1 Solving Systems of Equations and Inequalities that Involve Conics PROBLEM 4 PROBLEM 1 Standard 4, 9, 16, 17 PROBLEM 3 PROBLEM 2 PROBLEM 5 END SHOW PROBLEM.

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Solving Systems of Equations and Inequalities that Involve Conics

PROBLEM 4

PROBLEM 1

Standard 4, 9, 16, 17

PROBLEM 3

PROBLEM 2

PROBLEM 5

END SHOW

PROBLEM 6

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

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STANDARD 4:

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

STANDARD 8:

Students solve and graph quadratic equations by factoring, completing the square, or using the quadratic formula. Students apply these techniques in solving word problems. They also solve quadratic equations in the complex number system.

STANDARD 9:

Students demonstrate and explain the effect that changing a coefficient has on the graph of quadratic functions; that is, students can determine how the graph of a parabola changes as a, b, and c vary in the equation y = a(x-b) + c.

STANDARD 16:

Students demonstrate and explain how the geometry of the graph of a conic section (e.g., asymptotes, foci, eccentricity) depends on the coefficients of the quadratic equation representing it.

STANDARD 17:

Given a quadratic equation of the form ax + by + cx + dy + e = 0, students can use the method for completing the square to put the equation into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation.

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ALGEBRA II STANDARDS THIS LESSON AIMS:



ESTÁNDAR 4:

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

ESTÁNDAR 8:

Los estudiantes resuelven y grafican ecuaciones por factorización, completando el cuadrado, o usando la fórmula cuadrática. Los estudiantes aplican estas técnicas en resolución de problemas. Ellos también resuelven ecuaciones cuadráticas en el sistema de números complejos.

ESTÁNDAR 9:

Los estudiantes demuestran y explican los efectos que tiene el cambiar coeficientes en la gráfica de funciones cuadráticas; esto es, los estudiantes determinan como la gráfica de una parábola cambia con a, b, y c variando en la ecuación y=a(x-b) + c

ESTÁNDAR 16:

Los estudiantes demuestran y explican cómo la geometría de la gráfica de una sección cónica (ej. Las asímptotes, focos y excentricidad) dependen de los coeficientes de la ecuación cuadrática que las representa.

Estándar 17:

Dada una ecuación cuadrática de la forma ax +by + cx + dy + e=0, los estudiantes pueden usar el método de completar al cuadrado para poner la ecuación en forma estándar y pueden reconocer si la gráfica es un círculo, elipse, parábola o hipérbola. Los estudiantes pueden graficar la ecuación

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22



4

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Standard 4, 9, 16, 17Solve the following system of equations:

y = x – 2x – 3 2

y = -3x – 1

Graphing the parabola:

y = x – 2x – 3 2

y = x – 2x + 1 – 3 – 1 2

y = (x – 1) – 4 2

h= 1

k= - 4

Vertex: (1, - 4)

Axis of symmetry: x= 1

a= 1

Latus rectum: 1

Focus: ( 1, - 4 + )1

4 1

= ( 1, 3.75)

Directrix: y = -4 - 1

4( )1= - 4.25

Graphing the line:

y = -3x – 1

m = -3 =-3+1

b= -1

Now let’s check this result algebraically!

x= 1

y = -4.25

(-2, 5)

(1,- 4)

The solution is:

(-2, 5) and (1, - 4)

A = 1 and C= 0 This is a parabola



5

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y


y = x – 2x – 3 2

y = -3x – 1 x= 1

y = -4.25

(-2, 5)

(1,- 4)

Solving by substitution:

= x – 2x – 3 2 -3x – 1

+1 +1

-3x = x – 2x – 2 2

+3x +3x

0 = x + x – 2 2

0 = (x + 2)(x – 1)

-2 1

(2)(-1) 2 + -1= 1

x + 2= 0 x – 1= 0

-2 -2x = -2

+1 +1

x = 1

y = -3( ) – 1 -2

y = 6 – 1

y = 5

Using: x= -2

y = -3( ) – 1 1

y = -3 – 1

y = -4

Using: x= 1

(-2, 5) (1, - 4)PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


6


y = x + 3x + y – 4x – 6y + 9= 02 2 A = 1 and C= 1 This is a circle


x + ( ) – 4x – 6( ) + 9=0 2 2x + 3 x + 3

x + x + 6x + 9 – 4x – 6x – 18 + 9 = 02 2

2x – 4x = 02

2x(x – 2) = 0

2x = 0 x – 2 = 0

x = 0 +2 +2x = 2

y = ( ) + 3 0

y = 3

Using: x= 0

y = ( ) + 32

y = 5

Using: x= 2

(0, 3) (2, 5)

Now let’s verify by graphing!

x – 4x + + y – 6y + +9= 0 + +

2 24 49 9

(x – 2) + (y – 3) + 9 = 13

2 2

-9 -9

(x – 2) + (y – 3) = 42 2

Center: (2,3) Radius = 2

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

(2, 5)

(0, 3)



7


9x + 4y +18x –16y – 11 = 02 2 A = 9 and C= 4 This is an ellipse


9x + 4( ) + 18x –16( ) – 11=0 2 2

y = x +-32

72

x +-32

72

x +-32

72

9x + 4( ) + 18x –16( ) – 11=0 2 2-3x +72

-3x +72

9x + 4( ) + 18x –8( ) – 11=0 2 9x – 42x + 49

2

4-3x + 7

9x + 9x – 42x +49 + 18x + 24x – 56 – 11 = 02 2

18x – 18 = 02

18 (x+1)(x – 1 )= 018 18

x – 1 = 0+1 +1

x = 1

y = 2

Using: x= 1

(1, 2)

x + 1 = 0-1 -1

x = -1

-1

y = 5

Using: x= -1

(-1, 5)

y = ( ) +-32

72 1

y = ( ) +-

32

72

x – 1 = 0x + 1 = 0

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

(-1, 5)

(1, 2)



8


4x – 9y – 16x –18y – 29 = 02 2 A = 4 and C= -9 This is a hyperbola


4x – 9( ) – 16x –18( ) – 29=0 2 2

4x – 9x – 16x – 18x – 29 = 02 2

– 5x – 34x –29 = 02

y = x

x x

5x + 34x +29 = 02

Let’s solve this using the quadratic formula:

X=-b b - 4ac

2a

2+_

where: 0 = aX +bX +c2

We substitute values:

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

2( )

2+_

5

534 34 29

x=-34 1156 – (20)(29)

10

+_

-34 576x=

10

+_

10

-10x= = -1

-34 24x=

10

+_

+ -

x= -1

-34 +24x=

10

-34 – 24x=

10

10

-58x= = -5.8

x= -5.8

y = -5.8

Using: x= -5.8

(-5.8, -5.8)

-1

y = -1

Using: x= -1

(-1, -1)

-5.8

y = ( )

y = ( )

x

y


42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Standard 4, 9, 16, 17Solve by graphing the following system of equations:

y – 2x + 2 = 02 A = 0 and C= 1 This is a parabola

x + y – 2x – 4y + 1= 02 2 A = 1 and C= 1 This is a circle

Graphing the circle:

x – 2x + + y – 4y + +1= 0 + +

2 21 14 4

(x – 1) + (y – 2) + 1 = 52 2

-1 -1

(x – 1) + (y – 2) = 42 2

Center: (1,2) Radius = 2


x + y – 2x – 4y + 1= 02 2

y – 2x + 2 = 02

+2x +2x

2x = y + 22

2 2

x = y + 1212

x = (y – 0) + 1212

Vertex: (1,0)

Directrix:

x = 1 – 1

4 ( )12

x = 12

Focus:

( 1 + , 0 ) 1

4 ( )12

( , 0)12

1Axis of symmetry: y = 0

a = 12

Latus rectum:

12

1= 2

(1, 0)

(3, 2)


42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Standard 4, 9, 16, 17Solve by graphing the following system of inequalities:

y – 4x + 4 > 02 A = 0 and C= 1 This is a parabola

x + y – 2x + 4y – 4< 02 2 A = 1 and C= 1 This is a circle


x – 2x + + y + 4y + – 4< 0 + +

2 21 14 4

(x – 1) + (y + 2) – 4 < 52 2

+4 +4

(x – 1) + (y + 2) < 92 2

Center: (1,-2) Radius < 3

x + y – 2x + 4y – 4< 02 2

Testing (0,0)

0 + 0 – 2(0) + 4(0) – 4< 02 2

-4 < 0 True, so we shade the interior of the circle


42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y


y – 4x + 4 < 02 A = 0 and C= 1 This is a parabola



y – 4x + 4 < 02

+4x +4x

4x > y + 42

4 4

x > y + 1214

x > (y – 0) + 1214

Vertex: (1,0)

Directrix:

x = 1 – 1

4 ( )14

x = 0

Focus:

( 1 + , 0 ) 1

4 ( )14

(2,0)Axis of symmetry: y = 0

a = 14

Latus rectum:

14

1= 4

Testing (2,0)

0 – 4(2) + 4 < 02

-4 < 0 True, so we shade the interior of the parabola


42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y


y – 4x + 4 < 02 A = 0 and C= 1 This is a parabola



x – 2x + + y + 4y + – 4< 0 + +

2 21 14 4

(x – 1) + (y + 2) – 4 < 52 2

+4 +4

(x – 1) + (y + 2) < 92 2

Center: (1,-2) Radius < 3


x + y – 2x + 4y – 4< 02 2

y – 4x + 4 < 02

+4x +4x

4x > y + 42

4 4

x > y + 1214

x > (y – 0) + 1214

Vertex: (1,0)

Directrix:

x = 1 – 1

4 ( )14

x = 0

Focus:

( 1 + , 0 ) 1

4 ( )14

(2,0)Axis of symmetry: y = 0

a = 14

Latus rectum:

14

1= 4

Solution Region


1 Solving Systems of Equations and Inequalities that Involve Conics PROBLEM 4 PROBLEM 1 Standard 4, 9, 16, 17 PROBLEM 3 PROBLEM 2 PROBLEM 5 END SHOW PROBLEM.

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