Controls
Parent function f(x)
Readings
- Equation
- y = 2(x − 1)² − 1
- Where every point goes
- (x, y) → (x + 1, 2y − 1)
- Effect of a
- Stretched vertically ×2
- Effect of b
- No horizontal change
- Effect of h
- Moved right by 1
- Effect of k
- Moved down by 1
- Vertex
- (1, -1)
- Axis of symmetry
- x = 1
How to use this simulation
- Pick a parent function with “Parent function f(x)”: x², |x|, √x, sin x or 2ˣ. The faded grey curve is the parent y = f(x); the bold coloured curve is the transformed graph. The sliders start at y = 2(x − 1)² − 1: a = 2, h = 1, k = -1.
- Move k first. The whole graph slides up or down, nothing else changes. Then move h: it slides sideways, and the sign is the surprise, because h = 1 moves the graph to the right, not to the left.
- Move a. Large positive values stretch the graph away from the x-axis, values between 0 and 1 squash it towards the x-axis, and negative values flip it over the x-axis. Set a = 0 and the graph collapses onto the line y = k.
- Move b. It acts on the x direction and does the opposite of what you might guess: b = 2 squeezes the graph to half its width, b = 0.5 stretches it to double the width, and a negative b flips it over the y-axis. Try it on sin x to see the wave get faster.
- Watch the grey dot P walk along the parent graph while the orange dot P′, its image, walks along the new one, joined by a dashed line. That dashed line is the rule (x, y) → (x ÷ b + h, a·y + k) in action.
- Read the “Effect” boxes. Each slider has its effect written out in words, and the last two readings follow the vertex, midline or asymptote and the domain, period or range as you move.
One graph, four knobs
Think of the graph of x² as a rubber band shaped like a U. You can slide the whole band up, down, left or right. You can pull it taller or squash it flatter. You can flip it upside down. You can squeeze it narrower. All of that is still the same shape of band, just handled differently. That is exactly what function transformations are.
The beautiful part is that you do not need a new graph for every new equation. If you know the five or six basic graphs and the four knobs, you can sketch y = −2√(x + 3) + 1 or y = 3 sin(2x) − 1 in under a minute. This simulation lets you turn the knobs and see each effect separately.
The five parent functions
Each parent function is the simplest member of a family. The table gives three readings of each at x = −2, 0 and 2. A dash means the function is not defined there: √x only exists for x ≥ 0.
- x²: U-shaped, vertex at the origin, never below zero.
- |x|: V-shaped, vertex at the origin, never below zero.
- √x: half of a sideways U, starting at the origin, nothing for negative x.
- sin x: a smooth wave between −1 and 1 that repeats every 2π (about 6.28).
- 2ˣ: always positive, doubles for every step to the right, flattens towards the x-axis on the left.
| f(x) | f(−2) | f(0) | f(2) |
|---|---|---|---|
| x² | 4 | 0 | 4 |
| |x| | 2 | 0 | 2 |
| √x | — | 0 | 1.41 |
| sin x | -0.91 | 0 | 0.91 |
| 2ˣ | 0.25 | 1 | 4 |
k: moving up and down
Adding k to the output lifts every point by k. For y = x² + 3 the vertex rises from (0, 0) to (0, 3), and every y-value is 3 bigger than the parent’s. This one behaves exactly as it looks: plus means up, minus means down.
y = f(x) + kk > 0 moves the graph up, k < 0 moves it down
| x | x² | x² + 3 |
|---|---|---|
| -2 | 4 | 7 |
| -1 | 1 | 4 |
| 0 | 0 | 3 |
| 1 | 1 | 4 |
| 2 | 4 | 7 |
h: moving left and right (the sign trap)
Subtracting h from the input shifts the graph to the right by h. The graph of (x − 2)² has its vertex at x = 2: the output is zero when the input is 2. So x − 2 moves right, even though it has a minus sign. And x + 3 = x − (−3) moves left.
The cleanest way to remember it: find the input that makes the bracket zero. For (x − 2) it is x = 2, and for (x + 3) it is x = −3. That input is where the parent’s special point (the vertex, the start, the middle) lands.
y = f(x − h)h > 0 moves the graph right, h < 0 moves it left
a: stretching and flipping
Multiplying the output by a multiplies every height by a. If |a| > 1 the graph is stretched away from the x-axis. If 0 < |a| < 1 it is squashed towards it. If a is negative every height changes sign, so the graph flips over the x-axis. The points that sit on the x-axis (where y = 0) do not move at all.
y = a · f(x)|a| > 1 stretches, |a| < 1 squashes, a < 0 also reflects in the x-axis
| x | |x| | −3|x| |
|---|---|---|
| -2 | 2 | -6 |
| -1 | 1 | -3 |
| 0 | 0 | 0 |
| 1 | 1 | -3 |
| 2 | 2 | -6 |
b: squeezing and flipping sideways
Multiplying the input by b makes the function run b times faster, so the graph is compressed horizontally to 1/b of its width. With b = 2 the graph is half as wide. With b = 0.5 it is twice as wide. A negative b also reflects the graph over the y-axis.
The cleanest example is sine. sin x repeats every 6.28 (that is 2π). sin(2x) repeats every 3.14, twice as often. sin(x/2) repeats every 12.57, half as often. In general the period is 2π ÷ |b|.
y = f(b·x)|b| > 1 squeezes by 1/|b|, |b| < 1 stretches by 1/|b|, b < 0 also reflects in the y-axis
period of sin(bx) = 2π / |b|
Putting it together: the point rule
All four knobs together give y = a·f(b(x − h)) + k. Read it from the inside out: subtract h, multiply by b, apply f, multiply by a, add k. A point (x, y) on the parent graph then lands on (x ÷ b + h, a·y + k): the x-coordinate undoes b and shifts by h, the y-coordinate is multiplied by a and shifted by k.
At the start the simulation shows y = 2(x − 1)² − 1, with a = 2, b = 1, h = 1 and k = -1. The vertex moves from (0, 0) to (1, -1). Here is where five points of x² go.
(x, y) → (x/b + h, a·y + k)where a point of the parent graph ends up
| Point on x² | Lands on |
|---|---|
| (-2, 4) | (-1, 7) |
| (-1, 1) | (0, 1) |
| (0, 0) | (1, -1) |
| (1, 1) | (2, 1) |
| (2, 4) | (3, 7) |
What each parent does with the knobs
The same four knobs, read on each parent. The tracer always shows where the “important” point of the parent goes.
- x² and |x|: the vertex goes to (h, k) and the axis of symmetry is the line x = h. When a < 0 the graph opens downwards.
- √x: the starting point goes to (h, k). The graph exists for x ≥ h when b is positive, and for x ≤ h when b is negative.
- sin x: the midline is y = k, the highest and lowest points are k + |a| and k − |a|, and the period is 2π ÷ |b|.
- 2ˣ: the horizontal asymptote is y = k. The graph stays above it when a > 0 and below it when a < 0.
Try it in the simulation
Six short experiments, a minute each.
- Reset, then set k = 0 and h = 0. You now have y = 2x². Change a from 2 to 0.5 and then to −1. Which points never moved?
- Choose |x| and set h = −3. Where is the vertex? Predict before you look, using the “bracket is zero” trick.
- Choose sin x. Set b = 2, then b = 0.5. Read the period. Does it match 2π ÷ |b|?
- Choose √x and make b negative. The graph now runs to the left. What does the domain reading say?
- Choose 2ˣ and set k = −3. Where is the asymptote? Turn a negative and watch it flip below the line.
- Choose any function and set every slider to a different non-zero value. Use the point rule to predict where P′ is when P is at a clear point, then check the dashed line.
Solved problems
Eight common questions. Every number below is computed with the same functions as the simulation.
Problem 1: shift x² up by 3
The new equation is y = x² + 3. Every output is 3 bigger, so the vertex moves from (0, 0) to (0, 3). The table in the “k” section shows five points side by side.
Problem 2: where is the vertex of y = (x − 2)² and of y = (x + 3)²?
Set each bracket to zero. (x − 2) is zero at x = 2, so the first vertex is (2, 0). (x + 3) is zero at x = −3, so the second vertex is (−3, 0). Check with the table: the first graph reads 0 at x = 2 and the second reads 0 at x = −3. Both are U-shapes of the same width, just in different places.
| x | y = (x − 2)² | y = (x + 3)² |
|---|---|---|
| -4 | 36 | 1 |
| -3 | 25 | 0 |
| 0 | 4 | 9 |
| 2 | 0 | 25 |
| 3 | 1 | 36 |
Problem 3: describe y = −3|x|
Start from |x|. The factor −3 stretches it by 3 vertically and flips it over the x-axis, so the V opens downwards with its vertex still at the origin. At x = 2 the parent has height 2 and the new graph has height -6.
Problem 4: periods of y = sin(2x), y = sin(0.50x) and y = 3·sin(2x) + 1
Use period = 2π ÷ |b|. For b = 2 it is 3.14. For b = 0.5 it is 12.57. The parent has period 6.28.
For y = 3·sin(2x) + 1: amplitude 3, period 3.14, midline y = 1. The highest point is 1 + 3 = 4, the lowest is 1 − 3 = -2.
Problem 5: sketch y = −3√(x − 1) + 4
The starting point of √x moves to (1, 4). The factor −3 stretches by 3 and flips, so the curve runs down and to the right from (1, 4). Nothing exists for x < 1. The table checks it.
| x | y = −3√(x − 1) + 4 |
|---|---|
| 0 | — |
| 1 | 4 |
| 2 | 1 |
| 5 | -2 |
| 10 | -5 |
Problem 6: intercepts and asymptote of y = 2^(x + 1) − 4
The asymptote is y = −4: the graph climbs away from it on the right and settles towards it on the left. At x = −10 the height is -4, almost −4.
The y-intercept (x = 0) is -2. For the x-intercept set 2^(x + 1) = 4, so x + 1 = 2 and x = 1. Check: at x = 1 the graph reads 0.
Problem 7: a parabola with vertex (2, −3) through (4, 5)
Write y = a(x − 2)² − 3. Substitute the point: 5 = a(4 − 2)² − 3, so 8 = a × 4 and a = 2. The equation is y = 2(x − 2)² − 3.
Check: at x = 4 it gives 5. By symmetry it must also give 5 at x = 0, and it does (5).
Problem 8: does the order of moves matter?
Yes. “Double, then add 3” gives y = 2x² + 3; at x = 1 it reads 5. “Add 3, then double” gives y = 2(x² + 3); at x = 1 it reads 8. In y = a·f(x) + k, the stretch by a is applied first and the shift by k second.
Inside the brackets, factor before you read h. (2x − 3)² is (2(x − 1.50))², so it shifts by 1.50, not by 3. By contrast (2x − 6)² = (2(x − 3))² shifts by 3.
Common mistakes
These cost marks again and again.
- Reading (x − 2) as “move left 2”. It moves right 2.
- Reading f(2x) as “twice as wide”. It is half as wide. The x-direction moves are reciprocals.
- Reading 2(x − 3) and (2x − 3) as the same thing. Factor the b out first, then read h.
- Forgetting that a negative a flips the graph over the x-axis, and a negative b flips it over the y-axis. For x² and |x|, a flip in the y-axis is invisible because they are symmetric.
- Applying shifts and stretches in the wrong order. The outside is stretch (a) first, then shift (k).
- Moving the asymptote or midline by a instead of by k. Only k moves them.
Where transformations show up in real life
The same four knobs appear far outside the classroom.
- Sound: a note is a sine wave. a is the volume (amplitude), b is the pitch (frequency), h is where in the cycle it starts, and k shifts the whole signal up.
- Graphics and games: every sprite on screen is the same shape moved (h, k), scaled (a, b) or flipped (negative a or b).
- Finance: 2ˣ is compound growth. A negative a describes decay of a debt, and k moves the baseline.
- Physics: a mass on a spring follows a·sin(bt) + k. Tides and daylight hours are the same shape with different knobs.
- Data science: normalising data is a transformation, subtract the mean (h) and divide by the spread (b).
Exam corner
Transformation questions appear in IGCSE, A-level, SAT, ACT, JEE and HSC higher mathematics. The patterns are predictable.
- Describe the transformation from y = f(x) to y = f(x − 3) + 2 in words (right 3, up 2).
- Given a graph and a few points, write the equation.
- Apply a sequence of moves to a table of points with the point rule.
- State the new vertex, asymptote, domain, range or period.
- Multiple choice: spot which of four equations matches a described graph.
Revision summary
Four letters, four moves:
k: up (k > 0) or down (k < 0)
h: right (h > 0) or left (h < 0)
a: vertical stretch by |a|, flip over the x-axis if a < 0
b: horizontal squeeze by 1/|b|, flip over the y-axis if b < 0
(x, y) → (x/b + h, a·y + k)
Frequently asked questions
What is a function transformation?
A transformation changes a function’s graph by shifting, stretching, squeezing or reflecting it. The general form is y = a·f(b(x − h)) + k, where f is the parent function and a, b, h and k are the four knobs.
Why does f(x − 2) move the graph to the right?
Because the output now equals f’s value at an input 2 smaller. The special point that used to happen at x = 0 now happens when x − 2 = 0, which is x = 2. The graph has moved right.
What does a negative a do?
It flips the graph over the x-axis, because every height changes sign. A negative b flips it over the y-axis instead.
How do I find the period after a transformation of sine?
For y = a·sin(b(x − h)) + k the period is 2π ÷ |b|. The amplitude is |a| and the midline is y = k. h and k do not change the period.
In what order do I apply transformations?
Work from the inside out: shift by h, scale by b, apply f, scale by a, then shift by k. With the equation in factored form this order is automatic.
How do I read the vertex from a transformed parabola?
Write the equation as y = a(x − h)² + k. The vertex is (h, k). If the equation is not in that form, complete the square first.
Do horizontal shifts and stretches change the range?
No. The range depends only on a and k, because they act on the outputs. h and b act on the inputs, so they change the domain and the position, but not the set of y-values.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
Textbook
Class 9-10 Mathematics textbook
SSC
SSC Higher Mathematics: syllabus and preparation
HSC
HSC Higher Mathematics: syllabus and preparation
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SSC Mathematics: syllabus and preparation
Textbook
Class 9-10 Higher Mathematics textbook
Admission
Engineering admission test preparation
Admission
Jahangirnagar University mathematical and physical sciences unit
English
Paragraph: Artificial Intelligence
Bangla
রচনা: কৃত্রিম বুদ্ধিমত্তা (in Bangla)
SSC
SSC suggestions
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HSC suggestions
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SSC
SSC syllabus
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HSC syllabus
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Digital SAT Math guide
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Scholarships to study abroad
Tool
Compound interest calculator
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