the quadratic equation - qubee · the quadratic equation ... unit: ax.1 & ay. c„, ... ertica...

•

The Quadratic Equation Packet

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Graphing, Translation

Solving

Mr. Robi

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Date: Course: Mr. Robi Name:

Unit: ax.1 & ay. c„, Topic: tiNsi-vou et-iv ^ Ac ewe. (4 tr

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Part 2: Feedback (Check the one that applies and explain)

— 3: I understand today's lesson because

2: I somewhat understand today's lesson because

1: I did not understand today's lesson because

Class work completed (+10): Remained in class (+5): Cell-Phone: Out of Sight (+5):

Quadratic Functions

Points/Questions Notes

What are the properties of a quadratic function?

It is the simplest function.

A quadratic function is also called a degree function.

Graph of a quadratic function is called a .

Parabolas open up or • ( / face)

Parabola has a or a called

of a parabola is the lowest or the point on the graph of a quadratic function.

When the parabola opens up, it has a value, because the lowest point on the parabola is at the vertex.

When the parabola opens down, it has a value because the highest point on the parabola is at the vertex.

2 For what purpose are quadratic functions used feE•?

Construction --)

Defense 4

Games 4 (student real-life situation)

LI .>

4 ct

z ci4

O.

Pair with your shoulder partner and Share (A & B)

Share your answers w/ class.

A) Tell your friend three things you learned about parabola.

B) List two uses of quadratic functions in real-life situation.

•

How is a quadratic equation written in the standard-form?

f(x) = ax2 + bx + c

f(x) = 3x2 + 6x + 3 0. .... ___.) 6: L,

How do we call the different terms of a quadratic function?

f(x) = ax2 + bx + c f(x) = 3x2 + 6x + 3

is the quadratic term.

is the middle (linear) term

is the constant term.

Which term of the quadratic function is the most important?

The term is the most important. A quadratic constant term and/or the

the term. function can prevail without the linear term, but never without

How do you find the axis of symmetry of a parabola?

A quadratic function is symmetric. called the of symmetry. _

The line of symmetry is For a graph written in the

+ c, the axis of symmetry is

3x2 + 6x + 2, equals L:

standard-form, f(x) = ax2 + bx at:

x = -b/2a

Ex: Axis of symmetry of f(x) = a • , I la 1 —

x = -b/2a =

icrs

,.= z .,'.... at

Pi

Each team solves for the axis of symmetry and shares with the entire class.

f(x) = 2x2 + 6x + 8 f(x) = 2x2 — 1 OX + 3

•

•

•

Given a quadratic function in the standard form, how do you find coordinates of the vertex of the parabola?

Steps: 1 st) find the axis of symmetry. 2"d) find the y coordinate of the vertex.

f(x) = 3x2 - 6x + 4 f(x)=2x2+6x+8

40

4 v) 04 1.

0.

Graphing the parent function.

•

Find the the axis of Symmetry and of the Quadratic Functions —> x = -b 2a

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Translation of Quadratic Functions

• Vertical translation • Horizontal Translation • Composite Translation

6) X-2

2

What is the difference between graph #3 and #4?

What caused the difference between the two graphs?

2) 2,

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F F , 1C. • Graphing the Quadratic Terms — (Parent)

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•

• Name: Date: Algebra 2

Consolidating Graphs of the Quadratic Term with Different Leading Coefficients

In this graph, you will explore the behavior of graphs of quadratic functions as the leading coefficient changes.

Graph on the coordinate

1) y = X2

2) y = 2x2

3) y = 0.5x2

4) y = - x2

5) y = -2x2

6) y = -0.5x2

What happens to the graph as the positive leading coefficient increases?

What happens to the graph as the negative leading coefficient increases?

•

• Name:

Date: Algebra 2

Consolidating Graphs of the Quadratic Term with Different Leading Coefficients

•

In this graph, you will explore the behavior of graphs of quadratic functions as the leading coefficient changes.

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Graph on the coordinate

1) y = x2

2) y = 2x2

2-

3) y = 0.5x2

4) y = - x2

5) y = -2x2

6) y = -0.5x2

0 .r6i 62-

What did you note simt6tbehavior of the graphs as the positive leading coefficient increases?

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What did you note about the behavior of the Ar41444.3ehe negative leading coefficient increases?

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•

r F. 7a.

ertica ranslation 4 Moving the Quadratic Functions Up or Down

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•

Algebra 2

Assignment

• Sketch the graph of each function.

1) y = (x — 1) 2 + 3

AY

<-3 3 4 5 x

3) y = (x — 4) 2 + 3

2) y=—(x+1)2+4

4) y = (x — 1) 2 — 2

Date

Name ID: 1

Period

6)

1

0.5

-0.5

- I

-1.5

-2.5

-35

-40

-4.5

y = (x — 2) 2 — 4

2 4 S 6 x

5

y

5) y = (x — 2) 2 + 1

AY

•

2 0 4 7) x

3

2

6

5

4

3

11) y -(x - 3) 2 - 1

y

12) y = -(x + 1) 2 + 1

-0.5

-1.D

-3

-3.5

4

-4.5

-5

5

x

7) y = (x - 1) 2 - 4

8) y = (x + 2) 2 - 1

9) y = (x + 1) 2 - 1

10) y =(x + 1) 2 - 4

Quadratics Test 1 on

The Basics and Translations

•

•

•

Solving Quadratic Functions

By:

• Graphing. • Taking square-roots

• • Factoring • Completing the Square • Quadratic Formula

•

•

Algebra 2 Name ID: I

AREI 4b Solving QE by TakingSquareRoots Date Period

• Solve each equation by taking square roots.

1) 49x 2 = 36 2) —7m 2 = —42

3) —5x 2 = —240 4) n 2 — 1 = 15

5) r2 + 8 = 72

6) x 2 — 6 = 34

•

7) —4n 2 = —24 8) b 2 — 1 =48

9) n 2 + 8 = 65

10) 64r2 = 4

•

Algebra 2 Name ID: I

AREI 4b Solving QE by TakingSquareRoots Date Period

Solve each equation by taking square roots.

1) 49x 2 = 36 2) —7m 2 = —42

3) —5x 2 = —240 4) n 2 — 1 = 15

5) r 2 + 8 = 72 6) x 2 — 6 = 34

7) —4n 2 = —24

8) b 2 — 1 = 48

9) n 2 + 8 = 65 10) 64r2 = 4

•

4) n 2 - 1 = 15` ̀

-k n

• 5) r2 + 8 = 72 -3 -S

6) x2 - 6 =34 +b -00

9) n' +8= 65 -8 -a

10) 64r 2 = 4

--- 2

Algebra 2 Name ID: 1

AREI_4b_Solving QE by TakingSquareRoots Date Period

Solve each equation by taking square roots.

1) 49x 2 = 36 2) -7m 2 = -42 —.71- 7:1 4-t 7.- 3 6

1„,k4 (44 oft1.- G

tnkn.))

•

36 (tR

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S-t.0 - 1-5-1.0

7) -4/12 =-24 8) b 2 - 1 = 48 .04

t\-11 6 ‘47- 1

6 - ‘16

Name: Date: Course: Mr. Robi

•

•

•

Unit: Topic:

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Solving Quadratic Equations By Graphing

2)

3)(4 7: -1(-1- +

44)

(

5)

6 6) .-x

1

•Algebra 2 - ASSE3a

Factoring Quadratic Equations

Factor each completely.

1) n 2 + 8n — 20

Name

2) p2 — 17p + 70

Date

3) n 2 — 13n + 40 4) v2 — 5v — 24

• 5) p2 + 3 p — 28 6) a2 — 16a + 60

7) n 2 — 5n + 6 8) b 2 — 2b — 8

9) r 2 + 16r + 60

10) k2 — 14k + 45

•

Algebra 2 Name ID: 1

AREI_3b_Completing the Square Date Period

Solve each equation by completing the square.

1) r2 — 6r — 28 = 0 2) n 2 — 6n + 8 = 0

3) x2 — 16x + 28 = 0

4) x2 + 2x — 3 = 0

• 5) v 2 + 10v — 15 = 0

6) b 2 + 2b — 23 = 0

7) x2 + 2x — 8 = 0

8) m 2 — 16m + 33 = 0

9) n 2 + 10n + 24 = 0

10) n 2 — 6n — 16 = 0

•

Name ID: 1 Algebra 2 - AREI3a

Solving QE By Quadratic Formula Date Period

Solve each equation with the quadratic formula.

1) 2v2 + v — 10 = 0

2) 2r 2 — 3r + 1 =0

3) 2k2 — 3k — 9 = 0

4) 2p 2 + 5p — 3 = 0

• 5) 2x 2 — 5x — 25 = 0

6) 2a2 — 3a — 20 = 0

7) b 2 — 4b + 4 = 0

8) n 2 -I- 2n — 8 = 0

9) n 2 — 4n — 5 = 0

10) x2 — 4x + 3 = 0

•

•

Quadratics Test 2 on

The Basics and Translations and

• Solutions

•

the quadratic equation - qubee · the quadratic equation ... unit: ax.1 & ay. c„, ... ertica...

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