1 QUADRATIC FUNCTIONS PROBLEM 1 PROBLEM 2 PROBLEM 3 INTRODUCTION Standard 4, 9, 17 PROBLEM 4 PROBLEM 5 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All

1

QUADRATIC FUNCTIONS

PROBLEM 1 PROBLEM 2

PROBLEM 3

INTRODUCTION

Standard 4, 9, 17

PROBLEM 4

PROBLEM 5

END SHOW

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

PROBLEM 6

PROBLEM 7

http://www.mrperezonlinemathtutor.com/

2

STANDARD 4:

Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes

ESTÁNDAR 4:

Los estudiantes factorizan polinomios representando diferencia de cuadrados, trinomios cuadrados perfectos, y la suma de diferencia de cubos.

ALGEBRA II STANDARDS THIS LESSON AIMS:



3

x y (x,y)

-1

0

1

2

Evaluate the following equation for the given domain and then determine if it represents a function. y = 2x - 2x - 12 D={-1, 0, 1, 2}

2x - 2x - 12

22( ) -2( )- 1

22( ) -2( )- 1

22( ) -2( )- 1

22( ) -2( )- 1

-1-1

0 0

1 1

22

-1

3

-1

3

(-1,3)(0,-1)

(1,-1)

(2,3)42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

We can guess that is a parabola and observe that all over the x values, it has only one corresponding y value. So IT IS A FUNCTION.We can also perform the vertical line test and verify that it is a function because it crosses the curve a only one point.


REVIEW


4

y x (x-axis,y-axis)

-1

0

1

2

Evaluate the following equation for the given domain and then determine if it represents a function. x = 2y - 2y - 12 D={-1, 0, 1, 2}

2y - 2y - 12

22( ) -2( )- 1

22( ) -2( )- 1

22( ) -2( )- 1

22( ) -2( )- 1

-1-1

0 0

1 1

22

-1

3

-1

3

(3,-1)(-1,0)

(-1,1)

(3,2)42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

We can guess that is a parabola that opens to the right but most values in the domain have two values in the range. So it is not a FUNCTION.We can also perform the vertical line test and verify that it is NOT a function because it crosses the curve at MORE THAN ONE POINT.


REVIEW


5

Standard 4Features of the graph of a quadratic function (parabola):

axis of symmetry

x

y

vertex


root zero solution

root zero solution


6

Standard 4


Identify: Quadratic, Lineal and Constant terms.

Quadratic term

Lineal term

Constant term

y = 2x - 2x - 12

Quadratic term

Lineal term

Constant term

y = 4x +3x +82


7

Standard 4


Graph the following using the graph features of the quadratic function:

y = x + 4x + 12

y = ax + bx + c2

a = 1

b = 4

c = 1

Finding the axis of symmetry:

x = – b

2a

x = – ( )

2( )

4

1

x = – 4

2

x = – 2

Finding the vertex:

y = x + 4x + 12

y = ( ) + 4( ) + 12– 2 – 2

y = 4 – 8 + 1

y = – 3

Vertex:(– 2, – 3)Axis of symmetry:

Finding a point to the right of the vertex:

Using x = 0

y = ( ) + 4( ) + 120 0

y = 1

( 0, 1)

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

yx= -2

(– 2, – 3)

( 0, 1 )


8

Standard 4


Upward if a>0

Downward if a<0Direction of opening

(h,k)Vertex

x= hAxis of symmetry

y = a(x-h) + kForm of equation

VERTEX FORM EQUATION2


9

Standard 4


Graph the following using vertex form of the quadratic function:

y = x + 4x + 12

y = (x + 4x + )+ 1 – 242

2 42

2

y = (x + 4x + )+ 1 – 2 (2)2(2)2

y = (x + 4x + )+ 1 – 2 4 (4)

y = (x + 2) + 1 – 42

y = (x + 2) – 32

Rewriting the equation:

y = (x - -2) + (-3)2

y = a(x-h) + k2

h= -2k= -3a= 1

Vertex: (h,k) = (-2,-3)

Axis of symmetry: x= -2

Finding a point to the right of the vertex:

Using x = 0

y = ( ) + 4( ) + 120 0

y = 1

( 0, 1)

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

yx= -2

(– 2, – 3)

( 0, 1 )


10

Standard 4


Graph the following function by finding the vertex, axis of symmetry and roots:

y = x + 2x – 8 2

y = (x + 2x + ) – 8 – 222

2 22

2

y = (x + 2x + ) – 8 – 2 (1)2(1)2

y = (x + 2x + ) – 8 – 2 1 (1)

y = (x + 1) – 8 – 12

y = (x + 1) – 92


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

y = a(x-h) + k2

h= -1k= -9a= 1

Vertex: (h,k) = (-1,-9)


Finding the roots by factoring:

x + 2x – 8 = 02

– 8 + 2

(-1)(8) -1+ 8 = +7(-2)(4) -2+ 4 = +2(x – 2)(x + 4) = 0

x – 2 = 0 x + 4 = 0+2 +2

x = 2

-4 -4x = -4

(2,0) (-4,0)

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

x= -1

(– 1, – 9)

( 2, 0 )( -4, 0 )


11

Standard 4


y = x – 4 x + 32

y = (x – 4 x + ) + 3 – 242

2 42

2

y = (x – 4 x + ) + 3 – 2 (2)2(2)2

y = (x – 4 x + ) + 3 – 2 4 (4)

y = (x – 2) + 3 – 42

y = (x – 2) – 12


y = (x – 2) + (-1)2

y = a(x-h) + k2

h= 2k= -1a= 1

Vertex: (h,k) = (2,-1)

Axis of symmetry: x= 2

Finding the roots by Quadratic Formula:

x – 4 x + 3 = 02

X=-b b - 4ac

2a

2+_

where: 0 = aX +bX +c2

a = 1

b = - 4

c = 3



x=-( ) ( ) - 4( )( )

2( )

2+_

1

1 -4 -4 3

x= 4 16 – ( 4 )( 3 )

2

+_

4 16 – 12 x=

2

+_



x – 4 x + 3 = 02

X=-b b - 4ac

2a

2+_

where: 0 = aX +bX +c2

a = 1

b = - 4

c = 3

4 4 x=

2

+_

4 2 x=

2

+_

4 2 x=

2

+ 4 2 x=

2

_

x = 3 x = 1

LETS GO BACK TO THE PROBLEM!


13

Standard 4


y = x – 4 x + 32

y = (x – 4 x + ) + 3 – 242

2 42

2

y = (x – 4 x + ) + 3 – 2 (2)2(2)2

y = (x – 4 x + ) + 3 – 2 4 (4)

y = (x – 2) + 3 – 42

y = (x – 2) – 12


y = (x – 2) + (-1)2

y = a(x-h) + k2

h= 2k= -1a= 1

Vertex: (h,k) = (2,-1)

Axis of symmetry: x= 2


x – 4 x + 3 = 02a = 1

b = - 4

c = 3

x=-( ) ( ) - 4( )( )

2( )

2+_

1

1 -4 -4 3

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

x= 2

(2, – 1)

( 3, 0 )( 1, 0 )

x = 3

x = 1 (1,0)

(3,0)



14

Standard 4


y = x + 2x – 4 2

y = (x + 2x + ) – 4 – 222

2 22

2

y = (x + 2x + ) – 4 – 2 (1)2(1)2

y = (x + 2x + ) – 4 – 2 1 (1)

y = (x + 1) – 4 – 12

y = (x + 1) –52


y = (x - -1) + (-5)2

y = a(x-h) + k2

h= -1k= -5a= 1

Vertex: (h,k) = (-1,-5)


Finding the roots by completing the square:x + 2x – 4 = 02

+4 +4

x + 2x + = 4 + 2 22

2 22

2

x + 2x + = 4 + 2 (1)2(1)2

(x + 1) = 52

x + 1 = 5+_-1 -1

x = 5 – 1 +_

x = 5 – 1 x = 5 – 1 _

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

yx= -1

(– 1, – 5)

(1.2, 0)(-3.2, 0)


x = 1.2 x = -3.2


15

Standard 4


( ) = a(( ) – ( )) + ( )2

y = a(x-h) + k2

h= 3 k= -3

Vertex: (h,k) = (3,-3)

Find the equation of the graph below:

1-1 2-2 3 - 3 4-4 5 -5 6-6 7-7

1

-1

2

-2

3

-3

4

-4

5

-5

x

y

(6,0)(0,0)

8

(3, – 3)

0 0 3 -3

0 = a(-3) – 3

2

0 = a(9) – 3 +3 +33 = 9a9 9

a= 13

Using:

(0,0)(x,y)

y= (x – 3) + (– 3)213

y= (x – 3) – 3213

y = a(x-h) + k2

16

Standard 4Find the parabola that passes through points (1,0), (7,0), ( 10,9).

If Y = aX +bX +c2(x , y )1 1 (x , y )2 2 (x , y )3 3

and using the given points.

( ) = a( ) + b( ) + c2 ( ) = a( ) + b( ) + c2 ( ) = a( ) + b( ) + c21 1 7 7 10 100 0 9

0 = a + b + c 0 = 49a + 7b + c 9 = 100a + 10b + c

Solving the equations together:

0 = a + b + c

0 = 49a + 7b + c

9 = 100a + 10b + c

I

II

IIIUsing I and II:

0 = a + b + c

0 = 49a + 7b + c

(-1)

0 = -a – b – c0 = 49a + 7b + c0 = 48a + 6b IV

(-1)0 = 49a + 7b + c

9 = 100a + 10b + c

0 = -49a – 7b – c9 = 100a + 10b + c

9 = 51a + 3 b V

Solving IV and V:

(-2)0 = 48a + 6b9 = 51a + 3 b0 = 48a + 6b

-18 = -102a – 6b -54a = -18

-54a = -18-54 -54

a= 13

0 = 48a + 6bUsing IV:

0 = 48( ) + 6b13

0 = 16 + 6b-16 -16

-16 = 6b 6 6

b = - 83

Using I: 0 = a + b + c 13

– 83 + c = 0 – 7

3 + c = 0 c= 73

Using II and III:

17

Standard 4Find the parabola that passes through points (1,0), (7,0), ( 10,9).

If Y = aX +bX +c2(x , y )1 1 (x , y )2 2 (x , y )3 3

and using the given points.

( ) = a( ) + b( ) + c2 ( ) = a( ) + b( ) + c2 ( ) = a( ) + b( ) + c21 1 7 7 10 100 0 9

0 = a + b + c 0 = 49a + 7b + c 9 = 100a + 10b + c

Solving the equations together:

0 = a + b + c

0 = 49a + 7b + c

9 = 100a + 10b + cb = - 8

3

c= 73

a= 13

Y = aX +bX +c2

Then:Y = X X + 2 – 8

373

13

Y = (X – 8X + 7)213

Y = (X – 8X + ___ + 7 – ___ )213

82

2 82

2

(4)2(4)2

16 16

Y = (X – 8X + ___ + 7 – ___ )213

Y = (X – 8X + ___ + 7 – ___ )213

Y = ((X – 4) – 9)213

Y = (X – 4) – 3213

Documents

1 QUADRATIC FUNCTIONS PROBLEM 1 PROBLEM 2 PROBLEM 3 INTRODUCTION Standard 4, 9, 17 PROBLEM 4 PROBLEM 5 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All