input, and its square is the height.
  1. On f, height 4 meets the curve at (−2, 4) and (2, 4). The outer input can be −2 or 2.Both a negative and a positive number square to 4, so both intersections must be read.
  2. On g, output −2 occurs at point (−4, −2), so one starting input is −4.The first function must produce the chosen outer input −2.
  3. On g, output 2 occurs at point (0, 2), so another starting input is 0.The second matching outer input requires another backward path.
  4. Check forward: −4 → −2 → 4 and 0 → 2 → 4.Every claimed start must finish at the requested height after both graphs.
Answer
x = −4 or x = 0.
Check Algebra gives (x + 2)2 = 4. A number squaring to 4 is −2 or 2, so x + 2 = −2 or x + 2 = 2. Subtracting 2 gives x = −4 or x = 0, agreeing with both graph paths.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: After reading g(2) = 4, find where f has height 4.
That would search for an input giving output 4. Composition instead supplies 4 as an input.
✓ Instead: Find x = 4 on f and read its height 16.
Tips and tricks
  • Say horizontal, vertical, horizontal, vertical as you trace a graphical composition.
  • Use labeled points for exact values. If a graph reading is estimated, label the answer as an estimate.
  • Know cold: across on the x-axis, up or down to the curve, read the height, then carry the height to the next x-axis.
Trap. Using the first graph's output as a height on the second graph. Move it to the second x-axis before finding the outer output.