Unit 2 Day 10 FRED Functions – Part 2

Warm Up 11. Using the discriminant, determine the amount and type of solutions each equation will have. Then find the exact value of the solutions. a. x2 + 4x + 5 = 0 b. x2 – 2x + 1 = 0 c. 2x2 – 3x – 10 = 0 Discriminant = -4, 2 complex roots, Discriminant = 0, 1 real rational root, x = 1 Discriminant = 89, 2 real irrational roots,

Homework Answers – Packet p. 11

Homework Answers – Packet p. 11 Equation Effect to Harry’s graph 1. y=F(x) + 82 Translate up 82 2. y = F(x-13) Translate right 13 3. y = F(x + 9) Translate left 9 4. y = F(x) - 55 Translate down 55 5. y = F(x – 25) + 11 Translate right 25, up 11

Homework Answers – Packet p. 12 Equation Effect to Harry’s graph y = F(x + 51) Translate left 51 y = F(x) - 76 Translate down 76 y = F(x – 31) Translate right 31 y = F(x – 8) - 54 Translate right 8 and down 54 y = F(x + 100) - 12 Translate down 12 an left 100

Homework Answers – Packet p. 12 IV. D: {x │-1 ≤ x ≤ 3} R: {y │ -5 ≤ y ≤ 3} 2. D: {x │ -3 ≤ x ≤ 5} R: {y │ -3 ≤ y ≤ 2} V. D: {x │-2 ≤ x ≤ 2} R: {y │ 2 ≤ y ≤ 6} 2. D: {x │ -7 ≤ x ≤ -3} R: {y │ -3 ≤ y ≤ 1}

Homework Answers – Packet p. 1 10. 4n2(7n – 10)(n + 2) 11. b(3b – 2)(b – 1) 12. (7x + 10)(x – 6) 13. 3b(2n – 5)(5n – 2) 14. prime 15. r(9p + 10)(p + 7) 16. prime 17. x(4x + 3)(x + 10) 18. (10m – 1)(m + 9)

Checkpoint p. 30 Equation Effect to Fred’s graph Example: y=F(x) + 18 Translate up 18 units y = F(x) – 100 Translate down 100 units 2. y = F(x) + 73 Translate up 73 units 3. y = F(x) + 32 Translate up 32 units 4. y = F(x) – 521 Translate down 521 units

Translate left 30 units and up 18 units Checkpoint p. 31 Equation Effect to Fred’s graph Example: y=F(x + 18) Translate left 18 units y = F(x – 10) Translate right 10 units 2. y = F(x) + 7 Translate up 7 units 3. y = F(x + 48) Translate left 48 units 4. y = F(x) – 22 Translate down 22 units 5. y = F(x + 30) + 18 Translate left 30 units and up 18 units

Translate down 2 and right 6 Checkpoint p. 32 Equation Effect to Fred’s graph Example: y=F(x + 8) Translate left 8 units y = F(x) + 29 Translate up 29 units 2. y = F(x – 7) Translate right 7 3. y = F(x + 45) Translate left 45 4. y = F(x+5) + 14 Translate left 5 and up 14 5. y = F(x – 6) – 2 Translate down 2 and right 6 Use teacher notes (document) on Fred Functions Day 1 to check student answers.

An Exploration of Functions Plug the following functions into your calculator and describe how they change from y = x2. y = (x + 3)2 y = (x – 5)2 y = x2 – 4 y = x2 + 1 y = (x – 3)2 + 5 y = 3x2 y = ½x2 y = -x2 y = (-x – 3)2 y = (-x + 4)2 Translate Left 3 Translate Right 5 Translate down 4 Translate up 1 Translate Right 3, Up 5 Vertical stretch by 3 Vertical compression by ½ Reflect over x-axis Right 3, Reflect over y-axis Left 4, Reflect over y-axis

Summary x2 + a is f(x) = x2  shifted upward a units x2 – a is f(x) = x2  shifted downward a units (x + a) 2 is f(x) = x2  shifted left a units (x – a) 2 is f(x) = x2  shifted right a units –x2 is f(x) = x2  flipped upside down ("reflected about the x-axis") f(–x) is the mirror of f(x) = x2  ("reflected about the y-axis") a f(x) is f(x) = x2 when a > 1, is a “vertical stretch by a” a f(x) is f(x) = x2 when 0 < a < 1, is a “vertical stretch by a”

Vertex Form y = a(x – h)2 + k, where (h, k) is the vertex.

An Exploration of Functions What is the vertex? y = a(x – h)2 + k, where (h, k) is the vertex. y = (x + 3)2 y = (x – 5)2 y = x2 – 4 y = x2 + 1 y = (x – 3)2 + 5 y = 3x2 y = ½x2 y = -x2 y = (-x – 3)2 y = (-x + 4)2 (-3,0) (5,0) (0,-4) (0,1) (3,5) (0,0) (0,0) (0,0) (-3,0) (4,0)

Practice p. 38-39 Show answers under the document camera

Fred Functions Notes p. 33-37

Checkpoint p. 35

Checkpoint p. 36

Checkpoint p. 37 Reflect over x-axis, vertical stretch by 5 Equation Effect to Harry’s graph Example: y=-5H(x) Reflect over x-axis, vertical stretch by 5 a. y = 2H(x) Vertical stretch by 2 b. y = -2H(x) Reflect over x-axis, vertical stretch by 2 c. y = 1/2H(x) Vertical compression by 1/2

Checkpoint p. 37

Homework Packet p. 11-13