If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Solved Examples on the Transformations of Functions

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

Solve all questions
Show all work

(1.) ACT A point at (−5, 7) in the standard (x, y) coordinate plane is translated right 7 coordinate units and down 5 coordinate units.
What are the coordinates of the point after the translation?

$Point: (-5, 7) \\[3ex] x = -5, y = 7 \\[3ex] HOSH\:\: 7 \:\:units\:\: right:\;\;x-coordinate \:\:changes \\[3ex] -5 + 7 = 2 \\[3ex] (-5, 7) \rightarrow (2, 7) \\[3ex] x = 2, y = 7 \\[3ex] VESH\:\: 5 \:\:units\:\: down:\;\; y-coordinate \:\:changes \\[3ex] 7 - 5 = 2 \\[3ex] (2, 7) \rightarrow (2, 2)$
(2.) Complete the underlined:
Suppose that the graph of a function f is known.
Then the graph of y = f(x−2) may be obtained by a .............. shift of the graph of f .............. a distance of 2 units.

Suppose that the graph of a function f is known.
Then the graph of y = f(x−2) may be obtained by a horizontal shift of the graph of f right a distance of 2 units.
(3.) Which of the following functions has a graph that is the graph of $y = \sqrt{x}$ shifted up 6 units?

$A.\;\; y = \sqrt{x} - 6 \\[3ex] B.\;\; y = \sqrt{x + 6} \\[3ex] C.\;\; y = \sqrt{x} + 6 \\[3ex] D.\;\; y = \sqrt{x - 6} \\[3ex]$

$y = \sqrt{x} .......Parent\;\;Function \\[3ex] VESH\;\;6\;\;units\;\;up:\;\;y = \sqrt{x} + 6 ......Child\;\;function \\[3ex]$
(4.) ACT Point A is located at (3, 8) in the standard (x, y) coordinate plane.
What are the coordinates of A', the image of A after it is reflected across the y-axis?

A. (3, −8)
B. (−3, −8)
C. (−3, 8)
D. (8, 3)
E. (−8, 3)

Reflection across the y-axis is Horizontal Reflection (HORE)

$HORE:\;\;x-coordinate\;\;changes \\[3ex] A(3, 8) \rightarrow A'(-3, 8)$
(5.) ACT A triangle, △ABC, is reflected across the x-axis to have the image △A'B'C' in the standard (x, y) coordinate plane: thus, A reflects to A'.
The coordinates of point A are (c, d).
What are the coordinates of point A' ?

F. (c, −d)
G. (−c, d)
H. (−c, −d)
J. (d, c)
K. Cannot be determined from the given information.

Reflection across the x-axis is Vertical Reflection (VERE)

$VERE:\;\;y-coordinate\;\;changes \\[3ex] A(c, d) \rightarrow A'(c, -d)$
(6.) Fill in the blank:
Suppose that the graph of a function f is known.
Then the graph of y = f(−x) may be obtained by a reflection about the $\hspace{10mm}$ of the graph of the function y = f(x)

Suppose that the graph of a function f is known.
Then the graph of y = f(−x) may be obtained by a reflection about the y-axis of the graph of the function y = f(x)
(7.) Find the function that is finally graphed after the following transformations are applied to the graph of $y = \sqrt{x}$ in the order listed.
(a.) Shift down 3 units

$Parent\;\;Function:\;\; y = \sqrt{x} \\[3ex] 1st\;\;Transformation:\;\;VESH\;\;8\;\;units\;\;down \\[3ex] y = \sqrt{x} - 3 \\[5ex] 2nd\;\;Transformation:\;\;VERE \\[3ex] y = -(\sqrt{x} - 3) \\[5ex] 3rd\;\;Transformation:\;\;HORE \\[3ex] y = -(\sqrt{-x} - 3) = -\sqrt{-x} + 3 \\[5ex]$
(8.) ACT In the standard (x, y) coordinate plane, A' is the image resulting from the reflection of the point A(2, −3) across the y-axis.
What are the coordinates of A' ?

A. (−3, 2)
B. (−2, −3)
C. (−2, 3)
D. (2, 3)
E. (3, −2)

Reflection across the y-axis is Horizontal Reflection (HORE)
Only $x-value$ change
$y-value$ does not change

$Point: (2, −3) \\[3ex] x = 2, y = −3 \\[3ex] HORE:\;\;x-coordinate\;\;changes \\[3ex] (2, −3) \rightarrow (−2, −3)$
(9.) ACT In the standard (x, y) coordinate plane, the coordinates of the y-intercept of the graph of the function $y = f(x)$ are (0, −2).
What are the coordinates of the y-intercept of the graph of the function $y = f(x) - 3$?

$Point: (0, -2) \\[3ex] x = 0, y = -2 \\[3ex] f(x) - 3 = VESH\:\: 3\:\: units\:\: down:\:\: y-coordinate\:\: changes \\[3ex] -2 - 3 = -5 \\[3ex] (0, -2) \rightarrow (0, -5)$
(10.) True or False: The graph of $y = \dfrac{1}{9}g(x)$ is the graph of $y = g(x)$ compressed by a factor of 9.

$y = g(x)...Parent\;\;Function \\[3ex] y = \dfrac{1}{9}g(x)...Child\;\;Function \\[5ex]$ VECO: This is true because:
(1.) The graph of the child function is obtained by multiplying each y-coordinate of the parent function and
(2.) $0 \lt \dfrac{1}{9} \lt 1$
(11.) Write the function whose graph of:
(a.) y = x², is shifted to the right 6 units.
(b.) y = x³, is shifted up 9 units.
(c.) $y = 7\sqrt[4]{x}$, is reflected about the x-axis.
(d.) y = |x|, is vertically stretched by a factor of 6.

(a.) HOSH 6 units right: y = (x − 6)²
(b.) VESH 9 units up: y = x³ + 9
(c.) VERE: $y = -7\sqrt[4]{x}$
(d.) VEST by a factor of 6: y = 6|x|
(12.) ACT In the standard (x, y) coordinate plane, point A has coordinate (−8, −3).
Point A is translated 8 units to the right and 3 units up, and that image is labeled A'.
What are the coordinates of A' ?

A. (−16, −6)
B. (−11, −11)
C. (−8, −6)
D. (0, 0)
E. (16, 6)

A(−8, −3)
HOSH 8 units right
A(−8, −3) → A'(−8 + 8, −3)
→ A'(0, −3)
VESH 3 units up
A'(0, −3) → A''(0, −3 + 3)
→ A''(0, 0)
(13.) ACT In the standard (x, y) coordinate plane, given Parabola A with equation $y = 3x^2$, Parabola B is the image of Parabola A after a shift of 7 coordinate units to the left and 4 coordinate units down.
Parabola B has which of the following equations?

$F.\;\; y = 3(x - 4)^2 - 7 \\[3ex] G.\;\; y = 3(x - 7)^2 - 4 \\[3ex] H.\;\; y = 3(x - 7)^2 + 4 \\[3ex] J.\;\; y = 3(x + 7)^2 - 4 \\[3ex] K.\;\; y = 3(x + 7)^2 + 4 \\[3ex]$

Parabola A: Equation: y = 3x²
7 coordinate units to the left = HOSH 7 units left = 3(x + 7)²
4 coordinate units down = VESH 4 units down = 3(x + 7)² − 4
Parabola B: Equation: y = 3(x + 7)² − 4
(14.) Fill in the blank:
Suppose that the graph of a function f is known.
Then the graph of y = ​−f(​x) may be obtained by a reflection about the $\hspace{10mm}$ of the graph of the function y = f(x)

Suppose that the graph of a function f is known.
Then the graph of y = f(−x) may be obtained by a reflection about the x-axis of the graph of the function y = f(x)
(15.) Find the function that is finally graphed after the following transformations are applied to the graph of $y = \sqrt{x}$ in the order listed.
(b.) Shift up 7 units
(c.) Shift right 5 units

$Parent\;\;Function:\;\; y = \sqrt{x} \\[3ex] 1st\;\;Transformation:\;\;VERE \\[3ex] y = -\sqrt{x} \\[5ex] 2nd\;\;Transformation:\;\;VESH\;\;7\;\;units\;\;down \\[3ex] y = -\sqrt{x} + 7 \\[5ex] 3rd\;\;Transformation:\;\;HOSH\;\;5\;\;units\;\;right \\[3ex] y = -\sqrt{x - 5} + 7 \\[5ex]$
(16.) ACT The graph of y = |x − 6| is in the standard (x, y) coordinate plane.
Which of the following transformations, when applied to the graph of y = |x|, results in the graph of y = |x − 6|?

F. Translation to the right 6 coordinate units
G. Translation to the left 6 coordinate units
H. Translation up 6 coordinate units
J. Translation down 6 coordinate units
K. Reflection across the line x = 6

Parent Function: y = |x|
HOSH 6 units right
Transformed (Child) Function: y = |x − 6|
The answer is Option F.: Translation to the right 6 coordinate units
(17.) ACT A line segment has endpoints (a, b) and (c, d) in the standard (x, y) coordinate plane, where a, b, c, and d are distinct positive integers.
The segment is reflected across the x-axis.
After this reflection, what are the coordinates of the endpoints of the image?

$F.\;\; (-a, b)\;\;and\;\;(-c, d) \\[3ex] G.\;\; (a, -b)\;\;and\;\;(c, -d) \\[3ex] H.\;\; (-a, -b)\;\;and\;\;(-c, -d) \\[3ex] J.\;\; (a, b)\;\;and\;\;(c, d) \\[3ex] K.\;\; (a, 0)\;\;and\;\;(c, 0) \\[3ex]$

For any point, (x, y); reflection across the x-axis gives (x, −y)
This implies

$(a, b) \rightarrow (a, -b) \\[3ex] (c, d) \rightarrow (c, -d) \\[5ex] Option\;G:\;\; (a, -b)\;\;and\;\;(c, -d)$
(18.) Which of the following functions has a graph that is the graph of $y = f(x)$ stretched horizontally by a factor of 10?

$A.\;\; y = \dfrac{1}{10}f(x) \\[5ex] B.\;\; y = 10f(x) \\[3ex] C.\;\; y = f\left(\dfrac{1}{10}x\right) \\[5ex] D.\;\; y = f(10x) \\[3ex]$

$y = f(x) .......Parent\;\;Function \\[3ex] HOST\;\;by\;\;factor\;\;of\;\;10:\;\;y = f\left(\dfrac{1}{10}x\right) ......Child\;\;function$
(19.) ACT The point (3, 27) is labeled on the graph of f(x) = x³ in the standard (x, y) coordinate plane below.
The graph of f(x) will be translated 3 coordinate units to the left.
Which of the following points will be on the image of the graph after the translation?

F. (0, 27)
G. (3, 24)
H. (3, 27)
J. (3, 30)
K. (6, 27)

Based on the graph:

When we move the graph (green color) 3 units left, the point (3, 27) on the graph becomes (0, 27) on the transformed graph (blue color)
It is seen as moving from Point A to Point B
However, please note the equations of the parent function and the transformed function

$Parent\;\;Function:\;\; y = x^3 \\[3ex] Transformed\;\;Function:\;\;HOSH\;\;3\;\;units\;\;left:\;\; y = (x + 3)^3$
(20.) ACT In the standard (x, y) coordinate plane, the graph of the function $y = 5\sin(x) - 7$ undergoes a single translation such that the equation of its image is $y = 5\sin(x) - 14$.
Which of the following describes this translation?

F. Up 7 coordinate units
G. Down 7 coordinate units
H. Left 7 coordinate units
J. Right 7 coordinate units
K. Right 14 coordinate units

From −7 to −14
How do you get from −7 to −14?
What will you add to −7 to get −14?
Let the thing be p

$-7 + p = -14 \\[3ex] p = -14 + 7 \\[3ex] p = -7 \\[3ex]$ This represents a VESH of 7 units down
This means: Option G. Down 7 coordinate units

(21.) ACT In the standard (x, y) coordinate plane below, △ABC will be translated 10 units down and then the resulting image will be reflected over the y-axis.
What will be the coordinates of the final image of A resulting from both transformations?

A. (−5, 9)
B. (−1, 9)
C. (1, −9)
D. (5, −10)
E. (5, −9)

A(−5, 1)
VESH 10 units down
A(−5, 1) → A'(−5, 1 − 10)
→ A'(−5, −9)
HORE (Reflected over the y-axis)
A'(−5, −9) → A''(5, −9)
(22.) The point (−19, 9) is on the graph of y = f(x)
Complete the underlined. Write an ordered pair.
(a.) A point on the graph of y = k(x), where k(x) = −f(x) is ............
(b.) A point on the graph of y = k(x), where k(x) = f(−x) is ............
(c.) A point on the graph of y = k(x), where k(x) = f(x + 9) is ............
(d.) A point on the graph of y = k(x), where k(x) = f(x) − 1 is ............
(e.) A point on the graph of y = k(x), where k(x) = $\dfrac{1}{3}$f(x) is ............
(f.) A point on the graph of y = k(x), where k(x) = 6f(x) is ............

(−19, 9) on y = f(x)

(a.) −f(x) is VERE
(−19, 9) → (−19, −9)

(b.) f(−x) is HORE
(−19, 9) → [−(−19), 9]
→ (19, 9)

(c.) f(x + 9) is HOSH 9 units left
(−19, 9) → (−19 − 9, 9)
→ (−28, 9)

(d.) f(x) − 1 is VESH 1 unit down
(−19, 9) → (−19, 9 − 1)
→ (−19, 8)

(e.) $\dfrac{1}{3}$f(x) is VECO by a factor of $\dfrac{1}{3}$

$(-19, 9) \rightarrow \left(-19, \dfrac{1}{3} * 9\right) \\[5ex] \hspace{8ex} \rightarrow (-19, 3) \\[3ex]$ (f.) 6f(x) is VEST by a factor of 6
(−19, 9) → (−19, 6 * 9)
→ (−19, 54)
(23.) Write the function whose graph is the graph of:
(a.) $y = 7\sqrt{x}$ is reflected about the y-axis
(b.) $y = x^3$, is horizontally stretched by a factor by 2.

$(a.) \\[3ex] y = 7\sqrt{x} \\[3ex] Reflected\;\;about\;\;the\;\;y-axis = Horizontal\;\;Reflection \\[3ex] y = 7\sqrt{-x} \\[5ex] (b.) \\[3ex] y = x^3 \\[3ex] Horizontally\;\;stretched\;\;by\;\;a\;\;factor\;\;of\;\;2 \\[3ex] y = \left(\dfrac{1}{2}x\right)^3 \\[5ex]$
(24.) ACT Point A is located at (3, 8) in the standard (x, y) coordinate plane.
What are the coordinates of A', the image of A after it is reflected across the y-axis?

$A.\;\; (3, -8) \\[3ex] B.\;\; (-3, -8) \\[3ex] C.\;\; (-3, 8) \\[3ex] D.\;\; (8, 3) \\[3ex] E.\;\; (-8, 3) \\[3ex]$

For any point, (x, y); reflection across the y-axis gives (-x, y)
This implies

$A(3, 8) \rightarrow A'(-3, 8)$
(25.) ACT The function y = f(x) is graphed in the standard (x, y) coordinate plane below.

The points on the graph of the function y = 3 + f(x − 1) can be obtained from the points on y = f(x) by a shift of:

A. 1 unit to the right and 3 units up.
B. 1 unit to the right and 3 units down.
C. 3 units to the right and 1 unit up.
D. 3 units to the right and 1 unit down.
E. 3 units to the left and 1 unit down.

Parent Function: y = f(x)
HOSH 1 unit right
Transformed (Child) Function: y = f(x − 1)
VESH 3 units up
Transformed Function: y = f(x − 1) + 3
y = 3 + f(x − 1)
The answer is Option A.: 1 unit to the right and 3 units up.
(26.) The graph of the function $f\left(\dfrac{1}{19}x\right)$ can be obtained from the graph of $y = f(x)$ by one of the following actions:

A. horizontally stretching the graph of f(x) by a factor 19
B. horizontally compressing the graph of f(x) by a factor 19
C. vertically stretching the graph of f(x) by a factor 19
D. vertically compressing the graph of f(x) by a factor 19

$f\left(\dfrac{1}{19}x\right)$ means that it is something "horizontally" because the $\dfrac{1}{19}$ is inside with the x

$a = \dfrac{1}{19}$

Because: $0 \lt \dfrac{1}{19} \lt 1$; it is horizontally stretched by a factor of $1 \div \dfrac{1}{19}$

So, it is horizontally stretched by a factor of 19.
(27.)

(28.) ACT In the standard (x, y) coordinate plane, point A has coordinates (−7, −5).
Point A is translated 7 units to the left and 5 units down, and that image is labeled A'.
What are the coordinates of A'?

$F.\;\; (-14, -10) \\[3ex] G.\;\; (-12, -12) \\[3ex] H.\;\; (-7, -10) \\[3ex] J.\;\; (0, 0) \\[3ex] K.\;\; (14, 10) \\[3ex]$

Translated to the left or right is only for the x-coordinate
Translated up or down is only for the y-coordinate

$\underline{Translated\;\;7\;\;units\;\;to\;\;the\;\;left} \\[3ex] A(-7, -5) \rightarrow A'(-7 - 7, -5) \\[3ex] \hspace{4.5em} \rightarrow A'(-14, -5) \\[3ex] \underline{Translated\;\;5\;\;units\;\;down} \\[3ex] A'(-14, -5) \rightarrow A''(-14, -5 - 5) \\[3ex] \hspace{5.2em} \rightarrow A''(-14, -10)$
(29.) ACT In the standard (x, y) coordinate plane below, $\triangle$ABC will be translated 10 units down and then the resulting image will be reflected over the y-axis.
What will be the coordinates of the final image of A resulting from both transformations?

$A.\;\; (-5, 9) \\[3ex] B.\;\; (-1, 9) \\[3ex] C.\;\; (1, -9) \\[3ex] D.\;\; (5, -10) \\[3ex] E.\;\; (5, -9) \\[3ex]$

Translated up or down is only for the y-coordinate
For any point, (x, y); reflection across the y-axis gives (-x, y)
We are only concerned with Point A

$\underline{Translated\;\;10\;\;units\;\;down} \\[3ex] A(-5, 1) \rightarrow A'(-5, 1 - 10) \\[3ex] \hspace{3.7em} \rightarrow A'(-5, -9) \\[5ex] \underline{Reflected\;\;over\;\;y-axis} \\[3ex] A'(-5, -9) \rightarrow A''(5, -9)$
(30.) Which transformations are needed to graph the function $f(x) = 3(x + 1)^2 - 2$ ?

A. The graph of y = x² should be horizontally shifted to the left by 1 unit, vertically stretched by a factor of 3, and shifted vertically up by 2 units.

B. The graph of y = x² should be horizontally shifted to the right by 1 unit, vertically compressed by a factor of 3, and shifted vertically up by 2 units.

C. The graph of y = x² should be horizontally shifted to the left by 1 unit, vertically stretched by a factor of 3, and shifted vertically down by 2 units.

D. The graph of y = x² should be horizontally shifted to the right by 1 unit, vertically compressed by a factor of 3, and shifted vertically down by 2 units.

$f(x) = 3(x + 1)^2 - 2$

Parent Function: f(x) = x²
(x + 1)²: HOSH 1 unit left
3(x + 1)²: VEST by a factor of 3
3(x + 1)² - 2: VESH 2 units down
The correct option is C.
(31.)

(32.) ACT A point at (−3, 7) in the standard (x, y) coordinate plane is shifted down 3 units and right 7 units.
What are the coordinates of the new point?

$F.\;\; (-10, 10) \\[3ex] G.\;\; (0, 0) \\[3ex] H.\;\; (4, 4) \\[3ex] J.\;\; (4, 10) \\[3ex] K.\;\; (10, 10) \\[3ex]$

Let the point be A(-3, 7)
shifted down: only the y-coordinate is affected
shifted right: only the x-coordinate is affected

$\underline{shifted\;\;down\;\;3\;\;units} \\[3ex] A(-3, 7) \rightarrow A'(-3, 7 - 3) \\[3ex] \hspace{4em} \rightarrow A'(-3, 4) \\[5ex] \underline{shifted\;\;right\;\;7\;\;units} \\[3ex] A'(-3, 4) \rightarrow A''(-3 + 7, 4) \\[3ex] \hspace{4em} \rightarrow A''(4, 4)$
(33.) ACT A point with coordinates (a, b) is plotted in the standard (x, y) coordinate plane as shown below.
The point is then reflected across the y-axis.
Which of the following are the coordinates for the point after the reflection?

$A.\;\; (-a, b) \\[3ex] B.\;\; (a, -b) \\[3ex] C.\;\; (b, a) \\[3ex] D.\;\; (-b, a) \\[3ex] E.\;\; (b, -a) \\[3ex]$

For any point, (x, y); reflection across the y-axis gives (-x, y)
This implies

$(a, b) \rightarrow (-a, b)$
(34.)

(35.)

(36.) ACT A point at (−2, 8) in the standard (x, y) coordinate plane is shifted right 8 units and down 2 units.
What are the coordinates of the new point?

$F.\;\; (-10, 10) \\[3ex] G.\;\; (0, 0) \\[3ex] H.\;\; (6, 6) \\[3ex] J.\;\; (6, 10) \\[3ex] K.\;\; (10, 10) \\[3ex]$

Let the point be A(-2, 8)
shifted right: only the x-coordinate is affected
shifted down: only the y-coordinate is affected

$\underline{shifted\;\;right\;\;8\;\;units} \\[3ex] A(-2, 8) \rightarrow A'(-2 + 8, 8) \\[3ex] \hspace{4em} \rightarrow A'(6, 8) \\[3ex] \underline{shifted\;\;down\;\;2\;\;units} \\[3ex] A'(6, 8) \rightarrow A''(6, 8 - 2) \\[3ex] \hspace{3.5em} \rightarrow A''(6, 6)$
(37.)

(38.)

(39.)

(40.)

(41.)

(42.)

(43.)

(44.)

(45.)

(46.)