1. Unit 2 Day 10
FRED Functions – Part 2

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,

3. Homework Answers – Packet p. 11

4. 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

5. 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

6. 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}

7. 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)

8. 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

9. 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

10. 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.

11. 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

12. 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”

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

14. 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)

15. Practice p
Show answers under the document camera

16. Fred Functions Notes p

17. Checkpoint p. 35

18. Checkpoint p. 36

19. 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

20. Checkpoint p. 37

21. Homework Packet p