Combine shifts and read the domain and range

Explain it like I am five

Take the same clear drawing and slide it sideways, then raise or lower it. A corner of the drawing has two coordinates, like a street address and a floor number. Moving right changes the address. Moving up changes the floor. You can do those two moves in either order because neither move changes the coordinate used by the other. This is how you build a new graph from a familiar Toolkit function, a basic shape you recognize. Watch its corner, starting point, or gaps. These landmarks tell you where the whole shape goes. The Domain is the set of allowed input addresses. The Range is the set of heights the graph actually reaches.

The V moves right two and down four, taking its minimum height with it.

Reminder

Square root restrictions. A real square root needs a nonnegative inside and returns a nonnegative result: $square root of x + 2$ needs x + 2 ≥ 0, and adding 3 outside gives heights at least 3.
Adding to an inequality. Add or subtract the same number on both sides without reversing order: x + 2 ≥ 0 becomes x ≥ −2 after subtracting 2.
Division by zero. A denominator cannot be zero. For $1 over x − 1$ , solve x − 1 = 0 to find the forbidden input x = 1, then substitute: 1 − 1 = 0.
Interval notation. A bracket includes an endpoint: x ≥ −2 is [−2, ∞). A missing input splits an interval: x ≠ 1 is (−∞, 1) ∪ (1, ∞).

Why it works. For g(x) = f(x − h) + k and a known old point (u, v), choose x = u + h. Then x − h = u, so g(u + h) = f(u) + k = v + k. The new point is (u + h, v + k). Doing the vertical move first gives the same pair because adding k to v does not touch u. Domain and range move too: a square root still needs a nonnegative inside, and its smallest old output zero becomes k.

Ruleg(x) = f(x − h) + k sends (u, v) to (u + h, v + k). These shifts commute, meaning either order gives the same graph, because each changes a different coordinate.

The same idea, five ways

Say it

Say: move each old input by h and each old height by k. Then state where inputs are allowed and what heights occur.

Write it

Combining a Horizontal shift and a Vertical shift moves both coordinates, and carries the graph's starting points and restrictions with it.

In math

g(x) = f(x − h) + k
Old point (u, v); new point (u + h, v + k).
For $square root of x + 2$ + 3: Domain x ≥ −2, [−2, ∞), {x | x ≥ −2}.
Range y ≥ 3, [3, ∞), {y | y ≥ 3}.
For $1 over x − 1$ + 1: Domain x ≠ 1; Range y ≠ 1.
Graph words: translate the corner, endpoint, or asymptote guide lines by the same horizontal and vertical amounts.

A street address and a floor number can both change. Change the street address and the floor in either order to reach the same room.

See it

The corner of the V moves from (0, 0) to (2, −4), with the input and output changes handled separately.

The same idea, other ways

As a map and a floor number

Treat (u, v) as a street address and a floor number. Right 2 changes only the street address; down 4 changes only the floor. Doing either move first takes (0, 0) to the same new location (2, −4).

The corner of the V moves from (0, 0) to (2, −4), with the input and output changes handled separately.

As two numerical jobs

For |x − 2| − 4, the new input 2 sends 2 − 2 = 0 into absolute value. Absolute value returns |0| = 0, then subtracting 4 gives −4. That places the corner at (2, −4). The old point (1, 1) similarly becomes (3, −3).

As two numerical jobs

g(x) = f(x − h) + k

Connect the explanation to the displayed formula, then use the nearby worked numbers to check it.

As two shadows

A graph casts one shadow onto the horizontal axis for allowed inputs and another onto the vertical axis for attainable heights. For $square root of x + 2$ + 3, its starting point moves to (−2, 3): the horizontal shadow begins at −2 and the vertical shadow begins at 3.

The moved starting point locates both the beginning of the Domain and the beginning of the Range.

.1Absolute value corner

The absolute value graph is a V. Its corner is where the inside equals zero. Moving the V moves that corner and keeps both arms attached.

Formula: |x − h| + k.
Vertex: (h, k), the corner where the two arms meet.
Domain: (−∞, ∞). Range: [k, ∞) when there is no vertical reflection.

The corner moves right two and down four, while the arms keep their slopes.

Worked exampleUse the zero inside

Find the corner of |x − 2| − 4.
This asks you to find the new address or height of the given graph information.

The corner is (2, −4) because x − 2 becomes zero at 2.

Solve x − 2 = 0 to get x = 2.Absolute value is smallest at old input zero.
The height is |0| − 4 = −4.The outside shift lowers the old minimum zero by four.

Answer

Corner: (2, −4).

Check At inputs 1 and 3 the heights are both −3, one unit above the corner on each side.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The corner of |x − 2| − 4 is (−2, −4).

At input −2 the inside is −4 and the height is 4 − 4 = 0. The inside becomes zero at input 2, giving the true lowest height −4.

✓ Instead: Corner: (2, −4).

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.2Square root endpoint

A square root graph begins at an Endpoint, its included starting point. The beginning moves along with the curve, and the allowed inputs begin there too.

Formula: $square root of x − h$ + k.
Endpoint: (h, k).
Domain: x ≥ h, or [h, ∞). Range: y ≥ k, or [k, ∞).
The square root sign means the nonnegative root.

The curve starts at (−2, 3), so inputs begin at −2 and heights begin at 3.

Worked exampleA shifted square root from its starting point

A translated Square root function begins at (1, 2). Write its formula, domain, and range.
This asks you to find the new address or height of the given graph information.

The endpoint (1, 2) determines both shifts.

The old endpoint (0, 0) moved right 1 and up 2.Subtract the old coordinates from the new coordinates to find the two shifts.
Write g(x) = $square root of x − 1$ + 2.A right shift subtracts 1 inside and an upward shift adds 2 outside.
Require x − 1 ≥ 0, giving x ≥ 1. The square root output is nonnegative, giving g(x) ≥ 2.Real square roots need a nonnegative inside and return a nonnegative result.

Answer

g(x) = $square root of x − 1$ + 2.
Domain: [1, ∞).
Range: [2, ∞).

Check g(1) = 2 and g(2) = 3, agreeing with transformed old points (0, 0) and (1, 1).

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A square root beginning at (1, 2) has formula + 2.

That formula still begins at input 0. To begin at input 1, the inside must be x − 1, which becomes zero at 1.

✓ Instead: g(x) = + 2.
Domain: [1, ∞).
Range: [2, ∞).

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

.3Reciprocal asymptotes

A reciprocal graph has two branches and a forbidden input. An Asymptote is a line the curve approaches. Shifts move those guide lines along with the branches.

Formula: g(x) = $1 over x − h$ + k.
Domain: x ≠ h, because a denominator cannot be zero.
Range: y ≠ k, because $1 over x − h$ never equals zero.
Vertical asymptote: x = h. Horizontal asymptote: y = k.

The reciprocal guide lines move from the axes to x = 1 and y = 1.

Worked exampleMove a reciprocal graph

Shift the Reciprocal function one unit right and one unit up.
This asks you to find the new address or height of the given graph information.

The excluded lines move to x = 1 and y = 1.

Replace x by x − 1 in $1 over x$ .The right shift makes the old input appear one unit later.
Add 1 after dividing: g(x) = $1 over x − 1$ + 1.The up shift changes the completed output.
Exclude x = 1 and y = 1.The denominator would vanish at x = 1, and a reciprocal term cannot equal zero.

Answer

g(x) = $1 over x − 1$ + 1.
Domain: (−∞, 1) ∪ (1, ∞).
Range: (−∞, 1) ∪ (1, ∞).

Check Old point (1, 1) becomes (2, 2), and g(2) = 1 + 1 = 2.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The shifted reciprocal + 1 excludes input 0 and output 0.

At input 0 the denominator is −1, so the formula is defined and gives −1 + 1 = 0. The forbidden input and missing height have both moved to 1.

✓ Instead: g(x) = + 1.
Domain: (−∞, 1) ∪ (1, ∞).
Range: (−∞, 1) ∪ (1, ∞).

Tips and tricks

Use the displayed landmark or known column as a check before drawing any additional points.

Strategy: step by step

1. Recognize the original toolkit shape before changing it.
2. Read h from x − h. If you see x + 2, write h = −2.
3. Read k from the addition outside the function.
4. Move the landmark and at least two other points by (h, k).
5. Check allowed inputs using the original function's restrictions, then translate the heights to obtain the range.
6. Substitute a new point into the formula and compare with its transformed height.

Strategy

Combine shifts, then find Domain and Range from the starting graph

Is the starting function a Square root function?

YesRequire x − h ≥ 0. The endpoint becomes (h, k), the Domain is [h, ∞), and the Range is [k, ∞).

NoCheck for a Reciprocal function.

↓

Is there a reciprocal denominator x − h?

YesExclude x = h after showing h − h = 0. Its missing range height becomes k, so exclude y = k.

NoFor the Absolute value function or Quadratic function with shifts alone, all real inputs remain allowed and the lowest output becomes k.

↓

Are you using only isolated supplied points?

YesMove those known points. They do not determine an unknown function's complete Domain or Range.

NoUse the known toolkit function's full Domain and Range, together with its shifts.

1. Recognize the basic shape and recall its allowed inputs and outputs.
2. Read h from x − h and k from the outside addition. For x + 2, h = −2.
3. Move each old point to (u + h, v + k), including the landmark.
4. For a square root, require its inside ≥ 0 and solve. Set the inside to 0 to find the endpoint input, then substitute it to find its height.
5. For a reciprocal, set the denominator to 0 to find the forbidden input. Substitute that input into the denominator to show the zero, and exclude it.
6. Translate every original range height by k. A root's lowest height 0 becomes k; a reciprocal's missing height 0 becomes k.
7. Write each requested answer on its own line and check a moved point directly in the new formula.

Worked exampleDescribe a V moved right 2 and down 4

Describe how g(x) = |x − 2| − 4 comes from f(x) = |x|, and sketch it.
This asks you to find the new address or height of the given graph information.

The V moves right 2 and down 4.

Identify h = 2 from x − 2.The new input x = u + 2 compensates for subtracting 2 inside.
Identify k = −4.The subtraction after absolute value lowers every height by 4.
Move the corner (0, 0) to (2, −4).Add 2 to its horizontal coordinate and subtract 4 from its height.
Move (−1, 1) to (1, −3) and (1, 1) to (3, −3).Every point receives the same two shifts.
Join these points in a V opening upward.Shifts preserve the original V shape.

Answer

Right 2 and down 4.
Corner: (2, −4).
Domain: (−∞, ∞).
Range: [−4, ∞).

Check g(1) = |−1| − 4 = −3 and g(3) = |1| − 4 = −3, verifying both sides of the moved corner.

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: shift a known dot in two directions

(3, −2) is on f. Where is it on g(x) = f(x − 6) + 3?
This asks you to find the new address or height of the given graph information.

(9, 1).

The card records the exact result of the worked coordinate or symmetry calculation.

Solve x − 6 = 3 to get x = 9.The inside must reproduce old input 3.
Raise old height −2 by 3: −2 + 3 = 1.The addition outside changes the output.

Answer

(9, 1).

Check g(9) = f(3) + 3 = −2 + 3 = 1.

Rung 2Rung 2: identify the shifted corner

Find the corner and shifts for |x − 2| − 4.
This asks you to find the new address or height of the given graph information.

The corner is (2, −4).

Solve x − 2 = 0 to find x = 2.The old corner uses input zero.
Use height |0| − 4 = −4.The outside shift lowers zero by four.

Answer

Right 2 and down 4.
Corner: (2, −4).

Check Its equal heights at x = 1 and x = 3 show the V centered at x = 2.

Rung 3Rung 3: A V shifted left one and down three

Use f(x) = |x| to describe and sketch g(x) = |x + 1| − 3. This asks you to move each old sideways address one unit left and each old height three units down.

The V moves left one and down three, with its corner at (−1, −3).

Set x + 1 = 0 to find where the old corner input appears. Subtract 1 to get x = −1.The new corner must send old input 0 into absolute value. This solves for its new horizontal address.
At that input, g(−1) = |0| − 3 = −3.The outside subtraction lowers the old corner height 0 by 3.
Move old points (−1, 1), (0, 0), and (1, 1) to (−2, −2), (−1, −3), and (0, −2).Subtract 1 from each old input and 3 from each old output. These are separate coordinate changes.
Draw the same V through those moved points. Every real input is still allowed, and the smallest output is −3.A translation preserves the V shape. Absolute value stays nonnegative before the final downward shift.

Answer

Shift left 1 and down 3.
Corner: (−1, −3).
Domain: (−∞, ∞).
Range: [−3, ∞).

Check g(−1) = −3 verifies the corner. g(0) = |1| − 3 = −2 agrees with old point (1, 1) moved to (0, −2). The different formula |x − 2| − 4 moves right 2 and down 4, because its corner input solves x − 2 = 0 instead.

Rung 4Rung 4: write a square root shifted left and up

Write the function for $\sqrt{x}$ shifted left 2 and up 3. Include its domain and range.
This asks you to write the new rule from the specified moves and check its graph landmarks.

The endpoint moves left 2 and up 3.

Left 2 means replace x by x + 2.At new input u − 2, adding 2 inside returns old input u.
Add 3 outside: g(x) = $square root of x + 2$ + 3.Each completed square root height rises by 3.
Require x + 2 ≥ 0: x ≥ −2. The new height is at least 3.Real square roots require and return nonnegative numbers before the shift.

Answer

g(x) = $square root of x + 2$ + 3.
Endpoint: (−2, 3).
Domain: [−2, ∞).
Range: [3, ∞).

Check At x = 2, g(2) = + 3 = 5. This is old point (4, 2) moved to (2, 5).

Rung 5Rung 5: move guide lines and missing values

Shift $1 over x$ right 1 and up 1. State the new formula and the excluded input and output.
This asks you to find the new address or height of the given graph information.

The graph excludes input 1 and output 1.

Write g(x) = $1 over x − 1$ + 1.Right subtracts inside; up adds outside.
x = 1 is excluded.At 1 the denominator becomes zero.
y = 1 is excluded.To equal 1, the reciprocal term would have to equal zero, which is impossible.

Answer

g(x) = $1 over x − 1$ + 1.
Excluded input: 1.
Excluded output: 1.
Asymptotes: x = 1 and y = 1.

Check Given any y ≠ 1, solving y − 1 = gives x = 1 + , so every other output occurs.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: For g(x) = + 3, keep the old domain [0, ∞) and range [0, ∞).

The input addition allows the old input zero at new input −2. The outside addition changes old height zero to 3.

✓ Instead: Solve x + 2 ≥ 0 to get x ≥ −2. The root's nonnegative outputs become y ≥ 3. Domain: [−2, ∞). Range: [3, ∞).

✗ Not this: For g(x) = + 1, exclude input 0 and output 0 because that is what the old reciprocal excludes.

The shift moves the missing input and missing height along with the rest of the graph. At input 0, the formula is defined and gives −1 + 1 = 0.

✓ Instead: Exclude input 1 because 1 − 1 = 0 makes the denominator zero. Exclude output 1 because the reciprocal part can never be zero. The point (0, 0) is allowed.

Tips and tricks

Find the corner or endpoint by making its inside expression zero, then substitute that input to find its height.
Write Domain for allowed inputs and Range for attainable outputs on separate answer lines.
For reciprocal shifts, move the two asymptote guide lines and check the denominator at the proposed excluded input.
Commute means you may swap the order without changing the result. Here x gains h and y gains k, so neither job affects the other's number.

Trap. Moving the corner correctly but leaving the old domain and range. For + 3, the allowed inputs start at −2 and the heights start at 3. Check the inside restriction and the lowest height separately.