IB Mathematics: Analysis and Approaches Quiz: Function Transformations

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Function TransformationsQuestion 1 of 20

Let $f(x) = x^2$ . The graph of $f(x)$ is translated 2 units to the left and 3 units down to obtain the graph of $g(x)$ . Then, the graph of $g(x)$ is reflected in the x-axis to obtain the graph of $h(x)$ . Find the equation of $h(x)$ .

h(x) = (-x+2)^2 + 3

h(x) = -(x-2)^2 - 3

h(x) = -(x+2)^2 + 3

h(x) = -(x+2)^2 - 3

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Function Transformations

Practice Function Transformations in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Function Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$h(x) = (-x+2)^2 + 3$
$h(x) = -(x-2)^2 - 3$
$h(x) = -(x+2)^2 + 3$ (correct answer)
$h(x) = -(x+2)^2 - 3$

Explanation: When you encounter function transformations, work systematically through each transformation step by step, applying them to the original function in the order given. Starting with

f(x) = x^2

, let's find

g(x)

first. Translating 2 units left means replacing

x

with

(x+2)

, and translating 3 units down means subtracting 3 from the entire function. Therefore:

g(x) = (x+2)^2 - 3

Next, reflecting

g(x)

in the x-axis means multiplying the entire function by -1:

h(x) = -g(x) = -[(x+2)^2 - 3] = -(x+2)^2 + 3

Now let's examine why each answer choice is wrong or right: Choice A,

h(x) = (-x+2)^2 + 3

, incorrectly applies the horizontal translation. This represents a translation 2 units to the right, not left, and doesn't include the reflection properly. Choice B,

h(x) = -(x-2)^2 - 3

, has two errors: it translates right instead of left (using

x-2

instead of

x+2

), and it moves down instead of reflecting the vertical translation correctly after the x-axis reflection. Choice C,

h(x) = -(x+2)^2 + 3

, correctly shows the left translation with

(x+2)

, the downward translation of the original function, and the x-axis reflection that flips the sign of the entire expression. Choice D,

h(x) = -(x+2)^2 - 3

, correctly handles the horizontal translation and reflection but fails to account for how the x-axis reflection affects the vertical translation. Study tip: Remember that x-axis reflections flip the sign of the entire function, which affects how vertical translations appear in the final equation.

Question 2

The function $y=f(x)$ undergoes a sequence of transformations to become $g(x) = 2f(x+3) - 5$ . A point on the graph of $y=g(x)$ is $(1, 9)$ . Find the coordinates of the corresponding point on the graph of $y=f(x)$ .

$(-2, 7)$
$(4, 2)$
$(4, 7)$ (correct answer)
$(4, 13)$

Explanation: Let the original point on

y=f(x)

(x_0, y_0)

. The transformation

g(x) = 2f(x+3) - 5

maps this point to a new point

(x_{new}, y_{new})

. The horizontal transformation is

x \to x-3

and the vertical transformation is

y \to 2y-5

. Therefore,

x_{new} = x_0 - 3

and

y_{new} = 2y_0 - 5

. We are given the new point is

(1, 9)

. So, we need to solve for

(x_0, y_0)

. For the x-coordinate:

1 = x_0 - 3 \implies x_0 = 4

. For the y-coordinate:

9 = 2y_0 - 5 \implies 14 = 2y_0 \implies y_0 = 7

. The corresponding point on the graph of

y=f(x)

(4, 7)

Question 3

The graph of $y=f(x)$ is reflected in the x-axis, and then translated horizontally by 4 units to the left. The resulting graph is given by the equation $y=g(x)$ . Which equation correctly expresses $f(x)$ in terms of $g(x)$ ?

$f(x) = -g(x+4)$
$f(x) = -g(x-4)$ (correct answer)
$f(x) = g(-x-4)$
$f(x) = -g(x)+4$

Explanation: Function transformations in IB Mathematics require you to think carefully about the order of operations and how they affect the relationship between functions. When working backwards from a transformed function to find the original, you need to reverse each transformation step by step. Let's trace through what happens to

f(x)

to become

g(x)

. First,

f(x)

is reflected in the x-axis, giving us

-f(x)

. Then this reflected function is translated 4 units left, which means replacing

x

with

(x+4)

, resulting in

g(x) = -f(x+4)

. To find

f(x)

in terms of

g(x)

, we reverse these steps. From

g(x) = -f(x+4)

, we first undo the reflection by multiplying both sides by

-1

-g(x) = f(x+4)

. Then we undo the horizontal translation by replacing

x

with

(x-4)

f(x) = -g(x-4)

. This confirms answer B is correct. Looking at the wrong answers: A gives

f(x) = -g(x+4)

, which would represent translating right instead of left when reversing the transformation. C gives

f(x) = g(-x-4)

, incorrectly introducing a reflection about the y-axis that wasn't in the original problem. D gives

f(x) = -g(x)+4

, which confuses horizontal translation with vertical translation and gets the algebraic relationship completely wrong. Remember: when reversing transformations, work backwards through each step. Horizontal shifts of "left

a

" become "right

a

" when reversed, and reflections are their own inverse operations.

Question 4

$(-2, 7)$
$(4, 2)$
$(4, 7)$ (correct answer)
$(4, 13)$

Explanation: Let the original point on

y=f(x)

(x_0, y_0)

. The transformation

g(x) = 2f(x+3) - 5

maps this point to a new point

(x_{new}, y_{new})

. The horizontal transformation is

x \to x-3

and the vertical transformation is

y \to 2y-5

. Therefore,

x_{new} = x_0 - 3

and

y_{new} = 2y_0 - 5

. We are given the new point is

(1, 9)

. So, we need to solve for

(x_0, y_0)

. For the x-coordinate:

1 = x_0 - 3 \implies x_0 = 4

. For the y-coordinate:

9 = 2y_0 - 5 \implies 14 = 2y_0 \implies y_0 = 7

. The corresponding point on the graph of

y=f(x)

(4, 7)

Question 5

The graph of a function $f(x)$ is reflected in the y-axis, then stretched vertically by a factor of 3, and finally translated 2 units down. Which of the following equations represents the final transformed function, $g(x)$ ?

$g(x) = 3f(-x) - 2$ (correct answer)
$g(x) = 3(f(-x) - 2)$
$g(x) = 3f(-x-2)$
$g(x) = f(-3x) - 2$

Explanation: We apply the transformations step-by-step to

y=f(x)

Reflection in the y-axis: This transformation replaces $x$ with $-x$ . The function becomes $y = f(-x)$ .
Stretched vertically by a factor of 3: This transformation multiplies the entire function by 3. The function becomes $y = 3f(-x)$ .
Translated 2 units down: This transformation subtracts 2 from the entire function. The function becomes $y = 3f(-x) - 2$ . Therefore, the final equation is $g(x) = 3f(-x) - 2$ . Distractor B represents translating down before stretching, which would result in $g(x) = 3f(-x) - 6$ .

Question 6

The equation $f(x)=0$ has roots at $x=2$ and $x=-5$ . What are the roots of the equation $f(2x-1)=0$ ?

$x=-2, x=\frac{3}{2}$ (correct answer)
$x=3, x=-11$
$x=-\frac{3}{2}, x=2$
$x=\frac{1}{2}, x=-3$

Explanation: The roots of

f(x)=0

are the values of the input to

f

that make the output zero. For the equation

f(2x-1)=0

, the output will be zero when the input,

2x-1

, is equal to the original roots, 2 and -5. We set up two equations:

$2x-1 = 2 \implies 2x = 3 \implies x = \frac{3}{2}$
$2x-1 = -5 \implies 2x = -4 \implies x = -2$ The new roots are $x=-2$ and $x=\frac{3}{2}$ . Distractor B is a common error where the transformation $2x-1$ is applied directly to the roots. Distractor D results from a sign error when solving for x.

Question 7

The graph of $y=\sqrt{x}$ is transformed into the graph of $y=a\sqrt{x-h}+k$ . The transformed graph passes through the points $(5, 7)$ and $(10, 11)$ and has its domain as $x \ge 1$ . Find the value of $a$ .

1
2
3
4 (correct answer)

Explanation: The domain of

y=\sqrt{x}

x \ge 0

, and its endpoint is at

(0,0)

. The transformation

y=a\sqrt{x-h}+k

shifts the endpoint to

(h,k)

and changes the domain to

x \ge h

. Given that the domain of the transformed graph is

x \ge 1

, we have

h=1

. The function is

y = a\sqrt{x-1}+k

. Now, substitute the two given points into this equation to form a system of linear equations: For

(5, 7)

7 = a\sqrt{5-1} + k \implies 7 = a\sqrt{4} + k \implies 7 = 2a + k

. For

(10, 11)

11 = a\sqrt{10-1} + k \implies 11 = a\sqrt{9} + k \implies 11 = 3a + k

. Subtract the first equation from the second:

(11 - 7) = (3a - 2a) + (k - k)

which gives

4 = a

. So the value of

a

is 4.

Question 8

$f(x) = -g(x+4)$
$f(x) = -g(x-4)$ (correct answer)
$f(x) = g(-x-4)$
$f(x) = -g(x)+4$

f(x)

to become

g(x)

. First,

f(x)

is reflected in the x-axis, giving us

-f(x)

. Then this reflected function is translated 4 units left, which means replacing

x

with

(x+4)

, resulting in

g(x) = -f(x+4)

. To find

f(x)

in terms of

g(x)

, we reverse these steps. From

g(x) = -f(x+4)

, we first undo the reflection by multiplying both sides by

-1

-g(x) = f(x+4)

. Then we undo the horizontal translation by replacing

x

with

(x-4)

f(x) = -g(x-4)

. This confirms answer B is correct. Looking at the wrong answers: A gives

f(x) = -g(x+4)

, which would represent translating right instead of left when reversing the transformation. C gives

f(x) = g(-x-4)

, incorrectly introducing a reflection about the y-axis that wasn't in the original problem. D gives

f(x) = -g(x)+4

a

" become "right

a

" when reversed, and reflections are their own inverse operations.

Question 9

The graph of $y=f(x)$ is transformed by a vertical stretch with scale factor 2, followed by a translation by the vector $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ . If the point $(1, 4)$ is on the original graph, what are the coordinates of its image?

$(6, 3)$
$(4, 1)$
$(4, 7)$ (correct answer)
$(6, 7)$

Explanation: When you encounter function transformations, you need to apply them in the correct order to track how points move. Function transformations follow specific rules: vertical stretches multiply y-coordinates, and translations shift points by adding to the coordinates. Starting with the point

(1, 4)

on the original graph

y = f(x)

, let's apply each transformation step by step. First, the vertical stretch with scale factor 2 multiplies the y-coordinate by 2 while leaving the x-coordinate unchanged. So

(1, 4)

becomes

(1, 8)

. Next, the translation by vector

\begin{pmatrix} 3 \\ -1 \end{pmatrix}

means we add 3 to the x-coordinate and subtract 1 from the y-coordinate. Applying this to

(1, 8)

gives us

(1 + 3, 8 - 1) = (4, 7)

. Looking at the wrong answers: Answer A

(6, 3)

appears to have confused the transformations, possibly applying the stretch incorrectly or mixing up the order. Answer B

(4, 1)

gets the x-coordinate right but seems to have subtracted rather than added during the vertical stretch. Answer D

(6, 7)

correctly handles the y-transformations but incorrectly adds the translation vector's x-component twice, giving

1 + 3 + 2 = 6

instead of

1 + 3 = 4

. The correct answer is C

(4, 7)

. Strategy tip: Always apply transformations in the order given, and remember that vertical stretches affect only y-coordinates while translations affect both coordinates by adding the vector components. Write out each step to avoid mixing up the operations.

Question 10

$(6, 3)$
$(4, 1)$
$(4, 7)$ (correct answer)
$(6, 7)$

(1, 4)

on the original graph

y = f(x)

, let's apply each transformation step by step. First, the vertical stretch with scale factor 2 multiplies the y-coordinate by 2 while leaving the x-coordinate unchanged. So

(1, 4)

becomes

(1, 8)

. Next, the translation by vector

\begin{pmatrix} 3 \\ -1 \end{pmatrix}

means we add 3 to the x-coordinate and subtract 1 from the y-coordinate. Applying this to

(1, 8)

gives us

(1 + 3, 8 - 1) = (4, 7)

. Looking at the wrong answers: Answer A

(6, 3)

appears to have confused the transformations, possibly applying the stretch incorrectly or mixing up the order. Answer B

(4, 1)

gets the x-coordinate right but seems to have subtracted rather than added during the vertical stretch. Answer D

(6, 7)

correctly handles the y-transformations but incorrectly adds the translation vector's x-component twice, giving

1 + 3 + 2 = 6

instead of

1 + 3 = 4

. The correct answer is C

(4, 7)

Question 11

$g(x) = 3f(-x) - 2$ (correct answer)
$g(x) = 3(f(-x) - 2)$
$g(x) = 3f(-x-2)$
$g(x) = f(-3x) - 2$

Explanation: We apply the transformations step-by-step to

y=f(x)

Reflection in the y-axis: This transformation replaces $x$ with $-x$ . The function becomes $y = f(-x)$ .
Stretched vertically by a factor of 3: This transformation multiplies the entire function by 3. The function becomes $y = 3f(-x)$ .
Translated 2 units down: This transformation subtracts 2 from the entire function. The function becomes $y = 3f(-x) - 2$ . Therefore, the final equation is $g(x) = 3f(-x) - 2$ . Distractor B represents translating down before stretching, which would result in $g(x) = 3f(-x) - 6$ .

Question 12

$h(x) = (-x+2)^2 + 3$
$h(x) = -(x-2)^2 - 3$
$h(x) = -(x+2)^2 + 3$ (correct answer)
$h(x) = -(x+2)^2 - 3$

Explanation: When you encounter function transformations, work systematically through each transformation step by step, applying them to the original function in the order given. Starting with

f(x) = x^2

, let's find

g(x)

first. Translating 2 units left means replacing

x

with

(x+2)

, and translating 3 units down means subtracting 3 from the entire function. Therefore:

g(x) = (x+2)^2 - 3

Next, reflecting

g(x)

in the x-axis means multiplying the entire function by -1:

h(x) = -g(x) = -[(x+2)^2 - 3] = -(x+2)^2 + 3

Now let's examine why each answer choice is wrong or right: Choice A,

h(x) = (-x+2)^2 + 3

, incorrectly applies the horizontal translation. This represents a translation 2 units to the right, not left, and doesn't include the reflection properly. Choice B,

h(x) = -(x-2)^2 - 3

, has two errors: it translates right instead of left (using

x-2

instead of

x+2

), and it moves down instead of reflecting the vertical translation correctly after the x-axis reflection. Choice C,

h(x) = -(x+2)^2 + 3

, correctly shows the left translation with

(x+2)

, the downward translation of the original function, and the x-axis reflection that flips the sign of the entire expression. Choice D,

h(x) = -(x+2)^2 - 3

Question 13

1
2
3
4 (correct answer)

Explanation: The domain of

y=\sqrt{x}

x \ge 0

, and its endpoint is at

(0,0)

. The transformation

y=a\sqrt{x-h}+k

shifts the endpoint to

(h,k)

and changes the domain to

x \ge h

. Given that the domain of the transformed graph is

x \ge 1

, we have

h=1

. The function is

y = a\sqrt{x-1}+k

. Now, substitute the two given points into this equation to form a system of linear equations: For

(5, 7)

7 = a\sqrt{5-1} + k \implies 7 = a\sqrt{4} + k \implies 7 = 2a + k

. For

(10, 11)

11 = a\sqrt{10-1} + k \implies 11 = a\sqrt{9} + k \implies 11 = 3a + k

. Subtract the first equation from the second:

(11 - 7) = (3a - 2a) + (k - k)

which gives

4 = a

. So the value of

a

is 4.

Question 14

The function $f(x)$ is periodic with period 4. What is the period of the function $g(x) = f(2x - 1)$ ?

1
2 (correct answer)
4
8

Explanation: A function

f(x)

has period

P

f(x+P) = f(x)

for all

x

. We are given that

f(x)

has a period of 4, so

f(x+4) = f(x)

. We want to find the period

T

g(x) = f(2x-1)

, which means we need to find the smallest positive

T

such that

g(x+T) = g(x)

g(x+T) = f(2(x+T) - 1) = f(2x + 2T - 1)

. We want this to be equal to

g(x) = f(2x-1)

. So we need

f(2x-1+2T) = f(2x-1)

. Since the period of

f

is 4, the expression inside

f

must differ by a multiple of 4. We want the smallest positive

T

, so we set the added term equal to 4.

2T = 4

T = 2

. The period of

g(x)

is 2. Horizontal translations (like the -1) do not affect the period.

Question 15

The graph of $y = e^x$ is reflected in the line $y=2$ . What is the equation of the transformed graph?

$y = 4 - e^x$ (correct answer)
$y = 2 - e^x$
$y = e^{4-x}$
$y = -e^x + 2$

Explanation: A reflection of a point

(x, y)

across the horizontal line

y=c

results in a new point

(x, 2c-y)

. In this case,

c=2

. The transformation is

(x, y) \to (x, 4-y)

. Let the new coordinates be

(x', y')

. So,

x' = x

and

y' = 4-y

. We need to express the original coordinates in terms of the new ones:

x = x'

and

y = 4-y'

. Substitute these into the original equation

y = e^x

4 - y' = e^{x'}

Now, solve for

y'

y' = 4 - e^{x'}

Dropping the primes, the equation of the transformed graph is

y = 4 - e^x

Question 16

The function $y=f(x)$ has domain $[-4, 6]$ and range $[-2, 8]$ . Determine the domain and range of the function $g(x) = \frac{1}{2}f(4-2x) + 1$ .

Domain: $[-1, 4]$ , Range: $[0, 5]$ (correct answer)
Domain: $[-4, -1]$ , Range: $[0, 5]$
Domain: $[0, 5]$ , Range: $[-0.5, 4.5]$
Domain: $[-1, 4]$ , Range: $[-0.5, 4.5]$

Explanation: For the domain, we consider the argument of the function

f

. The original domain is

-4 \le x_{\text{old}} \le 6

. The new function has argument

4-2x

. So we set

-4 \le 4-2x \le 6

. Subtract 4 from all parts:

-8 \le -2x \le 2

. Divide by -2 and reverse the inequalities:

4 \ge x \ge -1

, which is

-1 \le x \le 4

. The new domain is

[-1, 4]

. For the range, we consider the transformations outside the function

f

. The original range is

-2 \le f(x) \le 8

. The new function is

g(x) = \frac{1}{2}f(...) + 1

. Multiply the range by

\frac{1}{2}

\frac{1}{2}(-2) \le \frac{1}{2}f(...) \le \frac{1}{2}(8) \implies -1 \le \frac{1}{2}f(...) \le 4

. Then add 1:

-1+1 \le \frac{1}{2}f(...) + 1 \le 4+1 \implies 0 \le g(x) \le 5

. The new range is

[0, 5]

Question 17

The function $f(x) = \frac{x+1}{x-3}$ has a vertical asymptote at $x=3$ and a horizontal asymptote at $y=1$ . Find the equations of the asymptotes of the function $g(x) = -f(2x) + 4$ .

$x = \frac{3}{2}, y = 3$ (correct answer)
$x = 6, y = 3$
$x = \frac{3}{2}, y = -5$
$x = 3, y = 3$

Explanation: The transformations from

f(x)

g(x)

are:

A horizontal stretch by a factor of $\frac{1}{2}$ (from $2x$ ).
A reflection in the x-axis (from the negative sign before $f$ ).
A vertical translation up by 4 units.

The vertical asymptote

x=3

is only affected by horizontal transformations. Applying the horizontal stretch by

\frac{1}{2}

, the new vertical asymptote is at

x = \frac{3}{2}

. The horizontal asymptote

y=1

is only affected by vertical transformations. Applying the reflection in the x-axis,

y=1

becomes

y=-1

. Then, applying the vertical translation up by 4,

y=-1

becomes

y = -1+4 = 3

. Thus, the new asymptotes are

x = \frac{3}{2}

and

y=3

Question 18

A function $f(x)$ has $f(3)=5$ . What is the value of $g(1)$ if $g(x) = 2f(x+2) - 3$ ?

2
7 (correct answer)
10
17

Explanation: We need to find the value of

g(1)

. We substitute

x=1

into the equation for

g(x)

g(1) = 2f(1+2) - 3

g(1) = 2f(3) - 3

We are given that

f(3)=5

. We can substitute this value into the expression:

g(1) = 2(5) - 3

g(1) = 10 - 3 = 7

. The value of

g(1)

is 7.

Question 19

The graph of $y=f(x)$ is transformed to the graph of $y=f(|x|)$ . This transformation has what effect on the graph of $y=f(x)$ ?

The entire graph is reflected in the y-axis.
The part of the graph for $y<0$ is replaced by a reflection of the part for $y>0$ in the x-axis.
The part of the graph for $x<0$ is replaced by a reflection of the part for $x>0$ in the y-axis. (correct answer)
The entire graph is reflected in the x-axis.

Explanation: When you encounter transformations involving absolute value functions, you need to carefully analyze what happens to the input versus the output of the function. The transformation from

y = f(x)

y = f(|x|)

affects the input to the function, not the output. The key insight is understanding what

|x|

does: it makes all negative x-values positive while leaving positive x-values unchanged. So when

x > 0

, we have

|x| = x

, meaning

f(|x|) = f(x)

— the right side of the graph stays the same. But when

x < 0

, we have

|x| = -x

(which is positive), so

f(|x|) = f(-x)

— we're now using the function values that originally corresponded to positive x-values. This means the part of the original graph for

x > 0

gets reflected across the y-axis to replace the part for

x < 0

. The correct answer is C. Looking at the wrong answers: A suggests the entire graph reflects in the y-axis, but only the left side changes while the right side remains identical. B incorrectly focuses on reflecting parts based on y-values rather than x-values — this transformation doesn't involve the x-axis at all. D suggests reflection in the x-axis, which would affect y-values, not the domain transformation we're seeing here. Strategy tip: For absolute value transformations, remember that

f(|x|)

always creates an even function (symmetric about the y-axis). The right side of the original graph survives, and its reflection replaces the left side.

Question 20

The function $y=f(x)$ is transformed to $y=f(x+c)-c$ . If the graph of the transformed function passes through the origin, which statement must be true about the original function $f(x)$ ?

Its y-intercept must be $c$ .
It must pass through the point $(-c, -c)$ .
It must pass through the origin.
It must pass through the point $(c, c)$ . (correct answer)

Explanation: When you encounter function transformations, the key is understanding how changes inside and outside the function notation affect the graph's position. The transformation

y = f(x + c) - c

combines two shifts:

f(x + c)

shifts the graph left by

c

units, and subtracting

c

shifts it down by

c

units. If this transformed function passes through the origin

(0, 0)

, then substituting these coordinates gives us:

0 = f(0 + c) - c = f(c) - c

, which means

f(c) = c

. This tells us the original function must pass through the point

(c, c)

, making answer D correct. Let's examine why the other options fail. Option A claims the y-intercept must be

c

. The y-intercept occurs where

x = 0

, so

f(0)

would need to equal

c

. However, our condition only tells us that

f(c) = c

, not

f(0) = c

. Option B suggests the original function passes through

(-c, -c)

. This would mean

f(-c) = -c

, but again, we only know

f(c) = c

. Option C claims the original function passes through the origin, meaning

f(0) = 0

. This isn't necessarily true since we established

f(c) = c

, not

f(0) = 0

. Remember this pattern: when a transformed function passes through a specific point, work backwards by substituting those coordinates into the transformation equation. This algebraic approach eliminates guesswork and reveals exactly what conditions the original function must satisfy.