Organises every graph transformation under one rule: changes written outside a function act on the output, vertically, and do what they say; changes written inside act on the input, horizontally, and do the opposite. Covers shifts, stretches, compressions and reflections, the order in which combined transformations are applied, and the even and odd symmetry tests.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 1 — Functions
§1.5 Transformation of Functions, pp. 101-136
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 101-136 — the pages these objectives are drawn from
Warm-up
You have graphed a parabola before. This lesson asks what happens to it when the rule is nudged.
Discussion prompt
Without computing anything, predict what the graph of the squaring rule with 3 added to the whole thing looks like, and then what the graph of the squaring rule with 3 added to x before squaring looks like.
Hint: For the second one, ask which input now produces the output 0.
Answer:
The first is the familiar parabola moved up 3. Every output is 3 larger, so every point rises by 3. Almost everyone predicts this correctly.
The second is the parabola moved left 3, and most people first guess right. The check: the new rule outputs 0 when x plus 3 is 0, which is at x equal to negative 3 — so the bottom of the parabola has moved to the left.
That single disagreement between intuition and fact is what this whole lesson is organised around. Outside changes behave as written; inside changes behave backwards.
Concept
A change written outside the function acts on the output: it moves the graph vertically and does exactly what it says. A change written inside acts on the input: it moves the graph horizontally and does the opposite of what it says.
transformation — A change to a function's formula that produces a predictable change in its graph. Transformations that act on the output are vertical and behave as written; those that act on the input are horizontal and behave in reverse.
\[ \underbrace{a}_{\text{outside}} \, f\bigl(\underbrace{b(x-h)}_{\text{inside}}\bigr) + \underbrace{k}_{\text{outside}} \]
There are twelve individual rules in this section if you count them separately, and two if you understand them. Everything vertical is outside and obvious; everything horizontal is inside and reversed. Committing to that organisation now saves memorising a table later.
Figure (svg): A card contrasting changes made outside the function, which affect the output vertically and do what they say, with changes made inside, which affect the input horizontally and do the opposite of what they say
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 101-106
Section
Section 1
Concept
Adding a constant outside the function moves the graph up by that amount. Adding a constant inside moves it left by that amount — the opposite of what the sign suggests.
\[ f(x)+k \text{ shifts up } k; \qquad f(x-h) \text{ shifts right } h \]
Writing the inside change as x minus h rather than x plus h is a deliberate convention, and a helpful one: in that form the number h is the shift, sign and all. Written as x plus 3, the shift is left 3; written as x minus negative 3, the same thing reads as a shift of negative 3, which is left 3. The second form does the sign flip for you.
Figure (svg): The squaring rule shown with a vertical shift up and a horizontal shift right, drawn on the same axes with the original dashed and each shifted version solid
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 101-110
Picture it
This argument is worth following once. After that the rule stops needing to be remembered.
Figure (svg): An explanation of why an inside shift moves the graph the opposite way, showing that to make the new function output what the old one did, the input must be reached three units earlier
The new rule reaches any given value of the old one at a smaller input, so every feature of the graph arrives earlier — which on a left-to-right axis means further left.
Worked example
Read the inside and the outside separately.
\[ \text{Describe the graph of } g(x)=\sqrt{x-2}+4 \text{ in terms of the square root rule.} \]
Identify the parent
Why: It is the square root rule.
Read the inside change
Why: Minus 2 inside, so it shifts right 2.
\[ \text{right } 2 \]
Read the outside change
Why: Plus 4 outside, so it shifts up 4.
\[ \text{up } 4 \]
State the new starting point
Why: The parent starts at the origin; the shift moves it.
\[ \text{starts at } (2, 4) \]
Figure (svg): The squaring rule shown with a vertical shift up and a horizontal shift right, drawn on the same axes with the original dashed and each shifted version solid
\[ \text{shift right } 2 \text{ and up } 4; \text{ domain } [2,\infty), \text{ range } [4,\infty) \]
Verify: check with the corner point
Why: The parent's corner is at the origin. The new corner should be at the point with x equal to 2 and y equal to 4, and substituting x equal to 2 gives the square root of 0 plus 4, which is 4. The corner is where it was predicted, so both shifts were read in the right direction.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 106-108
Sorting
Read where the change is written before deciding.
Sort into buckets
Sort each transformation by its direction.
Worked example
Going backwards is where the sign convention earns its keep.
\[ \text{Write the rule for the squaring function shifted left } 5 \text{ and down } 2. \]
Handle the horizontal shift first
Why: Left is the negative direction, so h is negative 5.
\[ h = -5 \]
Substitute into the standard form
Why: The form uses x minus h.
\[ (x - (-5)) ^{2} \]
Simplify the double negative
Why: Minus a negative is plus.
\[ (x + 5) ^{2} \]
Add the vertical shift outside
Why: Down 2 means subtracting 2.
\[ (x + 5) ^{2} - 2 \]
Figure (svg): The solution to Worked example write the formula from a description shown as a ladder of expressions, one row per legal move
\[ g(x)=(x+5)^2-2 \]
Verify: locate the vertex
Why: The vertex should have moved from the origin to the point where x is negative 5 and y is negative 2. Substituting x equal to negative 5 gives 0 minus 2, which is negative 2. Correct — and note that the plus 5 in the formula corresponds to a shift to the LEFT, which is the whole point of the section.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 108-110
Trap
\[ y=(x+4)^2 \;\Longrightarrow\; \text{shifted right } 4 \]
Read the plus sign inside as a positive shift
Why: A plus normally means adding, and adding normally means moving in the positive direction.
The graph is drawn with its vertex at the point where x equals 4.
Inside changes are reversed. A plus 4 inside shifts the graph LEFT 4, so the vertex sits where x is negative 4.
Check it in one substitution: the new rule outputs 0 when x plus 4 is 0, which is at x equal to negative 4. That is where the bottom of the parabola is.
When unsure, ask which input makes the inside zero. That input is where the parent's starting feature has moved to, and it settles the direction without any memorised rule.
Prediction
A function with domain from 0 to 4 is transformed by subtracting 3 inside.
Predict first
What happens to the domain and the range?
Correct: The domain becomes 3 to 7; the range is unchanged.
Why: An inside change is horizontal, so it moves the domain and leaves the outputs alone. Subtracting 3 inside shifts the graph right 3, so the domain moves from 0 through 4 to 3 through 7. The last option applies the shift in the wrong direction, which is the standard error.
Faded example
Write the absolute value rule shifted right 6 and up 1.
Fill in the blanks
g(x) = \bigl| x - 6 \bigr| + 1
Why: Right 6 is written as minus 6 inside, because inside changes are reversed. Up 1 is written as plus 1 outside, because outside changes are not. The corner of the V moves from the origin to the point where x is 6 and y is 1, which is a quick way to check the answer.
Socratic
The general shifted function is written using x minus h rather than x plus h.
Discussion prompt
What does that convention buy, given that both forms can express the same shifts?
Hint: In which form can you read the shift straight off the number?
Answer:
In the form with x minus h, the number h is the shift itself, including its sign: h equal to 3 is right 3, and h equal to negative 3 is left 3. No reversal is needed at the reading stage.
Written as x plus something, the reader has to remember to flip the sign, which is exactly the step people forget. The convention absorbs the reversal into the notation so it only has to be got right once, when the formula is written.
The same convention appears in the vertex form of a quadratic in §3.2 and in the standard forms of every conic in Chapter 10, always for this reason. It is worth getting used to now, because it recurs constantly.
Section
Section 2
Concept
Multiplying the output by a constant stretches the graph vertically by that factor. Multiplying the input by a constant compresses it horizontally by that factor — the reciprocal of what the number suggests.
\[ a\,f(x): \text{ vertical stretch by } a; \qquad f(bx): \text{ horizontal compression by } b \]
The horizontal case is the one worth thinking about rather than memorising. If the input is doubled before the rule runs, the rule reaches any given stage at half the input it used to, so the whole picture is squeezed towards the vertical axis by a factor of two. Same argument as the shift, same reversal.
Figure (svg): The squaring rule stretched vertically by a factor of three and reflected across the horizontal axis, drawn against the original
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 110-122
Picture it
Both of these act on the output, so both behave exactly as the formula reads.
Figure (svg): The squaring rule stretched vertically by a factor of three and reflected across the horizontal axis, drawn against the original
The multiplier 3 makes every height three times as large. The minus sign makes every height its own negative, which is a flip over the horizontal axis.
Worked example
Read the two multipliers separately, and reverse only the inside one.
\[ \text{Describe } g(x)=3f(2x) \text{ in terms of } f. \]
Find the outside multiplier
Why: The 3 is applied to the output.
Interpret it directly
Why: Every height is tripled.
\[ \text{vertical stretch by } 3 \]
Find the inside multiplier
Why: The 2 is applied to the input.
Interpret it in reverse
Why: The picture is squeezed by a factor of 2.
\[ \text{horizontal compression by } \frac{1}{2} \]
Figure (svg): The solution to Worked example identify the scalings shown as a ladder of expressions, one row per legal move
\[ \text{stretch vertically by } 3, \text{ compress horizontally by } \tfrac{1}{2} \]
Verify: check with one point
Why: If the graph of f passes through the point with x equal to 4 and y equal to 5, then the new rule at x equal to 2 computes 3 times f of 4, which is 15. So the point at 4 has moved to 2 and its height has tripled — halved horizontally and tripled vertically, exactly as described.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 112-116
Discrimination
The factor and its position together decide.
Sort into buckets
Sort each transformation.
Worked example
A negative sign is a scaling by negative one, so the same inside-outside rule applies.
\[ \text{Describe } g(x)=-f(-x) \text{ in terms of } f. \]
Read the outside sign
Why: The minus in front acts on the output.
Read the inside sign
Why: The minus on x acts on the input.
Combine them
Why: Reflecting in both axes in turn.
Recognise the combination
Why: Two perpendicular reflections make a half turn.
\[ \text{rotation by } 180 ^\circ \]
Figure (svg): The solution to Worked example reflections shown as a ladder of expressions, one row per legal move
\[ \text{a rotation of } 180^\circ \text{ about the origin} \]
Verify: test a point
Why: If f passes through the point with x equal to 2 and y equal to 5, the new rule at x equal to negative 2 gives the negative of f at 2, which is negative 5. So the point moved from (2, 5) to (negative 2, negative 5), which is a half turn about the origin. This combination is exactly the symmetry that defines an odd function, which the last section of this lesson takes up.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 118-122
Error analysis
A student describes the effect of doubling the input.
Annotate
On: \( g(x)=f(2x) \;\Longrightarrow\; \text{horizontal stretch by } 2 \)
The same reversal as the shifts, in multiplicative form. If you always test one point after describing a horizontal transformation, this error cannot survive.
Prediction
A function has range from negative 2 to 6, and every output is multiplied by 3.
Predict first
What is the new range?
Correct: From negative 6 to 18.
Why: An outside multiplier scales every output, endpoints included, so both ends are tripled. The domain is untouched because nothing was done to the input. Multiplying only the top endpoint, as the second option does, is the error of forgetting that a negative output also gets scaled.
Faded example
Describe the transformation taking f to the rule that halves the input and doubles the output.
Fill in the blanks
y = 2f(\tfrac22x): \text______, \text______
Why: The outside factor 2 stretches vertically by 2, read directly. The inside factor of one half is reversed into a horizontal stretch by 2, its reciprocal. Both directions stretch by the same amount here, so the graph is scaled up uniformly — the only case where an inside and an outside factor produce the same visual effect.
Explain it to yourself
An inside factor of b produces a horizontal scaling by one over b.
Discussion prompt
Explain why the reciprocal appears, using a specific input.
Hint: At which input does the new rule do what the old one did at 10?
Answer:
The new rule at an input x computes the old rule at b times x. So to make the new rule do what the old one did at 10, you need b times x to equal 10, which means x equals 10 over b.
With b equal to 2, that input is 5: the feature that was at 10 is now at 5. Everything has moved to half its former distance from the vertical axis, which is a compression by one half — the reciprocal of the 2 in the formula.
This is the same argument as the shift, in multiplicative form, and it is why the two inside rules feel alike. Whatever is done to the input, the graph does the undoing of it.
Section
Section 3
Concept
When several transformations are applied at once, the result depends on the order. The standard order matches the order the operations happen when the formula is evaluated.
There is a good reason the convention is what it is. In the standard form, the input has h subtracted and is then multiplied by b, while the output is multiplied by a and then has k added. Following the arithmetic through in that order is exactly what evaluating the formula does, so the convention is not arbitrary.
Figure (svg): Two orderings of the same pair of transformations applied to a point, showing that stretching before shifting and shifting before stretching land in different places
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 122-130
Picture it
The same stretch and the same shift, applied in the two possible orders.
Figure (svg): Two orderings of the same pair of transformations applied to a point, showing that stretching before shifting and shifting before stretching land in different places
Stretching first multiplies only the original height; shifting first means the shift gets stretched too. This is why the convention has to be stated rather than left to taste.
Worked example
Take the pieces in the standard order and track one point through.
\[ \text{Describe } g(x)=-2(x-1)^2+5 \text{ from the squaring rule.} \]
Read the inside
Why: Minus 1 inside shifts right 1.
\[ \text{right } 1 \]
Read the outside multiplier
Why: The 2 stretches heights.
\[ \text{vertical stretch by } 2 \]
Read the outside sign
Why: The minus flips over the horizontal axis.
Read the outside constant, last
Why: Plus 5 raises everything by 5.
\[ \text{up } 5 \]
Figure (svg): The solution to Worked example apply several transformations shown as a ladder of expressions, one row per legal move
\[ \text{vertex } (1,5), \text{ opening downward, twice as steep as } x^2 \]
Verify: locate the vertex directly
Why: Substituting x equal to 1 gives negative 2 times 0 plus 5, which is 5. So the vertex is at the point where x is 1 and y is 5, as predicted. Note that the plus 5 must be applied after the reflection: reflecting first and then raising gives a vertex at height 5, while raising first and then reflecting would give negative 5.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 124-127
Ranking
For the vertical work on a single formula.
Put in order
Why: Multiplying comes before adding, exactly as in evaluating the formula, so the stretch and the reflection — both multiplications — happen first, and the shift last. The stretch and the reflection can be done in either order between themselves, since multiplying by 2 and then by negative 1 gives the same result as the reverse; they are grouped together because both are multiplicative.
Worked example
The most reliable check on a combined transformation is to follow one point.
\[ \text{The graph of } f \text{ passes through } (3,4). \text{ Where is that point on } y=-2f(x-1)+5? \]
Apply the horizontal shift to the input
Why: Right 1 moves the x coordinate.
\[ x: 3 \to 4 \]
Apply the vertical stretch to the output
Why: Multiply the height by 2.
\[ y: 4 \to 8 \]
Apply the reflection
Why: Negate the height.
\[ y: 8 \to - 8 \]
Apply the vertical shift, last
Why: Add 5.
\[ y: - 8 \to - 3 \]
Figure (svg): Two orderings of the same pair of transformations applied to a point, showing that stretching before shifting and shifting before stretching land in different places
\[ (3,4) \longmapsto (4,-3) \]
Verify: confirm from the formula
Why: The new rule at x equal to 4 is negative 2 times f of 3, plus 5, which is negative 8 plus 5, giving negative 3. The two routes agree. Note the order in the vertical work: had the 5 been added before the doubling, the answer would have been negative 18 instead.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 128-130
Trap
\[ y=2f(x)+3: \; \text{shift up } 3, \text{ then stretch by } 2 \;\Longrightarrow\; \text{a point at height } 4 \text{ goes to } 14 \]
Apply the transformations in the order they are read aloud
Why: The formula mentions the 3 last, but the description was assembled shift-first.
The height 4 becomes 7 after the shift and then 14 after the stretch.
Stretch first, then shift. The height 4 doubles to 8, and then 3 is added, giving 11 rather than 14.
The formula itself settles it: evaluating 2 times f of x, plus 3, at a point where f is 4 gives 2 times 4 plus 3, which is 11. The convention is just the order of operations.
When in doubt, evaluate the formula at one point. The arithmetic decides the order, so the convention can always be reconstructed rather than recalled.
Prediction
A point at height 6 is to be doubled and raised by 4.
Predict first
How much do the two possible orders differ by?
Correct: By 4, since the shift itself gets doubled in one order.
Why: Doubling then adding gives 16; adding then doubling gives 20. The gap is 4, which is exactly the shift being multiplied by the stretch factor in the wrong order. The size of the discrepancy is always the shift times one less than the stretch factor, so it grows with both.
Fill the middle
The graph of f passes through the point where x is 2 and y is 3. Track it through the rule that triples the output and subtracts 1.
Fill in the blanks
(2,\,3) \longmapsto (2,\; 3\cdot 3 - 1) = (2,\; 8) \qquad \text8 y=3f(x)-1 \text___ x=2 \text___ ___
Why: Tripling 3 gives 9, and subtracting 1 gives 8. The x coordinate is unchanged because both transformations are outside and therefore vertical. Evaluating the formula directly gives the same 8, which is the check worth doing every time.
Edge cases
The convention says stretch before shift.
Discussion prompt
Are there stretch and shift pairs for which the order genuinely does not matter? Describe them.
Hint: What would make the shift immune to the stretching?
Answer:
Yes, in two cases. If the shift is zero there is nothing to move, so the orders trivially agree. And if the stretch factor is 1 there is no stretching, so again they agree.
There is a third: a horizontal shift combined with a vertical stretch commute, because they act on different coordinates entirely and never interfere.
So the convention only matters when a stretch and a shift act in the same direction. That observation is worth keeping, because it means the two independent groups — all the horizontal work and all the vertical work — can be done in either order relative to each other, which halves the amount of ordering that has to be tracked.
Section
Section 4
Concept
A function is even when replacing the input by its negative leaves the output unchanged, and odd when it negates the output. Each corresponds to a visible symmetry of the graph.
\[ \text{even: } f(-x)=f(x); \qquad \text{odd: } f(-x)=-f(x) \]
The names come from powers. An even power of a negative number is positive, so the rule raising to an even power is even; an odd power keeps the sign, so it is odd. Every function that is a sum of even powers is even, and every sum of odd powers is odd — which makes the test almost automatic for polynomials.
Figure (svg): Two graphs demonstrating symmetry: an even function mirrored across the vertical axis, and an odd function unchanged by a rotation of one hundred and eighty degrees about the origin
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 130-134
Picture it
Each picture shows a pair of points that the symmetry relates.
Figure (svg): Two graphs demonstrating symmetry: an even function mirrored across the vertical axis, and an odd function unchanged by a rotation of one hundred and eighty degrees about the origin
In the even case the two marked points sit at the same height. In the odd case one is as far above the axis as the other is below, and they are on opposite sides.
Worked example
Substitute negative x and compare with the original and its negative.
\[ \text{Is } f(x)=x^4-3x^2+1 \text{ even, odd, or neither?} \]
Substitute negative x
Why: Replace every x with its negative, in brackets.
\[ (-x) ^{4} - 3(-x) ^{2} + 1 \]
Simplify the even powers
Why: An even power kills the sign.
\[ x ^{4} - 3 x ^{2} + 1 \]
Compare with the original
Why: They are identical.
\[ f(-x) = f(x) \]
Conclude
Why: That is the definition of even.
Figure (svg): The solution to Worked example test a polynomial shown as a ladder of expressions, one row per legal move
\[ f(-x)=f(x), \text{ so } f \text{ is even} \]
Verify: check by inspecting the powers
Why: Every power present is even — the fourth, the second, and the constant, which is x to the zero. A polynomial whose terms all have even degree is always even, so the substitution was predictable. Had a single odd power been present, the answer would have been neither.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 131-132
Sorting
For polynomials, inspect the degrees of the terms present.
Sort into buckets
Sort each rule.
Worked example
The most common answer, and one worth being confident about.
\[ \text{Is } f(x)=x^3+x^2 \text{ even, odd, or neither?} \]
Substitute negative x
Why: In brackets, as always.
\[ (-x) ^{3} + (-x) ^{2} \]
Simplify each term
Why: The cube keeps the sign; the square kills it.
\[ -x ^{3} + x ^{2} \]
Compare with the original
Why: It is not the same, so not even.
Compare with the negative of the original
Why: The negative would be minus x cubed minus x squared.
Figure (svg): The solution to Worked example a function that is neither shown as a ladder of expressions, one row per legal move
\[ f(-x)\ne f(x) \text{ and } f(-x)\ne -f(x), \text{ so neither} \]
Verify: confirm with one input
Why: At x equal to 1 the rule gives 2, and at x equal to negative 1 it gives 0. Even would require both to be 2; odd would require the second to be negative 2. Neither holds, so a single pair of inputs settles it — which is a much faster way to rule out symmetry than the full algebra.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 132-133
Error analysis
A student tests a function, finds it is not even, and stops.
Annotate
On: \( f(-x) \ne f(x) \;\Longrightarrow\; f \text{ is odd} \)
Test both conditions separately. The two properties are special, and a function is entitled to have neither of them — which the great majority do.
Prediction
A graph is unchanged when the page is rotated by half a turn about the origin.
Predict first
What does that say about the function?
Correct: It is odd.
Why: A half turn about the origin sends the point with coordinates x and y to the point with coordinates negative x and negative y. Being unchanged by that means whenever the graph contains a point, it contains the doubly negated one, which is exactly the condition that f of negative x equals the negative of f of x.
Two truths and a lie
Two of these are true of even and odd functions and one is false.
Eliminate the wrong options
One of these claims is wrong.
Survives elimination: B
Why: B is the false claim. Even and odd are two separate conditions, not a dichotomy, and failing one says nothing about the other. Most functions fail both, so 'neither' is the ordinary case rather than an exception.
Counterexample
A classmate claims no function can be both even and odd.
Discussion prompt
Find one that is, and explain why it is the only one.
Hint: What would both conditions together require of every output?
Answer:
The zero function is both. Replacing x by negative x leaves 0 unchanged, satisfying even, and 0 is also its own negative, satisfying odd.
It is the only one. Both conditions together require f of negative x to equal both f of x and its negative, so f of x must equal its own negative, which forces every output to be zero.
So the classmate is very nearly right, and the exception is a genuine one rather than a technicality. It also shows why 'both' is a legitimate fourth answer to a symmetry question, even though it is almost never the one required.
Section
Section 5
Concept
The transformations can be read off directly only when the formula is in standard form, with the inside factored so that the input appears once, multiplied and shifted.
\[ y = a\,f\bigl(b(x-h)\bigr)+k \]
This is the step that turns a nearly-correct answer into a correct one. Written as f of 2x minus 6, the shift is not 6; factoring gives f of 2 times the quantity x minus 3, so the shift is 3. The horizontal factor has been applied to the shift as well, and the factoring is what accounts for it.
Figure (svg): A card contrasting changes made outside the function, which affect the output vertically and do what they say, with changes made inside, which affect the input horizontally and do the opposite of what they say
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 126-136
Picture it
Each of the four letters controls one transformation, and inside and outside behave differently.
Figure (svg): A card contrasting changes made outside the function, which affect the output vertically and do what they say, with changes made inside, which affect the input horizontally and do the opposite of what they say
Two of the parameters live inside and are reversed; two live outside and are not. That is the whole section in one diagram.
Worked example
The step that is skipped, and what it costs.
\[ \text{Describe } g(x)=f(2x-6) \text{ in terms of } f. \]
Resist reading the shift as 6
Why: The inside is not yet in standard form.
\[ \text{not } a\text{ shift of } 6 \]
Factor the horizontal coefficient out
Why: Take the 2 outside the bracket.
\[ f(2(x - 3)) \]
Read the factor
Why: The 2 compresses horizontally by one half.
\[ \text{compress by } \frac{1}{2} \]
Read the shift
Why: Now the bracket shows x minus 3.
\[ \text{shift right } 3 \]
Figure (svg): The solution to Worked example factor before reading the shift shown as a ladder of expressions, one row per legal move
\[ \text{compress by } \tfrac{1}{2}, \text{ then shift right } 3 \]
Verify: check where the parent's origin lands
Why: The parent does whatever it does at input 0. The new rule reaches that when 2x minus 6 is 0, which is at x equal to 3 — so the shift really is 3 and not 6. Testing the input that makes the inside vanish is the fastest possible check on any horizontal transformation.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 127-129
Faded example
Put the inside of f of the quantity 4x minus 20 into standard form.
Fill in the blanks
f(4x-20) = f\bigl(4(x-5)\bigr)
Why: Factoring 4 out of 4x minus 20 gives 4 times the quantity x minus 5. So the compression is by one quarter and the shift is right 5, not right 20. The check: 4x minus 20 vanishes at x equal to 5, which is where the parent's behaviour at 0 has moved to.
Worked example
For a quadratic, the standard form has to be produced before it can be read.
\[ \text{Describe } g(x)=x^2+6x+11 \text{ as a transformation of } x^2. \]
Take half the coefficient of x and square it
Why: Half of 6 is 3, and 3 squared is 9.
\[ \text{the magic number is } 9 \]
Add and subtract it
Why: This keeps the value unchanged.
\[ x ^{2} + 6 x + 9 - 9 + 11 \]
Group the perfect square
Why: The first three terms factor.
\[ (x + 3) ^{2} + 2 \]
Read the transformations
Why: Inside plus 3 is left 3; outside plus 2 is up 2.
\[ \text{left } 3,\text{ up } 2 \]
Figure (svg): The solution to Worked example complete the square to find the form shown as a ladder of expressions, one row per legal move
\[ g(x)=(x+3)^2+2: \text{ left } 3, \text{ up } 2, \text{ vertex } (-3,2) \]
Verify: check the vertex against the original
Why: Substituting x equal to negative 3 into the original gives 9 minus 18 plus 11, which is 2. The vertex is where predicted. Completing the square is exactly the operation that puts a quadratic into transformation-readable form, which is why §3.2 returns to it as the main technique for parabolas.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 129-132
Trap
\[ g(x)=f(3x+12) \;\Longrightarrow\; \text{shift left } 12, \text{ compress by } \tfrac{1}{3} \]
Read the constant inside as the shift
Why: The 12 is inside, so it is taken to be a horizontal shift of 12, reversed to the left.
The compression factor is read correctly, but the shift is taken straight from the constant.
Factor first. The inside becomes 3 times the quantity x plus 4, so the shift is 4, not 12.
The horizontal compression acts on the shift too, dividing it by 3. Reading the constant directly ignores that and overstates the shift by exactly the compression factor.
The check that never fails: find the input making the inside zero. Here 3x plus 12 is zero at x equal to negative 4, so the parent's behaviour at 0 has moved to negative 4 — a shift of 4 to the left.
Matching
Four formulas that look alike and describe different graphs.
Match the pairs
Why: The first and third differ only in whether the 2 has been factored through, and the shift differs accordingly — 2 against 4. The second and fourth put the same numbers outside, where no factoring question arises. Distinguishing the first from the third is exactly what this section is for.
Prediction
The inside of a transformed function is the expression 5x plus 15.
Predict first
What is the horizontal shift?
Correct: Left 3, after factoring out the 5.
Why: Factoring gives 5 times the quantity x plus 3, so the shift is 3 and the plus sign inside makes it a shift to the left. Reading 15 directly ignores the compression's effect on the shift. Checking where the inside vanishes gives x equal to negative 3, confirming the answer.
Step zero
You are given a quadratic written in expanded form and asked to describe it as a transformation of the squaring rule.
Discussion prompt
What must you do before any transformation can be read off, and why?
Hint: In what form does the input appear only once?
Answer:
Complete the square. In expanded form the input appears in two terms, and the transformations cannot be read from a formula where x appears more than once.
Completing the square rewrites it so x appears once, inside a single bracket, which is the standard form the four parameters are read from. The vertex form is not a different fact about the parabola; it is the same rule written in readable form.
This generalises. Whenever a transformation cannot be read off, the reason is almost always that the input appears in more than one place, and the fix is whatever algebra collects it into one — completing the square here, factoring the horizontal coefficient in the previous example.
Comparison
Fill the blanks from memory. Every rule in this lesson is one row of this table.
Comparison matrix
| outside the function | inside the function | |
|---|---|---|
| acts on | the output | the input |
| direction of movement | vertical | horizontal |
| behaves | as written | in reverse |
| adding a constant | moves up | moves left |
| multiplying by 2 | stretches to twice the height | compresses to half the width |
| which set it changes | the range | the domain |
The third row generates the fourth and fifth. If you remember only that inside changes are reversed, you can reconstruct every other row.
Pattern
Six steps, and the first is the one that is skipped most often.
Step 5 is worth doing every single time. It costs one substitution and catches every ordering error and every reversed sign in the section.
OpenStax Algebra and Trigonometry 2e, §3.5 Transformation of Functions §3.5
Check
Inside means reversed.
Check your understanding
The graph of the rule f(x + 7) is the graph of f moved how?
Answer: A
Why: The change is inside, so it is horizontal and reversed: adding 7 inside moves the graph left 7. Check by asking which input makes the inside zero, which is negative 7 — so the parent's behaviour at 0 has moved to the left.
Check
Factor first.
Check your understanding
For the rule f(3x - 9), what is the horizontal shift?
Answer: A
Why: Factoring gives 3 times the quantity x minus 3, so the shift is 3 and the minus sign inside makes it a shift to the right. The inside vanishes at x equal to 3, confirming it.
Check
Test both conditions.
Check your understanding
Which of these functions is odd?
Answer: A
Why: Every term has odd degree, so substituting negative x negates each term and therefore the whole function. That is exactly the odd condition, and the graph has half-turn symmetry about the origin.
Real world
Every audio equaliser, every image editor and every animation timeline is applying these transformations.
Discussion prompt
Turning up the volume on a sound wave and delaying it by half a second are two transformations of the same signal. Which is inside and which is outside, and what does that predict?
Hint: Which one changes the value of the signal, and which changes when it happens?
Answer:
Volume is an outside change: every amplitude is multiplied, so it is a vertical stretch. Turn the dial to 2 and the wave is twice as tall, exactly as the number says.
A delay is an inside change: it replaces the time input by the time minus half a second, so it shifts the wave to the right. And the sign convention bites here too — to delay a signal you SUBTRACT inside, which is the reversal this lesson is built on.
The prediction is that these two commute, since one is purely vertical and the other purely horizontal — and indeed it does not matter whether you turn up an audio track before or after delaying it. Two changes in the same direction, such as a delay and a time-stretch, do not commute, and audio software has to fix an order for exactly the reason this lesson does.
Commit first
State your confidence along with your answer.
Predict first
The graph of y equals f of the quantity negative x is the graph of f transformed how?
Correct: Reflected across the vertical axis.
Why: The negation is inside, so it is a horizontal transformation, and negating the input mirrors the graph left-to-right — a reflection in the vertical axis. Putting the minus sign outside instead would give a reflection in the horizontal axis, and doing both at once gives the half turn.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why adding a positive number inside a function moves its graph to the left, without asking them to memorise anything.
Hint: Ask which input the new rule needs in order to do what the old rule did at zero.
Answer:
The argument is one sentence: the new rule at an input x computes the old rule at x plus 3, so to get the old rule's behaviour at 0, you need x plus 3 to be 0, which means x equals negative 3.
Everything the old graph did at 0, the new graph does at negative 3, and the same holds at every other input. So every feature has moved 3 units earlier, which on a left-to-right axis is 3 units left.
A good explanation stresses that this is a compensation: the input has to be made smaller to cancel the plus 3 that is about to be added to it. That framing makes the multiplicative case follow immediately, since the input has to be made proportionally smaller to cancel a multiplication.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first, if genuinely understood, makes the second and fourth much easier, so it is often the most efficient thing to revisit even when the symptoms appeared elsewhere. The fourth is the one that separates a nearly-right answer from a right one on assessments.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw the standard form with its four parameters labelled, and mark which two are inside and which two are outside. Beside each, write what it does to the graph. Then sketch the squaring rule and, on the same axes, sketch it after a horizontal compression by one half, a reflection over the horizontal axis, and a shift up 2 — in that order — tracking the point where x is 2 all the way through.
If your tracked point ends up where a direct evaluation of the formula puts it, you have the order right, which is the half of this lesson that a table of rules will not give you.
Recap
Five things, and the first one generates most of the rest.
| if you remember one thing | it should be this |
|---|---|
| about the organising rule | inside changes are backwards, and that is the whole lesson |
| about horizontal factors | an inside factor of b compresses by one over b |
| about order | stretch and reflect first, then shift |
| about reading a formula | factor the inside before reading the shift |
Section 1.6 applies all of this to one particular toolkit function — the absolute value — whose V shape makes every transformation in this lesson visible at a glance.
OpenStax, Precalculus, §1.5 Transformation of Functions §1.5, pp. 101-136 — everything on these slides traces back here
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