Open Math›Year 11›Mathematics Advanced›Graph transformations›Reflections and translations

Reflections and translations

Apply reflections in the x- and y-axes and horizontal and vertical translations to functions; determine transformation parameters from equations written in the form

y - b = f(x - a)

Worked examples

Reflecting a function in the y-axis

Straightforward

Problem

The graph of

y = x^3 - 2

is reflected in the

y

-axis. Write the equation of the new graph.

Recall the rule for reflection in the

y

-axis: replace every

x

in the equation with

-x

\text{Replace } x \text{ with } -x: \quad y = (-x)^3 - 2

Simplify

(-x)^3

using the rule

(-x)^n = -x^n

for odd

n

(-x)^3 = -x^3

Write the final equation.

y = -x^3 - 2

Answer

y = -x^3 - 2

Identifying a translation from the equation $y - b = f(x - a)$

Moderate

Problem

The function

f(x) = x^2

is transformed to give

y + 5 = f(x - 3)

. Write the equation of the new graph, state the coordinates of the vertex, and state the domain and range.

Rearrange into the standard form

y = f(x - a) + b

by isolating

y

y = f(x - 3) - 5 = (x - 3)^2 - 5

Read off the translations:

a = 3

(right by 3) and

b = -5

(down by 5).

\text{The vertex of } y = x^2 \text{ is } (0, 0). \text{ It moves to } (0 + 3,\; 0 - 5) = (3, -5).

State the domain. A parabola is defined for all real

x

\text{Domain: } x \in \mathbb{R}

State the range. The parabola opens upward with vertex

y

-value

-5

, so

y

cannot go below

-5

\text{Range: } y \geq -5

Answer

y = (x - 3)^2 - 5

; vertex

(3, -5)

; domain

x \in \mathbb{R}

; range

y \geq -5

Applying a sequence of two transformations

Challenging

Problem

Starting from

y = \sqrt{x}

, apply (i) a reflection in the

x

-axis, then (ii) a translation of 4 units to the right. Write the equation of the resulting graph and state its domain and range.

Apply the reflection in the

x

-axis first. Replace

y

with

-y

, which is equivalent to multiplying the right-hand side by

-1

y = \sqrt{x} \xrightarrow{\text{reflect in } x\text{-axis}} y = -\sqrt{x}

Apply the translation of 4 units to the right to the result from Step 1. Replace

x

with

(x - 4)

y = -\sqrt{x} \xrightarrow{\text{translate right 4}} y = -\sqrt{x - 4}

State the domain. The expression under the square root must be non-negative.

x - 4 \geq 0 \implies x \geq 4

State the range. Since

\sqrt{x - 4} \geq 0

, multiplying by

-1

gives

y \leq 0

\text{Range: } y \leq 0

Answer

y = -\sqrt{x - 4}

; domain

x \geq 4

; range

y \leq 0

Practise

Q1·Straightforward

The graph of

y = x^3

is reflected in the

x

-axis. Which equation represents the new graph?

$y = x^3$$y = -x^3$$y = (-x)^3$$y = x^3 + 1$

Explanation

Reflecting in the

x

-axis multiplies the output by

-1

: if the original point is

(x, y)

, the reflected point is

(x, -y)

. Replacing

y

with

-y

y = x^3

gives

-y = x^3

, which rearranges to

y = -x^3

Q2·Straightforward

The graph of

y = \sqrt{x}

is reflected in the

y

-axis. Which equation represents the new graph?

$y = \sqrt{x}$$y = -\sqrt{x}$$y = \sqrt{-x}$$y = -\sqrt{-x}$

Explanation

Reflecting in the

y

-axis maps

(x, y)

(-x, y)

, so replace

x

with

-x

in the equation. Substituting into

y = \sqrt{x}

gives

y = \sqrt{-x}

. This graph exists for

x \leq 0

, which is the mirror image of

y = \sqrt{x}

Q3·Straightforward

The graph of

y = x^2

is translated 5 units to the right. Which equation represents the new graph?

$y = x^2 + 5$$y = x^2 - 5$$y = (x + 5)^2$$y = (x - 5)^2$

Explanation

To translate a graph

a

units to the right, replace

x

with

(x - a)

. Here

a = 5

, so

y = x^2

becomes

y = (x - 5)^2

. Note that shifting right uses subtraction inside the bracket, which is a common point of confusion.

Q4·Straightforward

The graph of

y = x^2

is translated 7 units upward. The new graph has its vertex at

(0, k)

. Find

k

Explanation

Translating

y = x^2

upward by 7 units gives

y = x^2 + 7

. The vertex of this parabola is at

(0, 7)

, so

k = 7

Q5·Moderate

The function

f(x) = x^2

is transformed to give

y - 3 = f(x - 5)

. What is the

x

-coordinate of the vertex of the resulting parabola?

Explanation

Rearranging:

y = f(x - 5) + 3 = (x - 5)^2 + 3

. This is

y = x^2

translated 5 units right and 3 units up. The vertex moves from

(0, 0)

(5, 3)

, so the

x

-coordinate of the vertex is

5

Q6·Moderate

The function

f(x) = x^2

is transformed to give

y + 6 = f(x - 1)

. What is the

y

-coordinate of the vertex of the resulting parabola?

Explanation

Rearranging:

y = f(x - 1) - 6 = (x - 1)^2 - 6

. The vertex is at

(1, -6)

, so the

y

-coordinate is

-6

. Adding a positive number on the left side of

y + 6 = \ldots

shifts the graph downward.

Q7·Moderate

The graph of

f(x) = \sqrt{x}

(domain

x \geq 0

) is transformed to

y = f(x - 4)

. At what value of

x

does the new graph have its

x

-intercept?

Explanation

The new graph is

y = \sqrt{x - 4}

, with domain

x \geq 4

. Setting

y = 0

\sqrt{x - 4} = 0 \implies x - 4 = 0 \implies x = 4

. The

x

-intercept is at

(4, 0)

, so

x = 4

Q8·Moderate

The graph of

y = f(x)

has a turning point at

(3, 7)

. After the transformation

y - 2 = f(x + 4)

, what are the coordinates of the new turning point?

Explanation

Rearranging:

y = f(x + 4) + 2 = f(x - (-4)) + 2

. This is a translation of 4 units to the left and 2 units upward. The turning point

(3, 7)

maps to

(3 - 4,\; 7 + 2) = (-1, 9)

Q9·Challenging

The graph of

y = x^2

is first reflected in the

x

-axis, then translated 5 units upward. What is the

y

-coordinate of the vertex of the resulting parabola?

Explanation

Step 1 — reflect in the

x

-axis:

y = x^2

becomes

y = -x^2

, with vertex at

(0, 0)

. Step 2 — translate 5 units upward:

y = -x^2 + 5

, with vertex at

(0, 5)

. The

y

-coordinate of the vertex is

5

Q10·Challenging

The graph of

y = \sqrt{x}

is translated 9 units to the right, then reflected in the

x

-axis. The resulting graph passes through the point

(13, k)

. Find

k

Explanation

Step 1 — translate 9 units right: replace

x

with

(x - 9)

to get

y = \sqrt{x - 9}

. Step 2 — reflect in the

x

-axis:

y = -\sqrt{x - 9}

. Substituting

x = 13

y = -\sqrt{13 - 9} = -\sqrt{4} = -2

. So

k = -2

Q11·Challenging

The function

f(x) = \dfrac{1}{x}

is translated 3 units to the left and then 4 units upward. What is the

y

-value of the horizontal asymptote of the resulting graph?

Explanation

Step 1 — translate 3 left: replace

x

with

(x + 3)

to get

y = \dfrac{1}{x + 3}

. Step 2 — translate 4 up:

y = \dfrac{1}{x + 3} + 4

. As

x \to \pm\infty

\dfrac{1}{x+3} \to 0

, so

y \to 4

. The horizontal asymptote is

y = 4

Q12·Challenging

The graph of

y = \sqrt{x}

is reflected in the

y

-axis to give graph

A

. Graph

A

is then translated 6 units to the left. The resulting graph passes through the point

(-7, k)

. Find

k

Explanation

Step 1 — reflect in the

y

-axis:

y = \sqrt{-x}

(domain

x \leq 0

). Step 2 — translate 6 units left: replace

x

with

(x + 6)

to get

y = \sqrt{-(x + 6)} = \sqrt{-x - 6}

(domain

x \leq -6

). Substituting

x = -7

y = \sqrt{-(-7) - 6} = \sqrt{7 - 6} = \sqrt{1} = 1

. So

k = 1