Quarry School

Vertical shifts: move every height by the same amount

Explain it like I am five

Picture a graph drawn on a clear sheet laid over a map. You can lift the sheet three squares without changing the drawing. Each dot stays over the same left-to-right address, but its height rises three squares. A Transformation changes a graph by moving, reflecting, or scaling it. This particular move is a Vertical shift. In a function, the input is your left-to-right address and the output is your height. Adding a number after the function finishes changes the output. That is an Outside change. You can use the same idea on a formula, a drawing, a table, or a measurement such as the area of an open vent.

−3−2−1123−1123456789range(0, 3)(−1, 4)(0, 3)(1, 4)
The shifted graph lies three units above the old graph at every shared input.
Reminder
  • Function notation and coordinates. f(−1) = 1 means the graph contains (−1, 1): input first, output second. In g(−1) = f(−1) + 3, use that known old output before adding.
  • Adding signed numbers. Adding a negative lowers the height: 1 + (−3) = −2. Adding a positive raises it: −5 + 3 = −2.
  • Domain and Range. Domain means allowed inputs; Range means actual outputs. x2 accepts every real x and produces y ≥ 0. Thus x2 + 3 still accepts every x and produces y ≥ 3.
Why it works. Suppose a point (u, v) is on the original graph. That means f(u) = v. If g(x) = f(x) + k, substituting u gives g(u) = v + k. The input u has not changed, and the output has increased by the same k as every other output. Therefore every dot moves the same vertical distance. Negative k moves downward because adding a negative number lowers the height. The graph keeps its shape because differences between any two heights stay the same.
Ruleg(x) = f(x) + k sends (u, v) to (u, v + k). Positive k moves up; negative k moves down. The domain stays the same; every range value gains k.
The same idea, five ways
Say it

Say: the new height is the old height plus k, at the same input.

Write it

A Vertical shift adds one fixed number to every output and leaves each input unchanged.

In math
  • g(x) = f(x) + k
  • Old point (u, v); new point (u, v + k).
  • For x2 + 3: Range y ≥ 3, or [3, ∞), or {y | y ≥ 3}.
  • Graph words: lift or lower the entire drawing by the same number of units.
Like

Lift a clear map without sliding it sideways. Every landmark keeps its street address and gains the same height.

See it
−222468range(0, 3)(−1, 4)(0, 3)(1, 4)
Every old height increases by three while its horizontal address stays fixed.
The same idea, other ways
As a picture

Lift the entire clear sheet. A dot that was two squares above the horizontal axis becomes five squares above it after an upward shift of three.

−222468newold
The lowest point rises from height 0 to height 3 without moving sideways.
As a machine

The old machine gives its answer first. A second worker adds three to that answer. The worker never touches the number that went into the machine.

2square4add 37firstsecond
The extra three is added after the original function has produced four.
With numbers

If f(−2) = 4, then g(−2) = f(−2) + 3 = 7. The point (−2, 4) becomes (−2, 7). Keeping the first coordinate is the signature of a vertical shift.

With numbers
g(x) = f(x) + k
Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.
A memory device with a reason

Remember: outside tells the truth. Outside +3 adds three to the height, so the graph goes up three. You can rebuild this by substituting any known input; there is no hidden change to x.

A memory device with a reason
g(x) = f(x) + k
Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.
You seeIt does when h and k are positive
f(x) + kUp k
f(x) − kDown k
f(x − h)Right h, explained in the next lesson
f(x + h)Left h, explained in the next lesson
.1Shift up

You raise every dot by the same number of squares. Even a dot below the axis rises; it does not have to cross the axis.

  • Formula: g(x) = f(x) + k with k > 0.
  • Point: (u, v) becomes (u, v + k).
  • Domain: unchanged. Range: every allowed output gains k.
−8−6−4−22468−224new center
The Cube root function keeps its bent shape while every height gains two.
Worked exampleA cube root moved up

Find the new height at x = 8 for g(x) = x3 + 2.
This asks for the new output at the stated input after the outside height change.

−8−6−4−22468−224newold
At input 8 the new height is 4, two above the original height.
  1. 83 = 2.23 = 8, so the cube root returns 2.
  2. g(8) = 2 + 2 = 4.The outside addition raises the old height by 2.
Answer
The old point (8, 2) becomes (8, 4).
Check Undo the shift: 4 − 2 = 2, the old cube root height.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For x3 + 2, move the old point (8, 2) to (10, 2).
That changes the input. The +2 comes after the cube root and therefore changes its height: 2 + 2 = 4.
✓ Instead: The old point (8, 2) becomes (8, 4).
Tips and tricks
  • Use the displayed landmark or known column as a check before drawing any additional points.
.2Shift down

Lower the whole drawing. The minus sign changes the height after the function has finished; it does not move the input.

  • Formula: g(x) = f(x) − k for a positive downward distance k.
  • Point: (u, v) becomes (u, v − k).
  • A negative new height is allowed whenever the old function and shift produce one.
−4−224−4−224(0, −3)
The Absolute value function's corner moves from height zero to height −3.
Worked exampleLower an absolute value height

Find g(1) when g(x) = |x| − 3.
The number in parentheses is the supplied input. This asks for the output produced at that input.

−4−224−4−22468new
At input 1 the new height is −2, three below 1.
  1. |1| = 1.Absolute value is distance from zero.
  2. g(1) = 1 − 3 = −2.The downward shift subtracts 3 from the height.
Answer
(1, 1) becomes (1, −2).
Check Adding 3 back to −2 gives the original height 1.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For |x| − 3 at input 1, calculate |1 − 3| = 2.
That puts the subtraction inside absolute value and changes the function. The stated rule first returns |1| = 1 and then subtracts 3 to give −2.
✓ Instead: (1, 1) becomes (1, −2).
Tips and tricks
  • Use the displayed landmark or known column as a check before drawing any additional points.
.3Read and transform an output row

A table stores dots in columns. For a vertical shift, change the output row and keep the input row paired with it.

  • Change outputs, not inputs.
  • A vertical shift uses addition, not multiplication.
  • If an output is zero, it also gains the shift.
input xoutput g(x)2−2406488
Read each column as one x input paired with its g(x) output.
Worked exampleRaising a table of outputs by 5

A function f is defined on the interval [−4, 2] and has range [−3, 6]. The table gives four of its points: f(−4) = −1, f(−1) = 6, f(0) = 2, f(2) = −3. Let g(x) = f(x) + 5. (a) Find the four corresponding points on the graph of g. (b) State the domain and range of g. (c) Check one point by substituting into the formula for g.

input xoutput f(x)−4−1−16022−3↓ evaluate: input given, read the output below it
The table of f: f(−4) = −1, f(−1) = 6, f(0) = 2 and f(2) = −3. Each output rises by 5 for g(x) = f(x) + 5.
  1. Read the number added outside f in g(x) = f(x) + 5. The shift is k = +5.The 5 is added after f produces its output, so it changes heights only. Its sign is positive, so the graph moves up 5 units.
  2. Keep the inputs −4, −1, 0 and 2 unchanged.g sends the same x into f, so every horizontal coordinate stays where it was.
  3. Add 5 to each height: (−4, −1) → (−4, −1 + 5) = (−4, 4); (−1, 6) → (−1, 11); (0, 2) → (0, 7); (2, −3) → (2, 2).g outputs the old height plus k, so (u, v) goes to (u, v + 5).
  4. Connect the moved points with the same shape as f. The whole graph is the graph of f lifted 5 units.Every point moves by the same amount, so the shape is unchanged.
  5. Domain of g: [−4, 2]. Range of g: [−3 + 5, 6 + 5] = [2, 11].The x-values are unchanged, so the domain stays the same. Every output gains 5, so each endpoint of the range gains 5.
  6. Check: g(−1) = f(−1) + 5 = 6 + 5 = 11, which matches the point (−1, 11).Substituting a point into the new formula confirms its new height.
Answer
(a) The points on g are (−4, 4), (−1, 11), (0, 7) and (2, 2). (b) The domain is [−4, 2] and the range is [2, 11]. (c) g(−1) = 6 + 5 = 11.
Check Each new height minus the old height is 5: 4 − (−1) = 5, 11 − 6 = 5, 7 − 2 = 5 and 2 − (−3) = 5. The lowest point (2, −3) becomes (2, 2) and the highest point (−1, 6) becomes (−1, 11). These agree with the range [2, 11].

Work to write

  1. k = +5, so the graph shifts up 5
  2. (−4, 4), (−1, 11), (0, 7), (2, 2)
  3. Domain of g: [−4, 2]
  4. Range of g: [2, 11]
  5. g(−1) = f(−1) + 5 = 11

(a) The points on g are (−4, 4), (−1, 11), (0, 7) and (2, 2). (b) The domain is [−4, 2] and the range is [2, 11]. (c) g(−1) = 6 + 5 = 11.

Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For f(x) − 3, change the table input 6 to 3 and keep its height 7.
The subtraction is outside f. The input stays 6, and the known height 7 becomes 7 − 3 = 4.
✓ Instead: g(6) = 4.
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 number added outside f. Include its sign, because −4 is a shift of k = −4.
  2. 2. Keep every horizontal coordinate, because the input sent into f has not changed.
  3. 3. Add k to every height, because g outputs the old height plus k.
  4. 4. Move several recognizable points and connect them with the original shape.
  5. 5. Substitute a point into the new formula to check its new height.
Strategy
Move a graph vertically from a formula or a known point
1
Is the added outside number positive?
YesMove every height up by that number.
NoCheck whether it is negative or zero.
↓
2
Is the added outside number negative?
YesMove every height down by its positive distance. For k = −3, calculate v − 3.
NoIf k = 0, every point stays in place.
↓
3
Is a table entry the only information you have about f?
YesTransform that known entry and the other supplied entries; leave missing outputs undetermined.
NoUse the known formula or familiar graph to select and check several points.
  1. 1. Find the added number after f finishes. Write it as k, including a minus sign.
  2. 2. Translate the request: you need the old height at the same input, followed by the outside addition.
  3. 3. For each old point (u, v), leave u alone and calculate v + k.
  4. 4. Move the whole familiar shape through the new points. Add k to its old range values; keep its domain.
  5. 5. Check by substituting u into g, or subtract k from the new height to recover v.
Worked exampleRaise a quadratic graph by three

Graph g(x) = x2 + 3 using f(x) = x2.
This asks you to find the new address or height of the given graph information.

−4−224−4−22468range(0, 3)
The original U rises three units; its new vertex is (0, 3).
  1. Choose old points (−1, 1), (0, 0), and (1, 1).Squaring −1, 0, and 1 gives 1, 0, and 1.
  2. Add 3 to each height: (−1, 4), (0, 3), and (1, 4).The +3 is outside the squaring operation.
  3. Draw an upward-opening curve with its lowest point at (0, 3).A shift moves the Quadratic function without changing its shape or direction.
  4. Domain: all real numbers. Range: y ≥ 3.Every real x can be squared, and x2 ≥ 0 means x2 + 3 ≥ 3.
Answer
  • Up 3.
  • Vertex: (0, 3).
  • Domain: (−∞, ∞).
  • Range: [3, ∞).
Check At x = 2 the old height is 4 and the new formula gives 22 + 3 = 7, exactly three more.
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: move one dot up

If f(4) = 2, find g(4) for g(x) = f(x) + 5.
This asks you to find the new address or height of the given graph information.

(4, 2) becomes (4, 7).
The card records the exact result of the worked coordinate or symmetry calculation.
  1. g(4) = f(4) + 5 = 2 + 5.Keep the input and use its known output.
  2. g(4) = 7.The old height 2 rises by 5.
Answer
(4, 2) becomes (4, 7).
Check 7 − 2 = 5, the requested upward distance.
Rung 2Rung 2: move a whole graph down

Write the formula for x2 shifted down 4.
This asks you to find the new address or height of the given graph information.

−4−224−4−22468(0, −4)
The vertex drops from (0, 0) to (0, −4).
  1. Keep x2 as the original function.A vertical shift leaves the input calculation unchanged.
  2. Subtract 4 outside: g(x) = x2 − 4.Every old height must become four less.
Answer
  • g(x) = x2 − 4.
  • Vertex: (0, −4).
Check g(0) = −4 and g(2) = 0, four below the old heights 0 and 4.
Rung 3Shifting a table of values down by 3

A function f is defined on the interval [−2, 3] and has range [−2, 5]. The table gives four of its points: f(−2) = 5, f(0) = 1, f(1) = −2, f(3) = 4. Let g(x) = f(x) − 3. (a) Find the four corresponding points on the graph of g. (b) State the domain and the range of g.

input xoutput f(x)−25011−234↓ evaluate: input given, read the output below it
Table of f: f(−2) = 5, f(0) = 1, f(1) = −2, f(3) = 4. Subtract 3 from each output to get the values of g(x) = f(x) − 3.
  1. Read the number added outside f in g(x) = f(x) − 3. It is k = −3.The −3 is added after f is evaluated, so it changes outputs, not inputs. Its sign is negative, so the graph moves down 3 units.
  2. Keep the x-values −2, 0, 1 and 3 unchanged.g still sends the same x into f, so no point moves left or right.
  3. Add −3 to each height: 5 + (−3) = 2, 1 + (−3) = −2, −2 + (−3) = −5, 4 + (−3) = 1. The new points are (−2, 2), (0, −2), (1, −5) and (3, 1).Each output of g is the old height f(x) plus k, and the rule sends (u, v) to (u, v + k).
  4. Plot (−2, 2), (0, −2), (1, −5) and (3, 1). Join them with the same shape as f, placed 3 units lower.Every point of f moves down by the same 3 units, so the shape does not stretch or tilt.
  5. The domain of g is [−2, 3]. The range is [−2 − 3, 5 − 3] = [−5, 2].A vertical shift does not change the domain. Each range value gains k = −3, so the lowest and highest heights both drop by 3.
Answer
(a) (−2, 2), (0, −2), (1, −5), (3, 1). (b) The domain of g is [−2, 3] and the range of g is [−5, 2].
Check Substitute x = 1 into the new formula: g(1) = f(1) − 3 = −2 − 3 = −5. This matches the point (1, −5). Substitute x = 3: g(3) = f(3) − 3 = 4 − 3 = 1. This matches the point (3, 1).

Work to write

  1. k = −3, so the graph shifts down 3 units
  2. x-values stay the same: −2, 0, 1, 3
  3. g(−2) = 2, g(0) = −2, g(1) = −5, g(3) = 1
  4. Points: (−2, 2), (0, −2), (1, −5), (3, 1)
  5. Domain of g: [−2, 3]
  6. Range of g: [−5, 2]
  7. Check: g(1) = f(1) − 3 = −2 − 3 = −5

(a) (−2, 2), (0, −2), (1, −5), (3, 1). (b) The domain of g is [−2, 3] and the range of g is [−5, 2].

Rung 4Rung 4: move a physical starting height

For an original practice model, h(t) = 14t − 5t2 is a ball's height in meters t seconds after launch from the ground. Use the same motion from a 10-meter platform. Find H(t) and H(2).
This asks you to find the new address or height of the given graph information.

2246810121416182022platformH(2)
The two height formulas differ by 10 at the same time; the point shown is H(2) = 18.
  1. Write H(t) = h(t) + 10.At the same time the ball is 10 meters higher when its starting platform is 10 meters higher.
  2. Substitute the old formula: H(t) = 14t − 5t2 + 10.The extra starting height is added to the output, not to the time.
  3. H(2) = 14·2 − 5·22 + 10 = 28 − 20 + 10 = 18.Square 2 first, then multiply, then combine the height terms.
Answer
  • H(t) = 14t − 5t2 + 10.
  • H(2) = 18 meters.
Check The ground-launched model gives h(2) = 8 meters, so 18 − 8 = 10 meters. The physical model is used only while the ball remains airborne.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: For g(x) = f(x) + 3, move old point (−1, 1) to (2, 1).
That adds 3 to the input coordinate. Here the old function finishes first, and the +3 changes its output.
✓ Instead: Keep the input −1. Add 3 to height 1: 1 + 3 = 4. The new point is (−1, 4).
✗ Not this: An upward shift cannot leave a point below the x-axis.
An upward shift adds the same amount to every signed height. A height can increase and still be negative.
✓ Instead: The old height −5 shifted up 3 becomes −5 + 3 = −2. It has risen by 3, while remaining below zero.
Tips and tricks
  • Write outside +3 beside an upward arrow, then keep the first coordinate unchanged in every moved point.
  • A vertical shift adds a fixed amount. Check new height minus old height: it must equal k at every input.
  • Check one point above and one point below the x-axis when a graph has both. An upward shift raises each signed height.
Trap. Changing x when the added number is outside. For f(x) + 3, finish finding the old height first and then add 3; the left-to-right address stays put.