Quarry School

Horizontal shifts: why the inside acts in the opposite direction

Explain it like I am five

Imagine a bus timetable. A stop that used to happen at eight now happens at six because the driver starts the whole route two hours earlier. At six, the new schedule must look up what the old schedule said for eight. That is why adding two inside a function moves the event earlier, toward the left. A Horizontal shift moves the graph sideways. An Inside change changes the input before the old function uses it. To locate the new dot, you ask which new input will feed the machine its old input. You solve that question backward. The sign in the formula describes what the machine does, while the graph shows where you must stand to undo it.

−2−1123456−112345678(3, 0)old vertexnew vertexsame old height 1
Subtracting three inside means every old height is reached at an input three greater.
Reminder
  • Solving a linear equation. Do the same undoing operation on both sides. x − 3 = 4 becomes x = 7 after adding 3; check 7 − 3 = 4.
  • Subtracting a negative. Adding 3 can be written as subtracting −3: x + 3 = x − (−3). In x − h, that means h = −3.
  • Known function values. If f(6) = 7, use it only when the inside expression equals 6. Thus f(9 − 3) = 7, while f(6 − 3) asks for an unspecified f(3).
input xoutput f(x)214367811
Read each column as one x input paired with its f(x) output.
input xoutput g(x)5173971111
Read each column as one x input paired with its g(x) output.
Why it works. Let f(u) = v. For g(x) = f(x − h), we want the input inside f to be u again so the old output v returns. Solve x − h = u by adding h to both sides: x = u + h. That is the new horizontal coordinate. Subtracting h inside therefore moves the dot right by h. For f(x + h), solve x + h = u to get x = u − h, a move left. The opposite direction comes from undoing the input change, not from an arbitrary memory rule.
Ruleg(x) = f(x − h) sends (u, v) to (u + h, v). Positive h moves right; negative h moves left. Solve new input − h = old input whenever the sign feels uncertain.
The same idea, five ways
Say it

Say: the new function at x uses the old function at x minus h. To keep the old height, move its input right h.

Write it

A Horizontal shift changes the input address at which each old height appears.

In math
  • g(x) = f(x − h)
  • x − h = u, so x = u + h.
  • Old point (u, v); new point (u + h, v).
  • For (x − 3)2: vertex (3, 0). For (x + 3)2: vertex (−3, 0).
  • Graph words: slide right when h > 0, left when h < 0, and keep all old heights.
Like

On a schedule that starts two hours earlier, new time 6 looks up old time 8. You add 2 before consulting the old timetable.

See it
−22462468(3, 0)old vertexnew vertexsame old height 1
The same old heights appear three units farther right because the new input must undo the inside subtraction.
The same idea, other ways
As a picture

The bottom of x2 occurs when its input is zero. The bottom of (x − 3)2 occurs when x − 3 is zero, which happens at x = 3. The whole curve follows that bottom to the right.

−2246246810(3, 0)
The new curve reaches its old lowest height three units later.
As a machine

A subtract-three worker stands before the old function. To feed the old function 2, give the worker 5. The worker subtracts 3 and hands the machine 2. The old point's address therefore changes from 2 to 5.

5subtract 32ff(2)firstsecond
A bigger starting input compensates for the subtraction before f.
As a schedule

Starting a vent schedule two hours earlier gives g(t) = f(t + 2). At new time 6, this looks up old time 8. Adding inside produces an earlier event because you need two fewer hours on the new clock.

As a schedule
g(x) = f(x − h)
Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.
As an equation

The dot with old input u must satisfy x − h = u. Adding h gives x = u + h. This works for positive and negative u and h, so it explains every dot, not only the lowest point.

As an equation
g(x) = f(x − h)
Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.
A memory device you can rebuild

Inside lies, outside tells the truth is a reminder about direction. The inside really tells the truth about the machine's input operation. You undo that operation to get the graph's position. Write the old-input equation whenever memory fails.

A memory device you can rebuild
g(x) = f(x − h)
Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.
Old input uOld output f(u)New input u + 3g(u + 3) for g(x) = f(x − 3)
2151
4373
6797
8111111
.1Shift right

An inside subtraction delays every old event. You must move farther right to give the old function its original input.

  • Formula: g(x) = f(x − h), h > 0.
  • Point: (u, v) becomes (u + h, v).
  • Range: unchanged, because all the same old heights still appear.
−2246246(3, 0)
The corner appears at x = 3 because 3 − 3 feeds zero to absolute value.
Worked exampleFind the new corner

Where is the corner of |x − 3|?
This asks you to find the new address or height of the given graph information.

−2246−4−22468(3, 0)
Zero inside puts the new corner at (3, 0).
  1. Solve x − 3 = 0, giving x = 3.The old corner of |x| happens at input zero.
  2. The output is |0| = 0.Nothing outside changes the height.
Answer
Corner: (3, 0).
Check |3 − 3| = 0, while |0 − 3| = 3, so zero is now reached at 3.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The corner of |x − 3| is (−3, 0).
The corner needs zero inside. At −3, the inside is −3 − 3 = −6 and the height is 6. At 3, the inside is 3 − 3 = 0.
✓ Instead: Corner: (3, 0).
Tips and tricks
  • Use the displayed landmark or known column as a check before drawing any additional points.
.2Shift left

An inside addition advances every old event. You start with a smaller input to cancel the addition before f receives it.

  • Formula: g(x) = f(x + h), h > 0.
  • Point: (u, v) becomes (u − h, v).
  • Domain: the allowed input addresses move left by h.
−6−4−22246810(−3, 0)
Adding three inside puts the lowest point three units left of zero.
Worked exampleAn old point appears earlier

If f(1) = 6, locate that height on g(x) = f(x + 3).
This asks you to find the new address or height of the given graph information.

The point (1, 6) becomes (−2, 6).
The card records the exact result of the worked coordinate or symmetry calculation.
  1. Set x + 3 = 1.f must receive the old input 1 to produce the known height 6.
  2. Subtract 3 to get x = −2.Subtraction undoes the inside addition.
Answer
The point (1, 6) becomes (−2, 6).
Check g(−2) = f(−2 + 3) = f(1) = 6.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For f(x + 3), move the old point (1, 6) right to (4, 6).
At 4 the inside is 7, not the known old input 1. Solving x + 3 = 1 gives x = −2; its inside is −2 + 3 = 1.
✓ Instead: The point (1, 6) becomes (−2, 6).
Tips and tricks
  • Use the displayed landmark or known column as a check before drawing any additional points.
.3Transform an input row

Keep each height and give it a new address. Do not invent a height for an input that your table never supplied.

  • For g(x) = f(x − 3), add 3 to known input coordinates.
  • The new input is what you give g. The old input is what g passes into f.
  • The output row is unchanged.
input xoutput g(x)5173971111
Read each column as one x input paired with its g(x) output.
Worked exampleUse a table without guessing

A table gives f(4) = 3. Find a known point on g(x) = f(x − 3).
This asks you to attach each known output to its transformed input and calculate any requested new height.

input xoutput f(x)214367811
Read each column as one x input paired with its f(x) output.
input xoutput g(x)5173971111
Read each column as one x input paired with its g(x) output.
  1. Read the displayed original table column by column. Each top input belongs to the output directly below it.The picture supplies the known values used in the following calculations; keep each column paired.
  2. Solve x − 3 = 4 to get x = 7.The required height 3 belongs to old input 4.
  3. g(7) = f(4) = 3.At x = 7 the inside expression is the known old input.
Answer
Known point: (7, 3).
Check Subtracting 3 from the new input 7 recovers old input 4.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Knowing f(4) = 3 lets you conclude g(4) = 3 for g(x) = f(x − 3).
g(4) requires f(1), which is not supplied by this table. The known height 3 appears at input 7 because 7 − 3 = 4.
✓ Instead: Known point: (7, 3).
Tips and tricks
  • Use the displayed landmark or known column as a check before drawing any additional points.
.4Interpret the units

Read the labels like a receipt. Adding to a gallons answer means extra gallons. Adding to a miles input means extra miles before you calculate gallons.

  • If f(x) is gallons for x miles, f(x) + 10 adds 10 gallons.
  • f(x + 10) asks for gallons for a trip 10 miles longer.
  • The mathematical left shift describes where the same output occurs; the context describes what quantity the formula changes.
x + 10 milesf: miles to gallonsf(x + 10) gallonsinputoutput
The ten is measured in miles because it enters before the miles-to-gallons machine.
Worked exampleSeparate gallons from miles

Interpret f(x) + 10 and f(x + 10) when f converts trip miles to gallons used.
This asks what the added number measures: more gallons after finding fuel use, or more miles before finding fuel use.

x milesadd 10 milesx + 10 milesff(x + 10) gallonsfirstsecond
Adding before f changes miles; adding after f changes gallons.
  1. f(x) + 10 is fuel for x miles plus 10 gallons.The addition is attached to the output, measured in gallons.
  2. f(x + 10) is fuel for x + 10 miles.The addition changes the input, measured in miles.
Answer
  • f(x) + 10: vertical shift up 10 gallons.
  • f(x + 10): horizontal shift left 10 miles on the mathematical graph.
Check With an original model f(x) = x25, the formulas become x25 + 10 and x+1025. Extra 10 miles uses 25 gallon, which differs from 10 extra gallons.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: If f converts miles into gallons, f(x) + 10 asks for a trip 10 miles longer.
f(x) is already the finished gallons answer. Adding outside adds 10 gallons. A longer trip instead changes the input before f: f(x + 10).
✓ Instead: f(x) + 10: vertical shift up 10 gallons.
f(x + 10): horizontal shift left 10 miles on the mathematical graph.
Tips and tricks
  • Use the displayed landmark or known column as a check before drawing any additional points.
Strategy: step by step
  1. 1. Find the expression inside f; it is the input the old function receives.
  2. 2. Set that expression equal to a known old input u, because this asks where the same old height appears.
  3. 3. Solve for the new input x. Add when the expression subtracts, and subtract when it adds.
  4. 4. Keep the old height v, because the function receives the same old input u.
  5. 5. Move several points and substitute their new inputs into g to check.
Strategy
Find where a known old height appears after a sideways shift
1
Are you moving a known old point?
YesSet the inside expression equal to that point's old input, then solve for the new input.
NoFor evaluation at a supplied new input, compute the inside expression first.
↓
2
Does the input expression subtract a positive number?
YesAdd that number to the old coordinate. This moves its point right.
NoIf it adds a positive number, subtract that number from the old coordinate. This moves its point left.
↓
3
When evaluating, is the required old input listed in the supplied table?
YesRead the output in that input's column.
NoState that the value cannot be determined from the table. Do not guess between entries.
  1. 1. Translate the request: you know an old input and height, and need the new address for that same height.
  2. 2. Name the old input u. Set the expression inside f equal to u.
  3. 3. Undo the inside addition or subtraction on both sides. In x − 3 = u, add 3 to obtain x = u + 3.
  4. 4. Keep the old height v. The new point is (u + h, v).
  5. 5. Substitute the new input into the inside expression. It must give the original input u.
  6. 6. If instead asked for g at a specified input, calculate its inside value first and look up f there.
Worked exampleShifting a parabola when the inside adds 3

Let f(x) = x2 − 4. Four points on the graph of f are (−2, 0), (0, −4), (1, −3) and (2, 0). Let g(x) = f(x + 3). (a) Find the four corresponding points on the graph of g. (b) State the direction and distance of the shift, and give the vertex of g.

−6−4−22−6−4−2246(−5, 0)(−3, −4)(−2, −3)(−1, 0)
Graph of g(x) = (x + 3)2 − 4. It is the graph of f(x) = x2 − 4 moved 3 units left, with the four shifted points marked.
  1. The expression inside f is x + 3. Write g(x) = f(x + 3) = f(x − (−3)), so h = −3.The expression inside f is the input the old function receives. Matching it to x − h shows the sign of h.
  2. For the point (−2, 0), set x + 3 = −2.This asks which new input x feeds the old input −2 into f, so that the same old height appears.
  3. Solve: x = −2 − 3 = −5. In the same way, x + 3 = 0 gives x = −3. x + 3 = 1 gives x = −2. x + 3 = 2 gives x = −1.The expression adds 3, so we subtract 3 to undo it. Each new input is the old input minus 3.
  4. Keep the heights: (−2, 0) → (−5, 0), (0, −4) → (−3, −4), (1, −3) → (−2, −3), (2, 0) → (−1, 0).At each new input, f receives the same old input u, so it returns the same old height v.
  5. The shift is 3 units left. The vertex (0, −4) of f moves to (−3, −4).With h = −3 the rule (u, v) → (u + h, v) moves every point 3 units left. This is the opposite of the + sign inside.
Answer
(a) (−5, 0), (−3, −4), (−2, −3), (−1, 0). (b) The graph of g is the graph of f shifted 3 units left. The vertex of g is (−3, −4).
Check Substitute the new inputs into g(x) = (x + 3)2 − 4. g(−5) = (−2)2 − 4 = 0. g(−3) = 02 − 4 = −4. g(−2) = 12 − 4 = −3. g(−1) = 22 − 4 = 0. All four heights match.

Work to write

  1. g(x) = f(x + 3) = f(x − (−3)), so h = −3
  2. Solve x + 3 = u, so x = u − 3
  3. (−2, 0) → (−5, 0); (0, −4) → (−3, −4); (1, −3) → (−2, −3); (2, 0) → (−1, 0)
  4. Shift 3 units left; vertex (−3, −4)
  5. Check: g(−3) = (0)2 − 4 = −4

(a) (−5, 0), (−3, −4), (−2, −3), (−1, 0). (b) The graph of g is the graph of f shifted 3 units left. The vertex of g is (−3, −4).

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: recover the old zero input

Where does the old point (0, 0) of x2 go on (x − 6)2?
This asks you to find the new address or height of the given graph information.

−4−22468−2246810(6, 0)
The old zero input is reached at new input 6.
  1. Set x − 6 = 0.The old height zero requires the old input zero.
  2. Add 6: x = 6, with the same height 0.You undo the subtraction before squaring.
Answer
(6, 0), a shift right 6.
Check (6 − 6)2 = 0.
Rung 2Rung 2: write a left shift

Write the formula for x2 shifted left 5.
This asks you to write the new rule from the specified moves and check its graph landmarks.

−8−6−4−22−4−22468(−5, 0)
The vertex appears at −5 when x + 5 is zero.
  1. An old input u needs new input u − 5.Left means five smaller on the horizontal number line.
  2. Replace x by x + 5: g(x) = (x + 5)2.Adding 5 inside recovers u from the new input u − 5.
Answer
  • g(x) = (x + 5)2.
  • Vertex: (−5, 0).
Check At x = −5, the inside is −5 + 5 = 0, so the lowest height is at the requested leftward address.
Rung 3Rung 3: shift every known table input

Create known points for g(x) = f(x − 3). Use the original f table displayed beside this question.
This asks you to attach each known output to its transformed input and calculate any requested new height.

input xoutput f(x)214367811
Read each column as one x input paired with its f(x) output.
input xoutput g(x)5173971111
Read each column as one x input paired with its g(x) output.
  1. Read the displayed original table column by column. Each top input belongs to the output directly below it.The picture supplies the known values used in the following calculations; keep each column paired.
  2. Add 3 to each old input: 2 + 3 = 5, 4 + 3 = 7, 6 + 3 = 9, 8 + 3 = 11.The new input must compensate for the subtraction inside f.
  3. Keep outputs 1, 3, 7, 11 in the same paired order.Each new input sends its corresponding old input to f.
Answer
  • Use the displayed transformed table
  • each column records one new input and its corresponding output.
Check g(9) = f(9 − 3) = f(6) = 7, matching the third column.
Rung 4Rung 4: a vent schedule starts earlier

The old vent schedule f(t) starts changing at 8 a.m. and first reaches 220 square feet at 10 a.m. Move the entire schedule two hours earlier. Give its formula and new times.
This asks you to find the new address or height of the given graph information.

input t in hoursoutput f(t) in square feet8010220
Read each column as one t in hours input paired with its f(t) in square feet output.
input t in hoursoutput g(t) in square feet608220
Read each column as one t in hours input paired with its g(t) in square feet output.
  1. Read the displayed original table column by column. Each top input belongs to the output directly below it.The picture supplies the known values used in the following calculations; keep each column paired.
  2. Write g(t) = f(t + 2).At any new time, look up what the old schedule did two hours later.
  3. Solve t + 2 = 8: t = 6.The old 8 a.m. change now happens at the input that feeds 8 to f.
  4. Solve t + 2 = 10: t = 8.The old 10 a.m. event moves two hours earlier as well.
Answer
  • g(t) = f(t + 2).
  • Change begins at 6 a.m.
  • The schedule first reaches 220 square feet at 8 a.m.
Check g(8) = f(10) = 220. A daily schedule can extend across midnight; if given only one day's data, translate the known time interval rather than assume extra entries.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: g(x) = f(x + 3) moves the graph right 3.
At a new input x, the old machine receives x + 3. To reach old input 0, solve x + 3 = 0, which gives x = −3.
✓ Instead: It moves left 3. For f(x) = x2, the old vertex (0, 0) becomes (−3, 0), and g(−3) = 0.
✗ Not this: For g(x) = f(x − 3), subtract 3 from the old table inputs 2, 4, 6, 8.
The formula subtracts 3 before f runs. The new address must be 3 larger so that this subtraction recovers each old address.
✓ Instead: Use the displayed shifted table. For example, the new column under 9 has output 7, because g(9) = f(9 − 3) = f(6) = 7. The displayed output row is unchanged.
✗ Not this: If f(4) = 3, then g(4) = 3 for g(x) = f(x − 3).
g(4) asks for f(1), which the supplied value f(4) does not tell you.
✓ Instead: The known height appears at new input 7: g(7) = f(4) = 3. g(4) remains unknown without a value for f(1).
Tips and tricks
  • Write new input expression = old input before choosing left or right. This equation explains the direction.
  • Inside lies is a memory cue about graph movement. The machine really performs the operation written inside; your new coordinate must undo it.
  • For a table, carry each old output to its new input column. Do not swap the task of moving an old point with the task of evaluating g at a specified input.
Trap. Reading f(x + 3) as right 3. Set x + 3 equal to an old input: x = old input − 3, so the graph moves left. The equation explains the direction when a memorized slogan feels unreliable.