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Algebra II

Algebra II. Unit 1. Functions and Relations. Unit 1.1 (Glencoe Chapter 2.1) Algebra II. 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. I. Function.

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Algebra II

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  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)} • 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)} • 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)} • 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] = -5 + 7 = 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 + 9 + 5m + 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 + 4 + 2x – 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. Please be sure you staying up to date on your work! For today: • 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. 13.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 0 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 → + ∞ f(x) → - ∞ as x → - ∞ b. Even c. 2

  18. CHECK IT OUT: page 351 • 46. even • 47. f(x) → - ∞ as x → + ∞ 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) 3x – 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 • Find [f ◦ g](x) and [g ◦ f](x) 2( - 5x – 1 ) + 7 - 10x – 2 + 7 = - 10x + 5 -5( 2x + 7 ) – 1 - 10x – 35 – 1 = - 10x – 36 ( x – 9 )2 + 2( x – 9 ) x2 – 18x + 81 + 2x – 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. 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 – 1 if - 1≤ x ≤ 1 - 2x if x < 1

  26. Domain: {x|x = all Reals} Range: {y|y ≥ -1} II. Graph Functions 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. 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. Remember: two statements two tables two pieces of the graph II. Graph Functions x – 1 for x ≤ 3 x y 3 2 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 - 1 if x > 3 – 1 for x > 3 x y 3 -1 (open circle because >) 4 -1 5 -1 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: -4 -3 -2 -1 0 1 2 3 4 • 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| = 35 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 = 0 is this possible? |5x – 6| = -9 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| = 0 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 ≤ 17 y – 2 > -3 or y + 4 ≤ -3 and

  37. II. Compound Inequalities B. Solve • 1. 10 ≤ 3y – 2 < 19 +2 +2 +2 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 -3 -2 -1 0 1 2 3 • 2. Consider: |d| > 3 -3 -2 -1 0 1 2 3 • So, if |a| < b then – b < a < b (a < b anda > - 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: If you finish: Work on your review! 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. 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 • 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 • 2. 25x2 – 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 + 18 GCF? √x2 = x and √25 = 5 AND 5 • x • 2 = 10 so (x – 5)2use the middle sign! (4x – 5)2 2(4x2 + 12x + 9) 2(2x + 3)2

  48. Your assignment: • Page 8 of your notes

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