1. Absolute value and reciprocal functions
Chapter 7

2. example
How many triangles are in the diagram below?

3. 7.1 – Absolute value
Chapter 7

4. |3| |–7| Absolute value Evaluate:
For a real number a, the absolute value is written as |a| and is a positive number. For example:
|5| = 5
|–5| = 5
Absolute value can be used to represent the distance of a number from zero on a real-number line.
Evaluate:
|3|
|–7|

5. example Evaluate the following:
a) |4| – |–6| b) 5 – 3|2 – 7| c) |–2(5 – 7)2 + 6|
|4| – |–6| = 4 – 6 = –2
b) 5 – 3|2 – 7| = 5 – 3|–5| = 5 – 3(5) = 5 – 15 = –10
c) Try it!

6. Pg #1, 6, 7(A,C,E), 11, 16
Independent Practice

7. handout
Answer the questions on the “Investigating Absolute Value Functions” worksheet to the best of your ability.

8. 7.2 – absolute value functions
Chapter 7

10. Absolute value functions
For what values of x is the function y = |x| equivalent to y = x?
 when x ≥ 0
When x < 0, what is the function represented by y = |x|?
 y = –x
We can write this as a piecewise function:

11. example Consider the absolute value function y = |2x – 3|.
Determine the y-intercept and the x-intercept.
Sketch the graph.
State the domain and range.
Express as a piecewise function.
The y-intercept is at x = 0.
y = |2x – 3|
y = |2(0) – 3|
y = |–3|
y = 3
The y-intercept is (0, 3).
The x-intercept is at y = 0.
0 = |2x – 3|
0 = 2x – 3
x = 3/2
The x-intercept is at (3/2, 0). b) x y -1 5 3
3/2 4

12. example Consider the absolute value function y = |2x – 3|.
Determine the y-intercept and the x-intercept.
Sketch the graph.
State the domain and range.
Express as a piecewise function. b) x y -1 5 3
3/2 4
D: {x | x E R}
R: {y | y ≥ 0, y E R}
d) The equation on the right is just y = 2x – 3.
What’s the one on the left?
It’s just –(2x – 3)!
The x-intercept is call an invariant point because it’s a part of both functions.

13. example Consider the absolute value function f(x) = |–x2 + 2x + 8|.
Determine the y-intercept and the x-intercepts.
Sketch the graph.
State the domain and range.
Express as a piecewise function
The y-intercept is at x = 0.
f(0) = |–(0)2 + 2(0) + 8|
= |8|
= 8
The x-intercepts are when y = 0.
0 = |–x2 + 2x + 8|
0 = –x2 + 2x + 8
0 = –(x – 4)(x + 2)
x = 4
x = –2
b) What’s the vertex of the function
f(x) = –x2 + 2x + 8
Use your calculator, or complete the square:
f(x) = –(x2 – 2x) + 8
f(x) = –(x2 – 2x + 1 – 1 ) + 8
f(x) = –(x2 – 2x + 1)
f(x) = –(x – 1)2 + 9
The vertex is (1, 9)

14. example Consider the absolute value function f(x) = |–x2 + 2x + 8|.
Determine the y-intercept and the x-intercepts.
Sketch the graph.
State the domain and range.
Express as a piecewise function
Recall:
y-intercept is (0, 8)
x-intercepts are (4, 0) and (–2, 0)
Vertex is (1, 9)
c) D: {x | x E R}
R: {y | y ≥ 0, y E R} d)

15. Pg , #2, 5, 7, 10,
Independent Practice

16. 7.3 - Absolute value equations
Chapter 7

17. Absolute value equations
When solving equations that involve absolute value equations you need to consider two cases:
Case 1: The expression inside the absolute value symbol is positive or zero.
Case 2: The expression inside the absolute value symbol is negative.

18. Example The solution is x = 10 or x = –4. Solve: |x – 3| = 7
Consider the equation as a piecewise function:
Case 1:
x – 3 = 7
 x = 10
Case 2:
–(x – 3) = 7
 x – 3 = –7
 x = –4
The solution is x = 10 or x = –4.

19. Try it
Solve |6 – x| = 2

20. example Solve |2x – 5| = 5 – 3x What is the x-intercept of y = 2x – 5?
Case 1: (x ≥ 5/2)
2x – 5 = 5 – 3x
5x = 10
x = 2
Case 2: (x < 5/2)
–(2x – 5) = 5 – 3x
–2x + 5 = 5 – 3x
x = 0
Consider:

21. example Solve: |3x – 4| + 12 = 9 |3x – 4| = –3
Is there any possible way that the absolute value of something is equal to –3?
 No solution.

22. example
Solve: |x – 10| = x2 – 10x

23. Pg , #4, 5, 6, 9, 11, 22, 23
Independent Practice

24. 7.4 – reciprocal functions
Chapter 7

25. example
Sketch the graphs of y = f(x) and its reciprocal function y = 1/f(x), where f(x) = x. Examine how the functions are related. x
y = x
y = 1/x –5
–1/5 –2
–1/2 –1
–1/10
–10
Undef.
1/10 10
1/2 2 1

26. Reciprocal functions What is the vertical asymptote?
An asymptote is a line whose distance from a curve approaches zero.
This graph has two pieces, that both approach the vertical asymptote, which is defined by the non-permissible value of domain of the function, and a horizontal asymptote, defined by the value that is not in the range of the function.
What is the vertical asymptote?
What is the horizontal asymptote?

27. example Consider f(x) = 2x + 5.
Determine its reciprocal function y = 1/f(x).
Determine the equation of the vertical asymptote of the reciprocal function.
Graph the function y = f(x) and its reciprocal function y = 1/f(x).
a) The reciprocal function is: c)
Characteristic
f(x) = 2x +5
f(x)=1/(2x + 5)
x-intercept/asymptotes
x-intercept at x = –5/2
Asymptote at x = –5/2
Invariant points
2x + 5 = 1
x = –2
(–2, 1)
(–2,1)
2x + 5 = –1
x = –3
(–3,–1)
b) The vertical asymptote is always the non-permissible values of the function.
2x + 5 = 0
2x = –5
x = –5/2
Invariant points are at y = 1 and y = –1.
There is a vertical asymptote at x = –5/2

28. example Consider f(x) = 2x + 5.
Determine its reciprocal function y = 1/f(x).
Determine the equation of the vertical asymptote of the reciprocal function.
Graph the function y = f(x) and its reciprocal function y = 1/f(x).
Characteristic
f(x) = 2x +5
f(x)=1/(2x + 5)
x-intercept/asymptotes
x-intercept at x = –5/2
Asymptote at x = –5/2
Invariant points
2x + 5 = 1
x = –2
(–2, 1)
(–2,1)
2x + 5 = –1
x = –3
(–3,–1)

29. example Consider f(x) = x2 – 4.
What is the reciprocal function of f(x)?
State the non-permissible values of x and the equation(s) of the vertical asymptote(s) of the reciprocal function.
What are the x-intercepts and y-intercepts of the reciprocal function?
Graph the functions. a)
c) How can I find the x-intercept of the the reciprocal?
Let f(x) = 0
There is no solution, so there is no x-intercept.
y-intercept:
What are the non-permissible values?
x2 – 4 = 0
(x – 2)(x + 2) = 0
x = 2
x = –2
The vertical asymptotes are at x = ±2
y-intercept is y = –1/4

30. Pg , #3, 5, 7, 8, 9, 10, 12
Independent practice