1. Algebra II
Unit 1

2. Functions and Relations
Unit 1.1
(Glencoe Chapter 2.1)
Algebra II

3. I. Function A relation in which x values are not repeated
Graph passes vertical line test
Page 60 #2, 4, 5, 6
#2 (fails vertical line test)
#4 Yes
#5 Yes
#6 No

4. I. Function Is it a function? 1. {(2, 7), (3, 7)} 2. {(1, 2), (1, -2)}
1. {(2, 7), (3, 7)}
2. {(1, 2), (1, -2)}
3. y = -2x + 1
how do you know?
4. x = y2 + 1
how can you tell?
(look at graph)
Yes No
Yes No

5. I. Function Domain and Range? 1. {(2, 7), (3, 7)} D:{2,3} R:{7}
1. {(2, 7), (3, 7)}
2. {(1, 2), (1, -2)}
3. y = -2x + 1
4. x = y2 + 1
D:{2,3} R:{7}
D:{1} R:{2, -2}
D: all reals
R: all reals
D: x  1
R: all reals

6. Take Five and You Try! Function? Domain and Range? 1. {(3, 6), (3, 7)}
1. {(3, 6), (3, 7)}
2. {(-1, 5), (1, 5)}
3. y = - ¾ x + 5
4. x = y2 – 2
No. D:{3} R:{6, 7}
Yes. D:{-1, 1} R:{5}
Yes. D: all reals
R: all reals
No. D: x  - 2
R: all reals

7. II. Evaluating Functions
Given: f(x) = 4x – 1, g(x) = x2 + 2x – 5
h(x) = x2 + 5x
1. f(3) =
2. 3f(-2)
3. g(-2) + f(2) =
4(3) – 1
= 11
3[4(-2) – 1]
= - 27
[(-2) 2 + 2(-2) – 5] + [4(-2) – 1]
= = 2

8. II. Evaluating Functions
Given: f(x) = 4x – 1, g(x) = x2 + 2x – 5
h(x) = x2 + 5x
4. g(3c2) =
5. h(m + 3) =
(3c2)2 + 2(3c2) – 5
9c4 + 6c2 – 5
(m + 3)2 + 5(m + 3)
m2 + 6m m + 15
m2 + 11m + 24

9. Take Five and You Try! Given: f(x) = 4x – 1, g(x) = x2 + 2x – 5
h(x) = x2 + 5x
6. g(x – 2) =
7. 7h(x) =
8. 5h(-1) =
(x – 2)2 + 2(x – 2) – 5
x2 – 4x x – 4 – 5
x2 – 2x – 5
7[x2 + 5x]
7x2 + 35x
5[(-1)2 + 5(-1)]
5[-4] = - 20

10. Take Five and You Try! Given f(x) = 2x + 1 and g(x) = 5x – 3
9. 4f(-1) + g(7) =
4[2(-1) + 1] + [5(7) – 3]
4[-1] + [32] 28

11. For today: Please be sure you staying up to date on your work!
Pg 60 – 61 #17 – 22, 28, 30, 32, 34
Pg 350 – 351 #28 – 34 even, 33

12. Operations on Functions: Composition of Functions
Unit 1.2
(Glencoe Chapter 7.1/7.7)
Algebra II

13. I. Graphs of Polynomial Functions
A. Even/ Odd Degree
Even: Quadratic Quartic deg. 2 deg. 4
Odd: Linear Cubic
deg. 1 deg. 3
Number of real roots  degree.
Limiting behavior (end behaviors) of graph determined by degree and leading coefficient.

14. I. Graphs of Polynomial Functions
Y = (x + 3)2
= x2 + 6x + 9
Only one zero!
B. Zeros
Real Zeros: x – intercepts
Non-Real Zeros: not x – intercept
#real + #non-real zeros = degree
Non-real roots are in pairs
Real Zeros located where curve crosses the x-axis

15. I. Graphs of Polynomial Functions
C. End Behavior
Describe what is happening at the ends of the graph
In terms of x (at infinity) and y (at infinity)
Page 350 #12 – 14
12. a a.
b b.
c c.
14.a. b. c.
f(x) → - ∞ as x →+ ∞
f(x) → + ∞ as x → - ∞
Odd 3
f(x) → + ∞ as x →+ ∞
f(x) → + ∞ as x → - ∞
even
f(x) → + ∞ as x →+ ∞
f(x) → - ∞ as x → - ∞
Odd 1

16. Take Ten: Page 351 #40 – 44 evens, 46 – 48
You may work with in your groups.

17. CHECK IT OUT: page 351 40. f(x) → + ∞ as x → + ∞ f(x) → + ∞ as x → - ∞
b. Even c. 4
42. f(x) → + ∞ as x → + ∞
f(x) → - ∞ as x → - ∞
b. Odd c. 5
44. f(x) → - ∞ as x → + ∞
b. Even c. 2

18. CHECK IT OUT: page 351 46. even 47. f(x) → - ∞ as x → + ∞
48. decrease…the graph appears to be turning downward at x = 30…so a decrease after 2000.

19. II. A. Arithmetic Operations on Functions
Given f(x) = x2 + 3x – 4 and g(x) = 3x – 2
(f + g)(x) means SUM: f(x) + g(x)
(f – g)(x) means DIFFERENCE: f(x) – g(x)
(f • g)(x) means PRODUCT: f(x) • g(x)
(f/g)(x) means QUOTIENT: f(x)
f (x) g(x)
g identify restrictions on denominator

20. II. A. Arithmetic Operations on Functions
Given f(x) = x2 + 3x – 4 and g(x) = 3x – 2
1. Find (f + g)(x)
2. Find (f – g)(x)
3. Find (f • g)(x)
4. Find (f/g)(x)
(x2 + 3x – 4) + (3x – 2)
x2 + 6x – 6
(x2 + 3x – 4) - (3x – 2)
x2 – 2
(x2 + 3x – 4)(3x – 2)
3x3 + 9x2 – 12x – 2x2 – 6x + 8
3x3 + 7x2 – 18x + 8
(x2 + 3x – 4)
(3x – 2) x – 2 ≠ 0 so x ≠ 2/3

21. II. B. Composition of Functions
1. Given f(x) = 2x + 7 and g(x) = - 5x – 1
Find [f ◦ g](x) and [g ◦ f](x)
2. Given f(x) = x2 + 2x and g(x) = x – 9
2( - 5x – 1 ) + 7
- 10x – = - 10x + 5
-5( 2x + 7 ) – 1
- 10x – 35 – 1 = - 10x – 36
( x – 9 )2 + 2( x – 9 )
x2 – 18x x – 18 = x2 – 16x + 63

22. Take Five and You Try! Given f(x) = x2 – 1, g(x) = - 4x2 and
h(x) = 3x – 2
3. Find [f ◦ h](x) and [h ◦ f](x)
4. Find [f ◦ g](x) and [g ◦ f](x)
5. Find (f – h)(x) and (f • h)(x)
2( x2 – 1 ) – 2
2x2 – 2 – 2 = 2x2 – 4
( 3x – 2 )2 - 1
9x2 – 12x + 4 – 1 = 9x2 12x + 3
( - 4x2 )2 – 1
16x4 – 1 = 16x4 – 1
-4( x2 – 1 )
- 4x2 + 4
( x2 – 1 ) – (3x – 2)
x2 – 3x + 1
( x2 – 1 ) (3x – 2)
3x3 – 2x2 – 3x + 2

23. Take Fifteen: Work in your Groups
Page 387 # 18, 30, 32, 34

24. Unit 1.3 (Chapter 2.6) Algebra II
Special Functions
Unit 1.3
(Chapter 2.6)
Algebra II

25. I. Three Functions to Consider
A. Absolute Value f(x) = |x|
B. Step or Greatest Integer
f(x) = x
2x if x > 1
C. Piecewise f(x) = x – if - 1≤ x ≤ 1
- 2x if x < 1

26. II. Graph Functions Domain: {x|x = all Reals} Range: {y|y ≥ -1}
A. Absolute Value
1. Use graphing calculator a table of values Or
2. Use x values to create a table of values
Graph f(x) = |2x| - 1

27. II. Graph Functions Domain: {x|x = all Reals}
Range: {y|y = all integers}
II. Graph Functions
B. Step or Greatest Integer
1. Use graphing calculator a table of values
2. Identify values between ends of “steps”
3. Identify which end of each step is open (open circle)
Graph f(x) = x - 2

28. II. Graph Functions Domain: {x|x = all reals} Range: {y|y ≤ 2}
Remember: two statements
two tables
two pieces of the graph
II. Graph Functions
x – 1 for x ≤ 3
x y
2 1
1 0
C . Piecewise
1. Create a table of values for each “piece”
2. Use the x values for each “piece” indicated by the statement attached to the “piece”.
3. Graph each table
4. Watch for the need for open vs. closed circles at an end of a “piece”
Graph g(x) = x – 1 if x ≤ 3
if x > 3
– 1 for x > 3
x y
(open circle because >)
Domain: {x|x = all reals}
Range: {y|y ≤ 2}

29. Today’s Practice:
Textbook Pages 93 – 94 # 16 – 20 and #24, 34, and 40. You will need graph paper!
You will need graph paper!
(answers are in the back)

30. Absolute Value Equations and Inequalities, Compound Inequalities
Unit 1.4
(Chapter 1.4, 1.6)
Algebra II

31. I. Absolute Value Equations
A. Absolute Value
1. always a positive value: do you know why?
Graph |3| on a number line:
2. Two values make the statement |x| = 3 correct: Both 3 and – 3

32. I. Absolute Value Equations
B. Absolute Value Equations
1. must be solved for both possible solutions: rewrite and solve each one
2. solutions must be checked to evaluate reasonableness
EX: |y + 3| = 8

33. I. Absolute Value Equations
C. Solve
1. |m – 5| = 12
m – 5 = 12 or m – 5 = - 12
m = 17, - 7
2. |-8x – 3| = 1
-8x – 3 = 1 or = 8x – 3 = - 1
x = - ½ , ¼
3. 7|4x – 13| = divide 7 first!
|4x – 13| = 5
4x – 13 = 5 or 4x – 13 = - 5
x = 9/2 , x = x = 2

34. I. Absolute Value Equations
D. Other cases
1. |5x – 6| + 9 = is this possible?
|5x – 6| = NO!
2. 2|8 + y| = 4y – 6 do both answers
|8 + y| = 2y – 3 work?
8 + y = 2y – 3 or 8 + y = -2y + 3
y = 11, - 5/3
3. |5x + 4| = will you have two
NO! answers?

35. Take Ten:
Textbook Page 31
#32 – 42 evens

36. II. Compound Inequalities
A. Differences
1. “And” means intersection (what is common to both)
2. “Or” means union (what is in either one)
Which is which?:
13< 2x + 7 ≤ y – 2 > -3 or y + 4 ≤ -3
and

37. II. Compound Inequalities
B. Solve
≤ 3y – 2 < 19
12 ≤ 3y < 21
4 ≤ y < 7 this is an and statement!
2. x + 3 < 2 or – x ≤ - 4
x < -1 or x ≥ 4

38. III. Absolute Value Inequalities
1. Consider: |d| < 3
2. Consider: |d| > 3
So, if |a| < b then – b < a < b (a < b and a > - b )
and if |a| > b then a > b or a < - b
FOR ALL REAL NUMBERS WHERE b > O!

39. III. Absolute Value Inequalities
Remember to check for
reasonableness of your
solutions!
A. Solve by rewriting
1. |2x – 2| ≥ 4
Rewrite: 2x – 2 ≥ 4 or 2x – 2 ≤ - 4
x ≥ 3 or x ≤ - 1
2. |4k – 8| < 20
Rewrite: 4k – 8 < 20 and 4k – 8 > - 20
k < 7 and k > -3

40. For Today’s Practice: Work on your review! If you finish:
Quiz tomorrow?
Page 30 # 8 – 13 all
Page 44 # 28 – 40 evens
Remember to check the reasonableness of your solutions!
And the answers are in the back.

41. Unit 1.5 (Chapter 5.4) Algebra II
Factoring Basics
Unit 1.5
(Chapter 5.4)
Algebra II

42. Remember: Product vs. Factors
3 • 5 is 3 fives to make 15!
The factors of 15 then are 3 and 5 because 3 • 5 makes 15!
Product: (x + 2)(x + 5)
x2 + 5x + 2x + 10
x2 + 7x + 10
Factor: x2 + 7x + 10
(x + 2)(x + 5)

43. I. Factor with GCF A. GCF only 1. 2xy2 – 4x2y 2. 3x2y – 6xy + 9xy2
2xy(y – 2x)
3xy(x – 2x + 3y)

44. II. Factor Trinomials (m + 8)(m – 3) (x + 2)(x – 9) (y + 9)(y + 3)
A. Factor to two binomials
Guess and Check/ Product – Sum
Check for GCF!
Inner and Outer Products must add to the middle term!
1. m2 + 5m – 24
2. x2 – 7x – 18
3. y2 + 12y + 27
(m + 8)(m – 3)
(x + 2)(x – 9)
(y + 9)(y + 3)

45. II. Factor Trinomials
B. GCF first, then factor to two binomials: Inner and Outer products must add to the middle term!
1. 2m2 + 20m + 48
2. 3x2 – 3x – 60
2(m2 + 10m + 24)
2(m + 6)(m + 4)
3(x2 – x – 20)
3(x + 4)(m – 5)

46. III. Factor Special Cases
A. Difference of Two Squares
If two squares with a minus: take roots and one is minus/one is plus!
1. 49m2 – 16
x2 – 9y2
3. 28x3 – 7x GCF?
(7m + 4)(7m – 4)
(5x + 3y)(5x – 3y)
7x(2x + 1)(2x – 1)

47. III. Factor Special Cases
B. Perfect Square Trinomial: first and last terms are square and middle is twice the product of their roots.
1. x2 – 10x + 25
2. 16x2 – 40x + 25
3. 8x2 + 24x GCF?
√x2 = x and √25 = 5
AND 5 • x • 2 = 10 so
(x – 5)2 use the middle sign!
(4x – 5)2
2(4x2 + 12x + 9)
2(2x + 3)2

48. Your assignment:
Page 8 of your notes