Transformations of Graphs

Where a point moves under f(x) + a, f(x + a), af(x), f(ax), −f(x) and f(−x), and finding the new equation.

In the exam for Edexcel IGCSE Maths 4MA1 · 22 real past-paper questions below

What a Graph Transformation Is

$y = f (x)$ is any curve. The exam changes the rule and asks where a point, like the top of the curve, moves to.

Rule

A change outside the bracket changes the y values: up, down, taller. A change inside the bracket changes the x values, and works the opposite way to how it looks.

Example

On every card here the curve has its highest point at $(4, 7)$ . Watch where that one point goes.

Key idea

You never need the curve's equation. Move the point with the rule, and the whole curve moves with it.

Translations: Up, Down, Left, Right

Adding a number slides the curve without changing its shape. Exams ask where the top or bottom point goes. Here the top of $y = f (x)$ is $(4, 7)$ .

Formula

f (x) + a : (x, y + a)

f (x + a) : (x - a, y)

Worked example

1
$f (x) - 3$ — The −3 is outside the bracket, so it changes the y value. The x value stays at 4.
2
$y : 7 - 3 = 4$ — Do to y what the number says: take 3 off 7.
3
$(4, 7) \to (4, 4)$ — Write the new point. It has dropped 3.
4
$f (x + 2)$ — The +2 is inside the bracket, so it changes the x value, and in the opposite way.
5
$x + 2 = 4 \Rightarrow x = 2$ — The new top is where the bracket equals the old x value, 4. Take 2 from both sides.
6
$(4, 7) \to (2, 7)$ — x is now 2 and y has not changed, so the point has moved 2 to the left.
7
$f (x - 5)$ — The −5 is inside the bracket, so again it changes x the opposite way.
8
$x - 5 = 4 \Rightarrow x = 4 + 5 = 9$ — Make the bracket equal 4, then add 5 to both sides to undo the −5.
9
$(4, 7) \to (9, 7)$ — x is now 9, so the point has moved 5 to the right.

A negative coordinate needs care. Keep it in brackets. Here the minimum of $y = f (x)$ is $(- 9, 15)$ and the curve becomes $y = f (x + 3)$ .

Worked example

1
$f (x + 3)$ — The +3 is inside the bracket, so it changes x the opposite way.
2
$x + 3 = (- 9)$ — The bracket must equal the old x value. Write −9 in brackets so the sign is not lost.
3
$x = (- 9) - 3 = - 12$ — Take 3 from both sides. −9 − 3 = −12, so the point moves 3 to the left.
4
$(- 9, 15) \to (- 12, 15)$ — y does not change, because the +3 is inside the bracket.

Common mistake

Moving $f (x + 2)$ to the right. The top is where the bracket equals what it did before: $x + 2 = 4$ gives $x = 2$ , so it moves left.

Stretches

Multiplying stretches the curve. Outside, the y values are multiplied. Inside, the x values are divided.

Formula

a f (x) : (x, a y)

f (a x) : (\frac{x}{a}, y)

Worked example

1
$2 f (x)$ — The 2 is outside the bracket, so it multiplies the y value. The x value stays at 4.
2
$y : 7 \times 2 = 14$ — Double the y value.
3
$(4, 7) \to (4, 14)$ — Write the new point. It is twice as high.
4
$f (2 x)$ — The 2 is inside the bracket, so it acts on the x value, the opposite way.
5
$2 x = 4 \Rightarrow x = 4 \div 2 = 2$ — Make the bracket equal the old x, 4, then divide both sides by 2. The x value is halved.
6
$(4, 7) \to (2, 7)$ — y does not change. The curve is squeezed towards the y-axis.
7
$f (\frac{x}{2})$ — Dividing by 2 inside is the other way round.
8
$\frac{x}{2} = 4 \Rightarrow x = 4 \times 2 = 8$ — Make the bracket equal 4, then multiply both sides by 2. The x value is doubled.
9
$(4, 7) \to (8, 7)$ — The curve is stretched away from the y-axis.
10
$\frac{1}{2} f (x) : y = 7 \times \frac{1}{2} = 3.5$ — A fraction outside the bracket shrinks y. The point is now (4, 3.5).

Common mistake

Doubling the x value for $f (2 x)$ . Inside works the opposite way: the curve is squashed towards the y-axis.

Reflections

A minus sign reflects the curve. Outside, it flips upside down. Inside, it flips left to right.

Formula

- f (x) : (x, - y)

f (- x) : (- x, y)

Worked example

1
$- f (x)$ — The minus is outside the bracket, so it changes the sign of the y value.
2
$y : - (7) = - 7$ — Put the old y in brackets and change its sign.
3
$(4, 7) \to (4, - 7)$ — The point flips upside down: a reflection in the x-axis.
4
$- f (x), minimum (6, - 3)$ — Now a point with a negative y value.
5
$y : - (- 3) = 3$ — A minus times a minus is a plus, so the minimum becomes a maximum at (6, 3).
6
$f (- x)$ — The minus is inside the bracket, so it changes the x value.
7
$- x = 4 \Rightarrow x = - 4$ — Make the bracket equal the old x, 4. Multiply both sides by −1.
8
$(4, 7) \to (- 4, 7)$ — y does not change: a reflection in the y-axis.

Key idea

“Describe fully” wants the mirror line too: reflection in the x-axis (the line y = 0), or in the y-axis (x = 0).

Finding the New Equation

Sometimes you are given $f (x)$ and asked for the equation after a move. Replace every x with the new bracket, then tidy up.

Worked example

1
$f (x) = x^{2} + 3 x$ — The rule. We want y = f(x − 2), so every x must become x − 2.
2
$f (x - 2) = (x - 2)^{2} + 3 (x - 2)$ — Replace each x with (x − 2). Keep the brackets so the 2 goes with the whole x term.
3
$(x - 2)^{2} = (x - 2) (x - 2)$ — Squaring means multiplying the bracket by itself.
4
$= x^{2} - 2 x - 2 x + 4$ — Multiply each term by each term: x·x, x·(−2), (−2)·x and (−2)·(−2) = +4.
5
$= x^{2} - 4 x + 4$ — Collect the two x terms: −2x − 2x = −4x.
6
$3 (x - 2) = 3 x - 6$ — Multiply both terms in the bracket by 3, the −2 included.
7
$x^{2} - 4 x + 4 + 3 x - 6$ — Put the two expanded parts together.
8
$x^{2} + (- 4 x + 3 x) + (4 - 6)$ — Group the like terms: the x terms and the numbers.
9
$= x^{2} - x - 2$ — −4x + 3x = −x and 4 − 6 = −2.

Example

Backwards: $y = x^{2} - 6 x + 10 = (x - 3)^{2} + 1$ is $y = x^{2}$ moved 3 right and 1 up, a translation by the vector $(1 3)$ .

New Equation: Stretches, Reflections

Exams also ask for the equation after a stretch or a reflection. Take $f (x) = x^{2} + 3 x$ . The number outside multiplies the whole rule. A number inside replaces each x.

Worked example

1
$2 f (x) = 2 (x^{2} + 3 x)$ — The 2 is outside, so it multiplies the whole rule, not just the first term.
2
$= 2 x^{2} + 6 x$ — Multiply each term in the bracket by 2.
3
$- f (x) = - (x^{2} + 3 x)$ — A minus outside multiplies the whole rule by −1.
4
$= - x^{2} - 3 x$ — Every term changes sign, so +3x becomes −3x.
5
$f (2 x) = (2 x)^{2} + 3 (2 x)$ — Replace each x with 2x. The brackets keep the 2 with the x.
6
$= 4 x^{2} + 6 x$ — (2x)² = 4x² because the 2 is squared too, and 3 × 2x = 6x.
7
$f (- x) = (- x)^{2} + 3 (- x)$ — Replace each x with −x. Brackets matter here.
8
$= x^{2} - 3 x$ — (−x)² = +x², because a negative times a negative is positive, and 3 × (−x) = −3x.

Common mistake

Writing $- f (x)$ as $- x^{2} + 3 x$ . The minus applies to every term, so the 3x changes sign too.

Combined Moves and Sketching

Real questions often join moves together. The x value and the y value are handled separately, so work them one at a time.

Worked example

1
$y = 3 f (x + 2) - 5, top (4, 7)$ — Inside is +2, so x changes. Outside are × 3 and − 5, so y changes.
2
$x + 2 = 4 \Rightarrow x = 2$ — The bracket equals the old x. Take 2 from both sides.
3
$y : 3 \times 7 = 21$ — Do the multiplying first, just as the rule is written: 3f(x) comes before the − 5.
4
$21 - 5 = 16$ — Now subtract 5.
5
$(4, 7) \to (2, 16)$ — Put the new x and y together.
6
$y = - f (x + 3), top (4, 7)$ — A reflection and a translation together.
7
$x + 3 = 4 \Rightarrow x = 1$ — Inside +3 moves the x value left 3.
8
$y : - (7) = - 7$ — The minus outside changes the sign of y.
9
$(4, 7) \to (1, - 7)$ — Put the new x and y together.

To draw a new graph, move the key points, then join them the same way. Key points are corners, turning points and where the graph meets the axes. Here $y = f (x)$ is straight lines through $(- 1, 0)$ , $(1, 4)$ and $(5, 0)$ .

Worked example

1
$y = 2 f (x) : (x, y) \to (x, 2 y)$ — A vertical stretch of scale factor 2 doubles every y value. x stays the same.
2
$(- 1, 0) \to (- 1, 0), (5, 0) \to (5, 0)$ — 2 × 0 = 0, so the points on the x-axis do not move.
3
$(1, 4) \to (1, 8)$ — The top corner is twice as high.
4
$y = f (- x) : (x, y) \to (- x, y)$ — A reflection in the y-axis changes the sign of every x value.
5
$(- 1, 0) \to (1, 0), (1, 4) \to (- 1, 4), (5, 0) \to (- 5, 0)$ — Change the sign of each x and keep y. Plot the three points and join them with straight lines.

Key idea

In a vertical stretch the points on the x-axis stay where they are. In a horizontal stretch the points on the y-axis stay where they are.

Describe Fully, Completed Square

“Describe fully” needs the right words: the type of move, then its size or direction. A translation needs a vector. A stretch needs a scale factor and the axis it is parallel to. A reflection needs its mirror line.

You see	Describe fully
$f (x) + a$	translation by the vector $(a 0)$
$f (x + a)$	translation by the vector $(0 - a)$
$a f (x)$	stretch, scale factor $a$ , parallel to the y-axis
$f (a x)$	stretch, scale factor $\frac{1}{a}$ , parallel to the x-axis
$- f (x)$	reflection in the x-axis
$f (- x)$	reflection in the y-axis

To spot the moves from an equation, complete the square. Describe fully the moves that take $y = x^{2}$ to $y = 3 x^{2} + 12 x + 5$ .

Worked example

1
$3 x^{2} + 12 x + 5 = 3 (x^{2} + 4 x) + 5$ — Take the 3 out of the x² and x terms only. The 5 stays outside.
2
$x^{2} + 4 x = (x + 2)^{2} - 4$ — Complete the square inside: half of 4 is 2, and take away 2² = 4 to put back what was added.
3
$3 [(x + 2)^{2} - 4] + 5$ — Put that back inside the 3[ ].
4
$= 3 (x + 2)^{2} - 12 + 5$ — Multiply the −4 by the 3 outside.
5
$= 3 (x + 2)^{2} - 7$ — Collect the numbers: −12 + 5 = −7.
6
$y = 3 f (x + 2) - 7, f (x) = x^{2}$ — Compare with f(x) = x². The 3 multiplies y, the +2 is inside, and the −7 is added to y.
7
$stretch, scale factor 3, parallel to the y-axis$ — The 3 multiplies every y value. Do this first, because the −7 is added after the multiplying.
8
$translation by (- 7 - 2)$ — The +2 inside moves the curve 2 left, and the −7 moves it 7 down.

Common mistake

Writing just “translation” or just “stretch”. Without the vector, or the scale factor and the axis, the answer scores no marks.

Which Method? Read the Question

Find where the number is: inside or outside the bracket.

You see	The point
$f (x) + a$	y + a: up a
$f (x + a)$	x − a: left a
$a f (x)$	y × a
$f (a x)$	x ÷ a
$- f (x)$	y becomes −y
$f (- x)$	x becomes −x
Equation of the new curve	put the new bracket for every x, or multiply the whole rule
Describe fully	translation vector, or scale factor and axis, or mirror line

Key idea

Outside changes y, the way it looks. Inside changes x, the opposite way.

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