section 9.1 quadratic functions and their graphs

Section 9.1

Quadratic Functions and Their Graphs

OBJECTIVES

A Graph a parabola of the form . y = ax2 + k

OBJECTIVES

B Graph a parabola of the form . = –

2y a x h + k

OBJECTIVES

C Graph a parabola of the form . y = ax2 + bx + c

OBJECTIVES

D Graph parabolas that are not functions of the form

x = ay2 + by + c x = a y – k 2

+ h and

OBJECTIVES

E Solve applications involving parabolas.

PROPERTIES

For Parabola 2g x = ax

The vertex is at the origin and the y-axis is its line of symmetry.

PROPERTIES

For Parabola g x = ax2

If a is positive, the parabola opens upward, if a is negative, the parabola opens downward.

PROPERTIES

For Parabola y = a x – h 2 + k

The graph is the same as y = ax 2 but moved h units horizontally and k units vertically.

PROPERTIES

For Parabola y = a x – h 2 + k

The vertex is at the point h, k , and the axis of symmetry is x = h.

PROPERTIES

For Parabola x = a y – k 2 + h

The graph is same as x = y 2 but moved h units horizontally and k units vertically.

PROPERTIES

For Parabola x = a y – k 2 + h

The vertex is at the point h,k , and the axis of symmetry is y = k.

Practice Test

Exercise #1

Chapter 9Section 9.1A

x

y

Graph the parabola y = – x2 – 4 .

Opens down. V 0, –4

1, –5 –1, –5

Section 9.2

Circles and Ellipses

OBJECTIVES

A Find the distance between two points.

OBJECTIVES

B Find the equation of a circle with a given center and radius.

OBJECTIVES

C Find the center and radius and sketch the graph of a circle when its equation is given.

OBJECTIVES

D Graph an ellipse when its equation is given.

DEFINITION

The Distance Formula

1 1 2 2

2 2

2 1 2 1

The distance between

, and , is

= – –

x y x y

d x x y y

RULE

Graphing Equation of a Circle

Equation of a circle with radius r and center at C h,k isx – h 2

+ y – k 2 = r 2

RULE

Equation of a circle with radius r and center at the origin 0, 0 is x 2 + y 2 = r 2

Graphing Equation of a Circle Centered at 0,0

Graphing Equation of an Ellipse Centered at (0,0)

x 2

a2 +

y 2

b2 = 1, where a2 > b2

Vertices : 0,a and 0, – a .If a and b are equal, the ellipse is a circle.

Graphing Equation of an Ellipse Centered at (h,k)

x – h 2

a2 +

y – k 2

b2 = 1, a2 > b2

Vertices : horizontally ±a units from h,k .

Practice Test

Exercise #9

Chapter 9Section 9.2B

x2 + y2 = r 2

Find an equation of the circle of radius 3 with its center at the origin, 0, 0 .

22 2 + = 3x y

x2 + y2 = 3

Practice Test

Exercise #10

Chapter 9Section 9.2C

Center at h,k in

x – h 2 + y – k 2

= r 2

h = –1, k = 2

Find the center and the radius and sketch the graph of

(x + 1)2 + (y – 2)2 = 9.

Center –1, 2 Radius 9 = 3

Center –1, 2 Radius 9 = 3

Find the center and the radius and sketch the graph of

(x + 1)2 + (y – 2)2 = 9.

x

y

C

Section 9.3

Hyperbolas and Identification of Conics

OBJECTIVES

A Graph hyperbolas.

OBJECTIVES

B Identify conic sections by examining their equations.

RULESGraphing Equation of a

Hyperbola with Center at (0,0)

x 2

a2 –

y 2

b2 = 1

Vertices : x = ± a, 0

y 0,b

– ,0a ,0a

0,–b

RULESGraphing Equation of a

Hyperbola with Center at (0,0)

2 2

2 2 – = 1

Vertices : = 0, ±

y x a b

y a

y 0,a

– ,0b ,0b

0,–a

x – h 2

a2 –

y – k 2

b2 = 1

Vertices : horizontally ±a units from h,k .

Graphing Equation of a Hyperbola with Center at (h,k)

Practice Test

Exercise #16

Chapter 9Section 9.3A

Graph x2

9 – y

2

25 = 1.

Hyperbola

Center = 0,0

a = 9 = 3

b = 25 = 5

Vertices = 3,0 – 3,0

x 2

a2 –

y 2

b2 = 1

x

y

V1 V2

Graph

x2

9 –

y2

25 = 1.

Center = 0,0 Vertices = 3,0 – 3,0

10

10

–10

–10

a = 3, b = 5

Section 9.4

Nonlinear Systems of Equations

OBJECTIVES

A Solve a nonlinear system by substitution.

OBJECTIVES

B Solve a system with two second-degree equations by elimination.

OBJECTIVES

C Solve applications involving nonlinear systems.

Practice Test

Exercise #18

Chapter 9Section 9.4A

Use the substitution method to solve the system.

x2 + y2 = 4 x + y = 2

y = 2 – x

x2 + 2 – x 2 = 4

x2 + 4 – 4x + x2 = 4

2x2 – 4x + 4 = 4

2x2 – 4x = 0

Use the substitution method to solve the system.

x2 + y2 = 4 x + y = 2

2x x – 2 = 0

2x = 0 or x – 2 = 0

x = 0 or x = 2

2x2 – 4x = 0

x = 0 or x = 2

Use the substitution method to solve the system.

x2 + y2 = 4 x + y = 2

y = 2 – 0

y = 2

0,2

y = 2 – 2

y = 0

2,0

Solutions: 0, 2 , 2, 0

x = 0 or x = 2

y = 2 – x y = 2 – x

Section 9.5

Nonlinear Systems of Inequalities

OBJECTIVES

A Graph second-degree inequalities.

OBJECTIVES

B Graph the solution set of a system of nonlinear inequalities.

PROCEDURE

1. Graph each of the inequalities on the same set of axes.

Graphing Nonlinear Inequalities

PROCEDURE

2. Find the region common to both graphs. The result is the solution set.

Graphing Nonlinear Inequalities

Practice Test

Exercise #22

Chapter 9Section 9.5A

y

x

2Graph the inequali y – x ty – 1.

Vertex: (0, –1)x y

1 – 2

–1 – 2

Boundary: y = – x2 – 1.