Function transformations

What you will learn

describe how the graph of $y = x^2$ changes when a constant is added or subtracted,
explain the effect of multiplying by a constant (vertical stretch/compression),
recognise a reflection in the $x$ -axis from a negative coefficient,
identify horizontal translations of the form $y = (x - h)^2$ ,
connect algebraic parameters to graphical features (vertex, orientation, width).

Worked example 0 Real-world example: adjusting a fountain arc

A water fountain follows the path $y = -x^2 + 4$ . The designer wants the jet to peak $2$ m higher.

The current peak is at $y = 4$ (when $x = 0$ ).
Adding $2$ to the whole function gives $y = -x^2 + 6$ .
The new peak is at $y = 6$ , exactly $2$ m higher.
Every point on the arc has moved up by $2$ units — a vertical translation.

Key idea: adding a constant $k$ to a function shifts its entire graph up by $k$ units.

1. The parent function $y = x^2$

The simplest parabola has its vertex at the origin $(0, 0)$ , opens upward, and is symmetric about the $y$ -axis.

The parent parabola y = x squared, with vertex at the origin.

All transformations below modify this parent graph.

2. Vertical translation: $y = x^2 + k$

Vertical translation

y = x^2 + k

$k > 0$ : graph shifts up $k$ units.
$k < 0$ : graph shifts down $|k|$ units.

The vertex moves from $(0, 0)$ to $(0, k)$ . Shape and width stay the same.

Worked example 1 Shifting up and down

Sketch $y = x^2 + 3$ and $y = x^2 - 2$ relative to $y = x^2$ .

$y = x^2 + 3$ : every $y$ -value increases by $3$ . Vertex at $(0, 3)$ .
$y = x^2 - 2$ : every $y$ -value decreases by $2$ . Vertex at $(0, -2)$ .
Both parabolas have the same width and open upward.

3. Vertical stretch and compression: $y = ax^2$

Vertical stretch / compression

y = ax^2

$|a| > 1$ : graph is narrower (stretched vertically).
$0 < |a| < 1$ : graph is wider (compressed vertically).
$a = 1$ : unchanged parent graph.

The vertex remains at $(0, 0)$ ; only the steepness changes.

Worked example 2 Comparing widths

Compare $y = 3x^2$ and $y = \tfrac{1}{2}x^2$ with the parent.

At $x = 2$ : parent gives $y = 4$ ; $y = 3x^2$ gives $y = 12$ (steeper); $y = \tfrac{1}{2}x^2$ gives $y = 2$ (flatter).
$y = 3x^2$ is a narrower parabola.
$y = \tfrac{1}{2}x^2$ is a wider parabola.
Both still have vertex at $(0, 0)$ and open upward.

4. Reflection: $y = -f(x)$

Reflection in the x-axis

y = -x^2

Multiplying the function by $-1$ flips every point across the $x$ -axis. The parabola now opens downward.

Worked example 3 Reflection

Describe the graph of $y = -x^2 + 5$ .

Start with $y = x^2$ : vertex at $(0, 0)$ , opens upward.
Reflect: $y = -x^2$ opens downward, vertex still $(0, 0)$ .
Translate up $5$ : $y = -x^2 + 5$ , vertex at $(0, 5)$ , opens downward.
The graph’s maximum value is $y = 5$ .

5. Horizontal translation: $y = (x - h)^2$

Horizontal translation

y = (x - h)^2

$h > 0$ : graph shifts right $h$ units.
$h < 0$ : graph shifts left $|h|$ units.

The vertex moves from $(0, 0)$ to $(h, 0)$ .

Worked example 4 Horizontal shift

State the vertex and sketch $y = (x - 4)^2 + 1$ .

Horizontal shift: $h = 4$ , so the vertex moves $4$ units right.
Vertical shift: $k = 1$ , so the vertex moves $1$ unit up.
Vertex is at $(4, 1)$ . The parabola opens upward with the same width as $y = x^2$ .

Summary of transformations

General vertex form

y = a(x - h)^2 + k

Parameter	Effect
$a > 0$	Opens upward
$a < 0$	Opens downward (reflection)
$\lvert a \rvert > 1$	Narrower (vertical stretch)
$0 < \lvert a \rvert < 1$	Wider (vertical compression)
$h$	Horizontal shift (vertex at $x = h$ )
$k$	Vertical shift (vertex at $y = k$ )

Vertex: $(h, k)$ . Axis of symmetry: $x = h$ .

Worked example 5 Identifying all transformations

Describe the transformations that turn $y = x^2$ into $y = -2(x + 1)^2 + 3$ .

Write in standard form: $a = -2$ , $h = -1$ , $k = 3$ .
$|a| = 2 > 1$ : vertical stretch (narrower).
$a < 0$ : reflection in the $x$ -axis (opens downward).
$h = -1$ : shift $1$ unit left.
$k = 3$ : shift $3$ units up.
Vertex at $(-1, 3)$ ; the parabola opens downward and is narrower than the parent.

Practice

Fluency

Tier 1: identify the transformation

State the vertex of $y = x^2 + 7$ .
State the vertex of $y = x^2 - 4$ .
Is $y = 5x^2$ narrower or wider than $y = x^2$ ?
Is $y = 0.3x^2$ narrower or wider than $y = x^2$ ?
Does $y = -x^2$ open upward or downward?
State the vertex of $y = (x - 6)^2$ .
State the vertex of $y = (x + 2)^2 - 1$ .
Write the equation of $y = x^2$ shifted $5$ units down.
Write the equation of $y = x^2$ shifted $3$ units right and $4$ units up.

Reasoning

Tier 2: mixed practice

Match each equation to its vertex: (a) $y = (x - 1)^2 + 2$ , (b) $y = (x + 3)^2 - 5$ , (c) $y = -x^2 + 4$ . Vertices: $(0, 4)$ , $(1, 2)$ , $(-3, -5)$ .
A parabola has vertex $(2, -3)$ and opens upward with the same width as $y = x^2$ . Write its equation.
Describe two different transformations that could move the vertex of $y = x^2$ to $(0, 9)$ .
The graph of $y = ax^2$ passes through $(1, 6)$ . Find $a$ .
Arrange from widest to narrowest: $y = 4x^2$ , $y = x^2$ , $y = \tfrac{1}{4}x^2$ , $y = 2x^2$ .
A parabola opens downward, is narrower than $y = x^2$ , and has vertex at $(1, 5)$ . Write a possible equation.

Reasoning

Tier 3: explain and apply

Explain why replacing $x$ with $(x - h)$ shifts the graph to the right rather than the left.
Sam says ” $y = -(x - 2)^2$ has vertex at $(-2, 0)$ .” Identify and correct the error.
A ball’s height is modelled by $y = -5(x - 1)^2 + 8$ , where $x$ is time in seconds. State the maximum height and when it occurs.
Two parabolas have equations $y = 2(x - 3)^2 + 1$ and $y = 2(x - 3)^2 - 4$ . How are they related? What is the vertical distance between their vertices?

Challenge

Reasoning

Harder reasoning

Find the equation in vertex form of a parabola that opens downward, passes through $(0, 0)$ and $(4, 0)$ , and has a maximum value of $8$ .
The graph of $y = a(x - h)^2 + k$ passes through $(0, 5)$ and $(6, 5)$ and has a minimum value of $-4$ . Find $a$ , $h$ , and $k$ .
A parabola $y = a(x - h)^2 + k$ has vertex $(3, 7)$ and passes through $(5, -1)$ . Find the values of $a$ , $h$ , and $k$ , and state whether the parabola opens upward or downward.
Explain why the graph of $y = (x - p)(x - q)$ has its axis of symmetry at $x = \dfrac{p + q}{2}$ , and find the vertex coordinates in terms of $p$ and $q$ .

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

$(0, 7)$
$(0, -4)$
Narrower (since $5 > 1$ ).
Wider (since $0.3 < 1$ ).
Downward.
$(6, 0)$
$(-2, -1)$
$y = x^2 - 5$
$y = (x - 3)^2 + 4$

Tier 2

(a) $(1, 2)$ , (b) $(-3, -5)$ , (c) $(0, 4)$ .
$y = (x - 2)^2 - 3$ .
Method 1: vertical translation $y = x^2 + 9$ . Method 2: vertical stretch $y = 9x^2$ (passes through $(1, 9)$ but vertex is still at origin, so only the translation gives vertex $(0, 9)$ ). Accept $y = x^2 + 9$ and $y = -(x)^2 + 9$ (reflection plus shift).
$a = 6$ . Method: $6 = a \times 1^2$ .
Widest to narrowest: $y = \tfrac{1}{4}x^2$ , $y = x^2$ , $y = 2x^2$ , $y = 4x^2$ .
One possible answer: $y = -2(x - 1)^2 + 5$ . Accept any equation with $a < -1$ , $h = 1$ , $k = 5$ .

Tier 3

Replacing $x$ with $(x - h)$ means we need a larger $x$ -value to produce the same output. For example, the vertex of $y = x^2$ is at $x = 0$ ; for $y = (x - 3)^2$ the output is $0$ when $x - 3 = 0$ , i.e. $x = 3$ . Every point shifts right by $h$ .
The vertex is at $(2, 0)$ , not $(-2, 0)$ . Sam confused the sign: $(x - 2)$ means shift right $2$ .
Maximum height is $8$ m, occurring at $x = 1$ s. The vertex $(1, 8)$ gives the peak because $a = -5 < 0$ (opens downward).
Both have the same shape ( $a = 2$ ) and the same axis of symmetry ( $x = 3$ ). The second is a vertical translation of the first, shifted $5$ units down. Vertical distance between vertices: $1 - (-4) = 5$ units.

Challenge

$y = -2(x - 2)^2 + 8$ . Method: axis of symmetry at $x = \tfrac{0+4}{2} = 2$ ; vertex $(2, 8)$ ; sub $(0, 0)$ : $0 = a(0-2)^2 + 8$ , so $4a = -8$ , $a = -2$ .
$a = 1$ , $h = 3$ , $k = -4$ . Method: axis of symmetry at $x = \tfrac{0+6}{2} = 3$ ; minimum is $k = -4$ ; vertex $(3, -4)$ ; sub $(0, 5)$ : $5 = a(9) - 4$ , $9a = 9$ , $a = 1$ .
$a = -2$ , $h = 3$ , $k = 7$ . Method: vertex gives $h = 3$ , $k = 7$ ; sub $(5, -1)$ : $-1 = a(5-3)^2 + 7$ , $4a = -8$ , $a = -2$ . Opens downward since $a < 0$ .
The axis of symmetry is the midpoint of the roots $p$ and $q$ , so $x = \tfrac{p+q}{2}$ . Substituting: $y = \bigl(\tfrac{p+q}{2} - p\bigr)\bigl(\tfrac{p+q}{2} - q\bigr) = \bigl(\tfrac{q-p}{2}\bigr)\bigl(\tfrac{p-q}{2}\bigr) = -\tfrac{(p-q)^2}{4}$ . Vertex: $\bigl(\tfrac{p+q}{2},\; -\tfrac{(p-q)^2}{4}\bigr)$ .

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What you will learn

1. The parent function y=x2y = x^2y=x2

2. Vertical translation: y=x2+ky = x^2 + ky=x2+k

3. Vertical stretch and compression: y=ax2y = ax^2y=ax2

4. Reflection: y=−f(x)y = -f(x)y=−f(x)

5. Horizontal translation: y=(x−h)2y = (x - h)^2y=(x−h)2

Summary of transformations

Practice

Challenge

Answer key

Tier 1

Tier 2

Tier 3

Challenge

1. The parent function $y = x^2$

2. Vertical translation: $y = x^2 + k$

3. Vertical stretch and compression: $y = ax^2$

4. Reflection: $y = -f(x)$

5. Horizontal translation: $y = (x - h)^2$