Transformations of Absolute Value Function: Insights and Applications

Transformations of $y = ∣ x ∣$
Vertical Translations	$Translation up k units, k > 0 y = ∣ x ∣ + k$
$Translation down k units, k < 0 y = ∣ x ∣ + k$
Horizontal Translations	$Translation to the right h units, h > 0 y = ∣ x - h ∣$
$Translation to the left h units, h < 0 y = ∣ x - h ∣$

Transformations of

y = ∣ x ∣

Vertical Translations

Translation up k units, k > 0 y = ∣ x ∣ + k

Translation down k units, k < 0 y = ∣ x ∣ + k

Horizontal Translations

Translation to the right h units, h > 0 y = ∣ x - h ∣

Translation to the left h units, h < 0 y = ∣ x - h ∣

Transformations of $y = f (x)$
Horizontal Translations	$Translation to the right by h units, h > 0 y = f (x - h)$
$Translation to the left by h units, h < 0 y = f (x - h)$
Vertical Translations	$Translation upwards by k units, k > 0 y = f (x) + k$
$Translation downwards by k units, k < 0 y = f (x) + k$

Transformations of

y = f (x)

Horizontal Translations

Translation to the right by h units, h > 0 y = f (x - h)

Translation to the left by h units, h < 0 y = f (x - h)

Vertical Translations

Translation upwards by k units, k > 0 y = f (x) + k

Translation downwards by k units, k < 0 y = f (x) + k

Transformations of $f (x)$
Vertical Stretch or Shrink	Vertical stretch, $a > 1$ $y = a f (x)$
Vertical shrink, $0 < a < 1$ $y = a f (x)$

Transformations of

f (x)

Vertical Stretch or Shrink

Vertical stretch,

a > 1

y = a f (x)

Vertical shrink,

0 < a < 1

y = a f (x)

Transformations of $f (x)$
Horizontal Stretch or Shrink	Horizontal stretch, $0 < b < 1$ $y = f (b x)$
Horizontal shrink, $b > 1$ $y = f (b x)$

Transformations of

f (x)

Horizontal Stretch or Shrink

Horizontal stretch,

0 < b < 1

y = f (b x)

Horizontal shrink,

b > 1

y = f (b x)

Transformations of $y = ∣ x ∣$
Vertical Stretch or Shrink	$Vertical stretch, a > 1 y = a ∣ x ∣$
$Vertical shrink, 0 < a < 1 y = a ∣ x ∣$
Horizontal Stretch or Shrink	$Horizontal stretch, 0 < b < 1 y = ∣ b x ∣$
$Horizontal shrink, b > 1 y = ∣ b x ∣$

Transformations of

y = ∣ x ∣

Vertical Stretch or Shrink

Vertical stretch, a > 1 y = a ∣ x ∣

Vertical shrink, 0 < a < 1 y = a ∣ x ∣

Horizontal Stretch or Shrink

Horizontal stretch, 0 < b < 1 y = ∣ b x ∣

Horizontal shrink, b > 1 y = ∣ b x ∣

Transformations of $f (x)$
Reflections	In the $x -$ axis $y = - f (x)$
In the $y -$ axis $y = f (- x)$

Transformations of

f (x)

Reflections

In the

x -

axis

y = - f (x)

In the

y -

axis

y = f (- x)

Transformations of $y = ∣ x ∣$
Reflections	$In the x - axis y = - ∣ x ∣$
$In the y - axis y = ∣ - x ∣$

Transformations of

y = ∣ x ∣

Reflections

In the x - axis y = - ∣ x ∣

In the y - axis y = ∣ - x ∣

$x$	$f (x) = ∣ 2 x - 5 ∣$	$f (x)$
$- 1$	$f (x) = ∣ 2 (- 1) - 5 ∣$	$7$
$0$	$f (x) = ∣ 2 (0) - 5 ∣$	$5$
$1$	$f (x) = ∣ 2 (1) - 5 ∣$	$3$
$2.5$	$f (x) = ∣ 2 (2.5) - 5 ∣$	$0$
$4$	$f (x) = ∣ 2 (4) - 5 ∣$	$3$
$5$	$f (x) = ∣ 2 (5) - 5 ∣$	$5$
$6$	$f (x) = ∣ 2 (6) - 5 ∣$	$7$

x

f (x) = ∣ 2 x - 5 ∣

f (x)

- 1

f (x) = ∣ 2 (- 1) - 5 ∣

7

0

f (x) = ∣ 2 (0) - 5 ∣

5

1

f (x) = ∣ 2 (1) - 5 ∣

3

2.5

f (x) = ∣ 2 (2.5) - 5 ∣

0

4

f (x) = ∣ 2 (4) - 5 ∣

3

5

f (x) = ∣ 2 (5) - 5 ∣

5

6

f (x) = ∣ 2 (6) - 5 ∣

7

$x$	$f (x) = \frac{1}{2} ∣ x - 4 ∣ - 2$	$f (x)$
$- 2$	$f (x) = \frac{1}{2} ∣ - 2 - 4 ∣ - 2$	$1$
$0$	$f (x) = \frac{1}{2} ∣ 0 - 4 ∣ - 2$	$0$
$2$	$f (x) = \frac{1}{2} ∣ 2 - 4 ∣ - 2$	$- 1$
$4$	$f (x) = \frac{1}{2} ∣ 4 - 4 ∣ - 2$	$- 2$
$6$	$f (x) = \frac{1}{2} ∣ 6 - 4 ∣ - 2$	$- 1$
$8$	$f (x) = \frac{1}{2} ∣ 8 - 4 ∣ - 2$	$0$
$10$	$f (x) = \frac{1}{2} ∣ 10 - 4 ∣ - 2$	$1$

x

f (x) = \frac{1}{2} ∣ x - 4 ∣ - 2

f (x)

- 2

f (x) = \frac{1}{2} ∣ - 2 - 4 ∣ - 2

1

0

f (x) = \frac{1}{2} ∣ 0 - 4 ∣ - 2

0

2

f (x) = \frac{1}{2} ∣ 2 - 4 ∣ - 2

- 1

4

f (x) = \frac{1}{2} ∣ 4 - 4 ∣ - 2

- 2

6

f (x) = \frac{1}{2} ∣ 6 - 4 ∣ - 2

- 1

8

f (x) = \frac{1}{2} ∣ 8 - 4 ∣ - 2

0

10

f (x) = \frac{1}{2} ∣ 10 - 4 ∣ - 2

1

The graph of $y = ∣ g (x) ∣$ is shown.

Determine

g (x)

if the function intersects the

y -

axis at

(0, - 4) .

The graph of y=|g(x)| is the graph of g(x) whose negative part is reflected in the x-axis. Looking at the given graph, we can see that y=|g(x)| intersects the y-axis at (0,4).

We know that g(x) intersects the y-axis at (0,4), which has the opposite sign of the current point of intersection, so we can conclude that we should reflect the left branch of y=|g(x)| in the x-axis.

With the reflection, we now have the graph of g(x), which is a straight line. From the graph we can see that the slope of g(x) is 1 and its y-intercept is -4. We can use this information to write the equation of g(x) in slope-intercept form. g(x)=x-4

The graph of $y = ∣ x - 4 ∣ + 1$ is shown.

Write an absolute value function whose graph forms a square with a side length of

22

units with the given graph.

We want to write an absolute value function whose graph forms a square with a side length of 2sqrt(2) with the given graph. Note that the angle between the two branches has a measure of 90 degrees.

Our first goal is to create an isosceles right triangle. Note that, due to the symmetry of the absolute value function, by drawing a horizontal line, we can have an isosceles triangle whose congruent sides are 2sqrt(2) units long. Since one of the angles is a right angle, the measure of the other two angles must be 45^(∘).

To decide where the horizontal line should be, we will find the length of thehypotenuse of the triangle using the Pythagorean Theorem.

Since the length of the hypotenuse is 4 units, the horizontal line should be y=3.

Let's now translate our graph down 3 units. We do this transformation by subtracting 3 from the output. y=|x-4|+1- 3 ⇕ y=|x-4|-2 Now the triangle will be formed with the x-axis.

Let's now reflect our graph in the x-axis. We do this by changing the sign of the output. y= - (|x-4|-2) ⇕ y=- |x-4|+2 Next we will graph the resulting function.

Note that these triangles are congruent. Therefore, the obtained triangle is an isosceles triangle with one right angle and two angles whose measure is 45^(∘). If we merge these isosceles triangles, they form a square.

Finally, we have to go back to the original position. To do so, we will translate both graphs up 3 units by adding 3 to the outputs.

Given Function	New Function
y=\|x-4\|-2+ 3 ⇕ y=\|x-4\|+1	y=- \|x-4\|+2+ 3 ⇕ y=- \|x-4\|+5

Let's see the final graph.

Therefore, the required function is y=- |x-4|+5.

Transformations of Absolute Value Functions

Catch-Up and Review

Investigating the Translations of Absolute Value Functions

Translations of Absolute Value Functions

Matching Graphs with Their Functions

Hint

Solution

Identifying and Correcting Error

Hint

Solution

Identifying the Translations of an Absolute Value Graph

Investigating the Transformations of an Absolute Value Function

Stretch and Shrink of a Function

Vertical Stretch and Shrink of a Function

Horizontal Stretch and Shrink of a Function

Stretch and Shrink of an Absolute Value Function

Draining and Refilling Water Tank

Hint

Solution

Reflection of a Function

Reflection of an Absolute Value Function

Construction via Reflection

Hint

Solution

Constructing an Absolute Value Function from a Linear Function

Transformations of Absolute Value Functions

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	14 Theory slides
	9 Exercises - Grade E - A
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