section 9.1 quadratic functions and their graphs
TRANSCRIPT
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Section 9.1
Quadratic Functions and Their Graphs
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OBJECTIVES
A Graph a parabola of the form . y = ax2 + k
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OBJECTIVES
B Graph a parabola of the form . = –
2y a x h + k
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OBJECTIVES
C Graph a parabola of the form . y = ax2 + bx + c
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OBJECTIVES
D Graph parabolas that are not functions of the form
x = ay2 + by + c x = a y – k 2
+ h and
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OBJECTIVES
E Solve applications involving parabolas.
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PROPERTIES
For Parabola 2g x = ax
The vertex is at the origin and the y-axis is its line of symmetry.
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PROPERTIES
For Parabola g x = ax2
If a is positive, the parabola opens upward, if a is negative, the parabola opens downward.
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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.
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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.
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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.
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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.
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Practice Test
Exercise #1
Chapter 9Section 9.1A
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x
y
Graph the parabola y = – x2 – 4 .
Opens down. V 0, –4
1, –5 –1, –5
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Section 9.2
Circles and Ellipses
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OBJECTIVES
A Find the distance between two points.
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OBJECTIVES
B Find the equation of a circle with a given center and radius.
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OBJECTIVES
C Find the center and radius and sketch the graph of a circle when its equation is given.
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OBJECTIVES
D Graph an ellipse when its equation is given.
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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
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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
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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
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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.
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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 .
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Practice Test
Exercise #9
Chapter 9Section 9.2B
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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
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Practice Test
Exercise #10
Chapter 9Section 9.2C
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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
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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
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Section 9.3
Hyperbolas and Identification of Conics
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OBJECTIVES
A Graph hyperbolas.
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OBJECTIVES
B Identify conic sections by examining their equations.
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RULESGraphing Equation of a
Hyperbola with Center at (0,0)
x 2
a2 –
y 2
b2 = 1
Vertices : x = ± a, 0
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y 0,b
– ,0a ,0a
0,–b
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RULESGraphing Equation of a
Hyperbola with Center at (0,0)
2 2
2 2 – = 1
Vertices : = 0, ±
y x a b
y a
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y 0,a
– ,0b ,0b
0,–a
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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)
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Practice Test
Exercise #16
Chapter 9Section 9.3A
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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
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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
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Section 9.4
Nonlinear Systems of Equations
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OBJECTIVES
A Solve a nonlinear system by substitution.
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OBJECTIVES
B Solve a system with two second-degree equations by elimination.
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OBJECTIVES
C Solve applications involving nonlinear systems.
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Practice Test
Exercise #18
Chapter 9Section 9.4A
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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
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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
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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
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Section 9.5
Nonlinear Systems of Inequalities
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OBJECTIVES
A Graph second-degree inequalities.
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OBJECTIVES
B Graph the solution set of a system of nonlinear inequalities.
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PROCEDURE
1. Graph each of the inequalities on the same set of axes.
Graphing Nonlinear Inequalities
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PROCEDURE
2. Find the region common to both graphs. The result is the solution set.
Graphing Nonlinear Inequalities
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Practice Test
Exercise #22
Chapter 9Section 9.5A
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y
x
2Graph the inequali y – x ty – 1.
Vertex: (0, –1)x y
1 – 2
–1 – 2
Boundary: y = – x2 – 1.
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