1. WARM UP If f(x) = 2x, then If f(x) = x - 3,
If f(x) = x , write the equation for the inverse relation
If f(x) = 2 , then

2. REFLECTIONS, ABSOLUTE VALUE & OTHER TRANSFORMATIONS

3. OBJECTIVES
Given a function, transform it by reflection and by applying absolute value to the function or its argument.
Use the graphs of familiar functions to to demonstrate how y = |f(x)| transformation affects the range and y = f|(f)|
Understand the concept of even and odd functions

4. TERMS & CONCEPTS Reflection Even function Reflection across the y-axis
Odd function
Reflection across the x-axis
Step discontinuity
Displacement
Greatest integer function, f(x) = |x|
Piecewise function
Absolute value transformations

5. OBJECTIVES
In Section 1-3, you learned that if y = f (x), then multiplying x by a constant causes a horizontal dilation. Suppose that the constant is −1. Each x-value will  be 1/(−1) or −1 times what it was in the pre-image.
The graph shows that   the resulting   image is a horizontal reflection of the graph across the y-axis. The  new graph is the same size and shape, simply a mirror image of the original.   Similarly, a vertical dilation by a factor of −1 reflects the graph vertically across  the x-axis.
Horizontal reflection y = f(-x)
Vertical reflection y = -f(x)
In this section you will learn special transformations of functions that reflect their graphs in various ways and what happens to the absolute value of a function or of the independent variable x.

6. EXAMPLE 1
The pre-image function y = f (x) in Figure 1-6a is f (x) = x2 − 8x + 17, where 2 ≤ x ≤ 5.
Write an equation for the reflection of this function across the y-axis.
Write an equation for the reflection of this function across the x-axis.
c. Plot the pre-image and the two reflections on the same screen

7. SOLUTION
The pre-image function y = f (x) in Figure 1-6a is f (x) = x2 − 8x + 17, where 2 ≤ x ≤ 5. a. Write an equation for the reflection of this function across the y axis. A reflection across the y – axis is a horizontal dilation by a factor of -1. So,
y = f (−x) = (−x)2 − 8(−x) + 17
Substitute –x for x
y = x2 + 8x + 17
Domain: 2 ≤ − x ≤ 5
−2 ≥ x ≥ − 5 or − 5 ≤ x ≤ − 2
Multiply all three sides of the inequality by −1. The inequalities reverse.

8. SOLUTION
The pre-image function y = f (x) in Figure 1-6a is f (x) = x2 − 8x + 17, where 2 ≤ x ≤ 5. b. Write an equation for the reflection of this function across the x-axis.
For a reflection across the x-axis, find the opposite of f(x)
y = -f(x)
y = x2 + 8x + 17
The domain remains 2 < x < 5
b. Plot the pre-image and the two reflections on the same screen
y1 = x2 − 8x + 17 / (x ≥ 2 and x ≤ 5)
Divide by the Boolean variable to restrict the domain.
y2 = x2 + 8x + 17 / (x ≥ −5 and x ≤ −2)
y3 = −x2 + 8x − 17 / (x ≥ 2 and x ≤ 5)

9. GRAPH OF THE SOLUTION y2 and y3. The graphs are shown below
You can check the algebraic solutions by plotting y4 = y1(−x) and y5 = −y1(x) using thick style. The graphs should overlay
Horizontal reflection y = f(-x)
y2 and y3.  
Vertical reflection y = -f(x)