Interactive Classroom by liuhongmei

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									Lesson 9-1 Graphing Quadratic Functions
Lesson 9-2 Solving Quadratic Equations by Graphing
Lesson 9-3 Solving Quadratic Equations by
           Completing the Square
Lesson 9-4 Solving Quadratic Equations by Using the
           Quadratic Formula
Lesson 9-5 Exponential Functions
Lesson 9-6 Growth and Decay
Five-Minute Check (over Chapter 8)
Main Ideas and Vocabulary
California Standards
Key Concept: Quadratic Function
Example 1: Graph Opens Upward
Example 2: Real-World Example: Graph Opens Downward
Key Concept: Axis of Symmetry of a Parabola
Example 3: Vertex and Axis of Symmetry
Example 4: Standards Example: Match Equations and Graphs
• Graph quadratic functions.
• Find the equation of the axis of symmetry and the
  coordinates of the vertex of a parabola.


• quadratic function      • vertex
• parabola                • symmetry
• minimum                 • axis of symmetry
• maximum
Standard 21.0 Students graph quadratic functions
and know that their roots are the x-intercepts.
           Graph Opens Upwards

Use a table of values to graph y = x2 – x – 2.
Graph these ordered pairs and connect
them with a smooth curve.

Answer:
Use a table of values to graph
y = x2 + 2x + 3.

A.                B.




                                          A.    A
C.                D.
                                          B.    B
                                     0%   C.
                                           0%   C
                                                0%   0%


                                          D.    D



                                 A




                                          B




                                                C




                                                     D
                      Graph Opens Downward

A. ARCHERY The equation y = –x2 + 6x + 4
represents the height y of an arrow x seconds after it
is shot into the area. Use a table of values to graph
y = –x2 + 6x + 4.
Graph these ordered pairs and connect them with a
smooth curve.
                     Answer:
                       Graph Opens Downward

B. ARCHERY The equation y = –x2 + 6x + 4 represents
the height y of an arrow x seconds after it is shot into
the area. What are the mathematical domain and
range of the function?


Answer: domain: all real numbers; range: {y | y ≤ 13}
                       Graph Opens Downward

C. ARCHERY The equation y = –x2 + 6x + 4 represents
the height y of an arrow x seconds after it is shot into
the area. Describe reasonable domain and range
values for this situation.


Answer: The arrow is in the air for about 6.6 seconds, so
        a reasonable domain is D: {x | 0 < x < 6.6}. The
        height of the arrow ranges from 0 to 13 feet, so
        a reasonable range is R: {y | 0 < y < 13}.
A. Use a table of values to graph
y = –x2 + 4.

A.                  B.



                                    1.   A       0%




C.                  D.              2.   B
                                    3.   C
                                    4.   D
                                             A   B    C   D
B. What are the domain and range of the function
y = –x2 + 4?

A. D: {x | x is a real number.}
   R: {y | y  4}

B. D: {x | x is a real number.}              0%
   R: {y | –2  y  2}              1.       A
                                    2.       B
C. D: {x | –2  x  2}              3.       C
   R: {y | y is a real number.}
                                    4.       D

D. D: {x | x  4}
                                         A   B    C   D




   R: {y | y is a real number.}
            Vertex and Axis of Symmetry

A. Consider the graph of y = –2x2 – 8x – 2. Write the
equation of the axis of symmetry.
In y = –2x2 – 8x – 2, a = –2 and b = –8.
                            Equation for the axis of
                            symmetry of a parabola
                            a = –2 and b = –8


Answer: The equation of the axis of symmetry is x = –2.
            Vertex and Axis of Symmetry

B. Consider the graph of y = –2x2 – 8x – 2. Find the
coordinates of the vertex.
Since the equation of the axis of symmetry is x = –2
and the vertex lies on the axis, the x-coordinate for the
vertex is –2.
y = –2x2 – 8x – 2                Original equation
y = –2(–2)2 – 8(–2) – 2          x = –2
y = –8 + 16 – 2                  Simplify.
y=6                              Add.
Answer: The vertex is (–2, 6).
            Vertex and Axis of Symmetry

C. Consider the graph of y = –2x2 – 8x – 2. Identify
the vertex as a maximum or minimum.




Answer: Since the coefficient of the x2 term is negative,
        the parabola opens downward and the vertex
        is a maximum point.
            Vertex and Axis of Symmetry

D. Consider the graph of y = –2x2 – 8x – 2. Graph
the function.
You can use the symmetry of the
parabola to help you draw its graph.
On a coordinate plane, graph the
vertex and the axis of symmetry.
Choose a value for x other than –2.
For example, choose –1 and find the
y-coordinate that satisfies the equation.
y = –2x2 – 8x – 2             Original equation
y = –2(–1)2 – 8(–1) – 2       x = –1
y=4                           Simplify.
             Vertex and Axis of Symmetry

D. Consider the graph of y = –2x2 – 8x – 2. Graph
the function.
Graph (–1, 4).
Since the graph is symmetrical
about its axis of symmetry
x = –2, you can find another
point on the other side of the
axis of symmetry. The point
at (–1, 4) is 1 unit to the right of
the axis. Go 1 unit to the
left of the axis and plot the point
(–3, 4).
            Vertex and Axis of Symmetry

D. Consider the graph of y = –2x2 – 8x – 2. Graph
the function.

Repeat this for several other points.
Then sketch the parabola.

                              Answer:
A. Consider the graph of y = 3x2 – 6x + 1. Write the
equation of the axis of symmetry.

A. x = –6
                                              0%


B. x = 6
                                         1.       A
                                         2.       B
C. x = –1                                3.       C
                                         4.       D
                                          A   B    C   D


D. x = 1
B. Consider the graph of y = 3x2 – 6x + 1. Find the
coordinates of the vertex.

A. (–1, 10)
                                              0%


B. (1, –2)
                                         1.       A
                                         2.       B
C. (0, 1)                                3.       C
                                         4.       D
                                          A   B    C   D

D. (–1, –8)
C. Consider the graph of y = 3x2 – 6x + 1. Identify the
vertex as a maximum or minimum.

A. minimum
                                               0%



B. maximum                               1.        A
                                         2.        B
C. neither                               3.        C
                                         4.        D
                                           A   B    C   D



D. cannot be determined
D. Consider the graph of y = 3x2 – 6x + 1. Graph the
function.
A.                 B.



                                        1.       A0%
                                        2.       B
C.                 D.                   3.       C
                                        4.       D

                                             A    B    C   D
                   Match Equations and Graphs

Which is the graph of y = –x2 – 2x –2?

A                  B




C                  D
                     Match Equations and Graphs

Read the Item
You are given a quadratic function, and you are asked
to choose its graph.
Solve the Item
Find the axis of symmetry of the graph y = –x2 – 2x – 2.

                         Equation for the axis of
                         symmetry
                         a = –1 and b = –2
                    Match Equations and Graphs

The axis of symmetry is –1. Look at the graphs. Since
only choices C and D have x = –1 as their axis of
symmetry, you can eliminate choices A and B. Since
the coefficient of the x2 term is negative, the graph
opens downward. Eliminate choice C.




Answer: D
Which is the graph of y = –x2 + 2x?
A.                 B.




C.                 D.
                                               A.    A
                                               B.    B
                                          0%   0%   0%   0%
                                               C.    C




                                      A
                                               D.    D




                                               B




                                                    C




                                                         D
Five-Minute Check (over Lesson 9-1)
Main Ideas and Vocabulary
California Standards
Example 1: Two Roots
Example 2: A Double Root
Example 3: No Real Roots
Example 4: Factoring
Example 5: Rational Roots
Example 6: Real-World Example
• Solve quadratic equations by graphing.
• Estimate solutions of quadratic equations by
  graphing.



• quadratic equation
• roots
• zeros
• double root
Standard 21.0 Students graph quadratic functions
and know that their roots are the x-intercepts.
Standard 22.0 Students use the quadratic formula or
factoring techniques or both to determine whether
the graph of a quadratic function will intersect the
x-axis in zero, one, or two points.
            Two Roots

Solve x2 – 3x – 10 = 0 by graphing.
Graph the related function f(x) = x2 – 3x – 10.
            Two Roots

Make a table of values
to find other points to
sketch the graph.




To solve x2 – 3x – 10 = 0 you need to know where the
value of f(x) is 0. This occurs at the x-intercepts. The x-
intercepts of the parabola appear to be –2 and 5.
             Two Roots

Check                     Solve by factoring.
x2 – 3x – 10 = 0          Original equation
(x – 5)(x + 2) = 0        Factor.
x – 5 = 0 or x + 2 = 0    Zero Product Property
x = 5      x = –2       Solve for x.

Answer: The solutions of the equation are –2 and 5.



                                    Animation: Solving Quadratic
                                       Equations By Graphing
Solve x2 – 2x – 8 = 0 by graphing.

A. {–2, 4}


B. {2, –4}


C. {2, 4}
                                        A.   A
                                        B.   B
D. {–2, –4}         0%   0%   0%   0%
                                        C.   C
                                        D.   D
                A




                         B




                              C




                                   D
            A Double Root

Solve x2 + 8x = –16 by graphing.
First, rewrite the equation so one side is equal to zero.
      x2 + 8x = –16           Original equation
x2 + 8x + 16 = –16 + 16 Add 16 to each side.
x2 + 8x + 16 = 0              Simplify.
(x + 4)(x + 4) = 0            Factor.
x + 4 = 0 or x + 4 = 0        Zero Product Property
   x = –4            x = –4
           A Double Root

Answer: The solution is –4.
Solve x2 + 2x = –1 by graphing.

A. {1}


B. {–1}
                                           0%
                                  1.   A
C. {–1, 1}
                                  2.   B
                                  3.   C
D.                                4.   D
                                       A   B    C   D
            No Real Roots

Solve x2 + 2x + 3 = 0 by graphing.
Graph the related function
f(x) = x2 + 2x + 3.




Answer: The graph has no x-intercept. Thus, there are
        no real number solutions for the equation.
Solve x2 + 4x + 5 = 0 by graphing.

A. {1, 5}


B. {–1, 5}
                                              0%
                                     1.   A
C. {5}                               2.   B
                                     3.   C
                                     4.   D
D.                                        A   B    C   D
            Factoring

Use factoring to determine how many times the graph
of f(x) = x2 + 3x – 10 intersects the x-axis. Identify
each root.
The graph intersects the x-axis when
f(x) = 0.
x2 + 3x – 10 = 0     Original equation
(x – 2)(x + 5) = 0   Factor.
Answer: Since the trinomial factors into
        two distinct factors, the graph
        of the function intersects the
        x-axis 2 times. The roots are
        x = 2 and x = –5.
Use factoring to determine how many times the graph
of f(x) = x2 + 2x – 15 intersects the x-axis. Identify
each root.

A. 2 times; x = 3 and x = 5

B. 2 times; x = 3 and x = –5

C. 2 times; x = –3 and x = 5
                                                 A.    A
D. 2 times; x = –3 and x = –5                    B.    B
                                            0%   C.
                                                 0%    C
                                                      0%   0%


                                                 D.    D


                                        A




                                                 B




                                                      C




                                                           D
           Rational Roots

Solve x2 – 4x + 2 = 0 by graphing. If integral roots
cannot be found, estimate the roots by stating the
consecutive integers between which the roots lie.
Graph the related function f(x) = x2 – 4x + 2.

                  Notice that the value
                  of the function changes
                  from negative to positive
                  between the x values of
                  0 and 1 and between 3
                  and 4.
           Rational Roots




The x-intercepts of the
graph are between 0 and 1
and between 3 and 4.


Answer: One root is between 0 and 1, and the other
        root is between 3 and 4.
Solve x2 – 2x – 5.
A. One root is between 0 and
   1, and the other root is
   between 4 and 5.
B. One root is between –1 and
   0, and the other root is
   between 3 and 4.
C. One root is between –1 and
   –2, and the other root is             A.    A
   between 3 and 4.                      B.    B
D.                                  0%   C.
                                         0%    C
                                              0%   0%


                                         D.    D


                                A




                                         B




                                              C




                                                   D
MODEL ROCKETS Shelly built a model rocket for
her science project. The equation y = –16t2 + 250t
models the flight of the rocket, launched from
ground level at a velocity of 250 feet per second,
where y is the height of the rocket in feet after t
seconds. For how many seconds was Shelly’s
rocket in the air?
You need to find the solution of the equation
0 = –16t2 + 250t. Use a graphing calculator to graph the
related function y = –16t2 + 250t. The x-intercept is
between 15 and 16 seconds.
Answer: between 15 and 16 seconds
GOLF Martin hits a golf ball with an
upward velocity of 120 feet per
second. The function y = –16t2 + 120t
models the flight of the golf ball, hit
at ground level, where y is the height
of the ball in feet after t seconds.
How long was the golf ball in the air?

A. between 7 and 8 seconds
                                                   A. A
B. between 6 and 7 seconds
                                         0%   0%
                                                   B. 0% B   0%
C. between 8 and 9 seconds                         C. C

                                     A




                                              B




                                                     C




                                                             D
D. between 0 and 1 second                          D. D
Five-Minute Check (over Lesson 9-2)
Main Ideas and Vocabulary
California Standards
Example 1: Irrational Roots
Key Concept: Completing the Square
Example 2: Complete the Square
Example 3: Solve an Equation by Completing the
           Square
Example 4: Real-World Example: Solve a Quadratic
           Equation in Which a ≠ 1
• Solve quadratic equations by finding the square root.
• Solve quadratic equations by completing the square.



• completing the square
Standard 14.0 Students solve a quadratic equation
by factoring or completing the square. (Key)
            Irrational Roots

Solve x2 + 6x + 9 = 5 by taking the square root of
each side. Round to the nearest tenth if necessary.
x2 + 6x + 9 = 5     Original equation
   (x + 3)2 = 5     x2 + 6x + 9 is a perfect square
                    trinomial.
                    Take the square root of each side.
                    Take the square root of each side.
                    Definition of absolute value
            Irrational Roots

                     Subtract 3 from each side.
                     Simplify.

Use a calculator to evaluate each value of x.

                or



Answer: The solution set is {–5.2, –0.8}.
Solve x2 + 8x + 16 = 3 by taking the square root of
each side. Round to the nearest tenth if necessary.

A. {–4}


B. {–2.3, –5.7}


C. {2.3, 5.7}                                    A. A
                                       0%   0%
                                                 B. 0% B   0%
                                                 C. C
D. Ø
                                   A




                                            B




                                                   C




                                                           D
                                                 D. D
               Complete the Square

Find the value of c that makes x2 – 12x + c a perfect
square.
Method 1               Use algebra tiles.




 Arrange the tiles                          To make the
 for x2 – 12x + c so                        figure a
 that the two sides                         square, add
 of the figure are                          36 positive
 congruent.                                 1-tiles.


x2 – 12x + 36 is a perfect square.
           Complete the Square

Method 2       Complete the square.
Step 1

Step 2         Square the result               (–6)2 = 36
               of Step 1.
Step 3         Add the result of               x2 –12x + 36
               Step 2 to x2 – 12x.
Answer: c = 36 Notice that x2 – 12x + 36 = (x – 6)2.

                                     Animation: Completing the Square
Find the value of c that makes x2 + 14x + c a perfect
square.
A. 7


B. 14
                                               0%


                                      1.           A
C. 156                                2.           B
                                      3.           C
D. 49                                 4.           D
                                           A   B    C   D
            Solve an Equation by Completing
            the Square
Solve x2 – 18x + 5 = –12 by completing the square.
Isolate the x2 and x terms. Then complete the square
and solve.

    x2 – 18x + 5 = –12        Original equation
x2 + 18x – 5 – 5 = –12 – 5    Subtract 5 from each side.
        x2 – 18x = –17        Simplify.

  x2 – 18x + 81 = –17 + 81
            Solve an Equation by Completing
            the Square
 (x – 9)2 = 64             Factor x2 –18x + 81.
  (x – 9) = ±8             Take the square root of each
                           side.
x – 9 + 9 = ±8 + 9         Add 9 to each side.
        x=9±8              Simplify.
x = 9 + 8 or x = 9 – 8     Separate the solutions.
 = 17            =1        Simplify.
Answer: The solution set is {1, 17}.
Solve x2 – 8x + 10 = 30.

A. {–2, 10}

                                0%
B. {2, –10}
                           1.       A
C. {2, 10}                 2.       B
                           3.       C
                           4.       D
D. Ø                        A   B    C   D
                       Solve a Quadratic Equation in
                       Which a ≠ 1
CANOEING Suppose the rate
of flow of an 80-foot-wide
river is given by the equation
r = –0.01x2 + 0.8x where r is the rate in miles per
hour, and x is the distance from the shore in feet.
Joacquim does not want to paddle his canoe
against a current faster than 5 miles per hour. At
what distance from the river bank must he paddle
in order to avoid a current of 5 miles per hour?
Explore You know the function that relates distance
        from shore to the rate of the river current. You
        want to know how far away from the river bank
        he must paddle to avoid the current.
                     Solve a Quadratic Equation in
                     Which a ≠ 1
Plan    Find the distance when r = 5. Use completing
        the square to solve –0.01x2 + 0.8x = 5.
Solve   –0.01x2 + 0.8x = 5        Equation for the
                                  current
                                  Divide each
                                  side by –0.01.

        x2 – 80x = –500           Simplify.
                        Solve a Quadratic Equation in
                        Which a ≠ 1
x2 – 80x + 1600 = –500 + 1600



      (x – 40)2 = 110            Factor x2 – 80x + 1600.

                                 Take the square root of
                                 each side.
                                 Add 40 to each side.
                                 Simplify.
                       Solve a Quadratic Equation in
                       Which a ≠ 1
Use a calculator to evaluate each value of x.




Examine The solutions of the equation are about 7 ft
        and about 73 ft. The solutions are distances
        from one shore. Since the river is about 80 ft
        wide, 80 – 73 = 7.

Answer: He must stay within about 7 feet of either
        bank.
CANOEING Suppose the rate of flow of a 60-foot-
wide river is given by the equation r = –0.01x2 + 0.6x
where r is the rate in miles per hour, and x is the
distance from the shore in feet. Joacquim does not
want to paddle his canoe against a current faster
than 5 miles per hour. At what distance from the river
bank must he paddle in order to avoid a current of 5
miles per hour?
A. 6 feet                                           A.   A
B. 5 feet                                           B.   B
                                                    C.   C
C. 1 foot                       0%   0%   0%   0%
                                                    D.   D
D. 10 feet
                            A




                                     B




                                          C




                                               D
Five-Minute Check (over Lesson 9-3)
Main Ideas and Vocabulary
California Standards
Key Concept: The Quadratic Formula
Example 1: Solve Quadratic Equations
Concept Summary: Solving Quadratic Equations
Example 2: Real-World Example: Use the Quadratic
           Formula to Solve a Problem
Key Concept: Using the Discriminant
Example 3: Use the Discriminant
• Solve quadratic equations by using the Quadratic
  Formula.
• Use the discriminant to determine the number of
  solutions for a quadratic equation.



• Quadratic Formula
• discriminant
Standard 19.0 Students know the quadratic
formula and are familiar with its proof by completing
the square. (Key)
Standard 20.0 Students use the quadratic formula
to find the roots of a second-degree polynomial and
to solve quadratic equations. (Key)
Standard 22.0 Students use the quadratic formula
or factoring techniques or both to determine whether
the graph of a quadratic function will intersect the
x-axis in zero, one, or two points
            Solve Quadratic Equations

A. Solve x2 – 2x – 35 = 0. Round to the nearest tenth
if necessary.
Method 1        Factoring
x2 – 2x – 35 = 0            Original equation
(x –7)(x + 5) = 0           Factor x2 –2x – 35.
x –7 = 0 or x + 5 = 0       Zero Product Property
   x=7         x = –5       Solve for x.
           Solve Quadratic Equations

Method 2     Quadratic Formula
                                 Quadratic Formula


                                 a = 1, b = –2,
                                 and c = –35

                                 Multiply.
            Solve Quadratic Equations

                                       Add.

                                       Simplify.



                or                     Separate the
                                       solutions.
    =7                  = –5

Answer: The solution set is {–5, 7}.
            Solve Quadratic Equations

B. Solve 15x2 – 8x = 4. Round to the nearest tenth if
necessary.
Step 1    Rewrite the equation in standard form.
15x2 – 8x = 4               Original equation
15x2 – 8x – 4 = 4 – 4       Subtract 4 from each side.
15x2 – 8x – 4 = 0           Simplify.
          Solve Quadratic Equations

Step 2   Apply the Quadratic Formula.

                               Quadratic Formula

                               a = 15, b = –8,
                               and c = –4

                               Multiply.
Solve Quadratic Equations

                     Add.

  or                 Separate the
                     solutions.

                     Simplify.
            Solve Quadratic Equations

Check the solutions by using the CALC menu on a
graphing calculator to determine the zeros of the related
quadratic function.




Answer: To the nearest tenth, the set is {–0.3, 0.8}.
A. Solve x2 + x – 30 = 0. Round to the nearest tenth if
necessary.
A. {6, –5}


B. {–6, 5}


C. {6, 5}
                                                   A. A
                                         0%   0%
                                                   B. 0% B   0%
D. Ø                                               C. C

                                     A




                                              B




                                                     C




                                                             D
                                                   D. D
B. Solve 20x2 – 4x = 8. Round to the nearest tenth if
necessary.
A. {0.5, 0.7}


B. {–0.5, –0.7}


C. {–0.5, 0.7}
                                                  A. A
                                        0%   0%
                                                  B. 0% B   0%
D. Ø                                              C. C

                                    A




                                             B




                                                    C




                                                            D
                                                  D. D
                        Use the Quadratic Formula to
                        Solve a Problem
SPACE TRAVEL Two possible future destinations of
astronauts are the planet Mars and a moon of the
planet Jupiter, Europa. The gravitational acceleration
on Mars is about 3.7 meters per second squared. On
Europa, it is only 1.3 meters per second squared.
Using the information and equation from Example 2
in the textbook, find how much longer baseballs
thrown on Mars and on Europa will stay above the
ground than similarly thrown baseballs on Earth.
In order to find when the ball hits the ground, you must
find when H = 0. Write two equations to represent the
situation on Mars and on Europa.
                      Use the Quadratic Formula to
                      Solve a Problem
Baseball Thrown on Mars    Baseball Thrown on Europa




These equations cannot be factored, and completing
the square would involve a lot of computation.
                       Use the Quadratic Formula to
                       Solve a Problem
To find accurate solutions, use the Quadratic Formula.
                       Use the Quadratic Formula to
                       Solve a Problem
Since a negative number of seconds is not reasonable,
use the positive solutions.



Answer: A ball thrown on Mars will stay aloft 5.6 – 2.2
        or about 3.4 seconds longer than the ball
        thrown on Earth. The ball thrown on Europa
        will stay aloft 15.6 – 2.2 or about 13.4 seconds
        longer than the ball thrown on Earth.
SPACE TRAVEL The gravitational acceleration on Venus is
about 8.9 meters per second squared, and on Callisto, one of
Jupiter’s moons, it is 1.2 meters per second squared. Suppose a
baseball is thrown on Callisto with an upward velocity of 10
meters per second from two meters above the ground. Find
how much longer the ball will stay in air than a similarly-thrown
ball on Venus. Use the equation                         where H is
the height of an object t seconds after it is thrown upward, v is
                                                         is
the initial velocity, g is the gravitational pull, and h A the initial
                                                    1.
height.                                             2.   B
A. about 10 seconds                      0%
                                                    3.   C
B. about 25.2 seconds                               4.   D

C.   about 12.5 seconds
                                     A   B    C   D
D.   about 14.5 seconds
           Use the Discriminant

A. State the value of the discriminant for
3x2 + 10x = 12. Then determine the number of
real roots of the equation.

Step 1    Rewrite the equation in standard form.

    3x2 + 10x = 12             Original equation
3x2 + 10x – 12 = 12 – 12       Subtract 12 from each
                               side.
3x2 + 10x – 12 = 0             Simplify.
            Use the Discriminant

Step 2     Find the discriminant.

b2 – 4ac = (10)2 – 4(3)(–12)        a = 3, b = 10, and
                                    c = –12

         = 244                      Simplify.

Answer: The discriminant is 244. Since the discriminant
        is positive, the equation has two real roots.
            Use the Discriminant

B. State the value of the discriminant for
4x2 – 2x + 14 = 0. Then determine the number
of real roots of the equation.

b2 – 4ac = (–2)2 – 4(4)(14)    a = 4, b = –2, and
                               c = 14
        = –220                 Simplify.

Answer: The discriminant is –220. Since the
        discriminant is negative, the equation has no
        real roots.
A. State the value of the discriminant for the equation
x2 + 2x + 2 = 0. Then determine the number of real
roots for the equation.

A. –4; no real roots
                                              0%




B. 4; 2 real roots                      1.        A
                                        2.        B
                                        3.        C
C. 0; 1 real root                       4.        D
                                          A   B    C   D




D. cannot be determined
B. State the value of the discriminant for the equation
–5x2 + 10x = –1. Then determine the number of real
roots for the equation.

A. –120; no real roots
                                              0%




B. 120; 2 real roots                    1.        A
                                        2.        B
                                        3.        C
C. 0; 1 real root                       4.        D
                                          A   B    C   D




D. cannot be determined
Five-Minute Check (over Lesson 9-4)
Main Ideas and Vocabulary
California Standards
Key Concept: Exponential Function
Example 1: Graph an Exponential Function with
           a>1
Example 2: Graph an Exponential Function with
           0<a<1
Example 3: Real-World Example: Use Exponential
           Functions to Solve Problems
Example 4: Identify Exponential Behavior
• Graph exponential functions.
• Identify data that displays exponential behavior.



• exponential function
Preparation for Algebra II Standard 12.0 Students
know the laws of fractional exponents, understand
exponential functions, and use these functions in
problems involving exponential growth and decay. (Key)
           Graph an Exponential Function with a > 1

A. Graph y = 3x. State the y-intercept.




Graph the ordered pairs and
connect the points with a smooth
curve.


Answer: The y-intercept is 1.
               Graph an Exponential Function with a > 1

B. Use the graph to determine the approximate value
of 31.5.
The graph represents all real
values of x and their corresponding
values of y for y = 3x.

Answer: The value of y is about 5
        when x = 1.5.

Use a calculator to confirm this
value.
31.5 ≈ 5.196
A. Graph y = 5x.

A.                 B.




C.                 D.
                            0%   0%   0%   0%
                                 A.   A




                        A




                                 B




                                      C




                                           D
                                 B.   B
                                 C.   C
                                 D.   D
B. Use the graph to determine the
approximate value of 50.25.

A. about 2.5

B. about 5

C. about 2
                                          A.   A
                                          B.   B
D. about 1.5
                      0%   0%   0%   0%
                                          C.   C
                                          D.   D
                  A




                           B




                                C




                                     D
           Graph Exponential Functions
           with 0 < a < 1
A.

               Graph the
               ordered pairs
               and connect the
               points with a
               smooth curve.




Answer: The y-intercept is 1.
            Graph Exponential Functions
            with 0 < a < 1
B.




Answer: The value of y is about
        8 when x = –1.5.

Use a calculator to confirm
this value.
A. Graph

A.         B.




C.         D.       0%   0%
                         A.
                              0%
                              A
                                   0%




                A




                         B




                              C




                                   D
                         B.   B
                         C.   C
                         D.   D
B. Use the graph to determine
  the approximate value of

A. about 1


B. about 3                              1.   A
                           0%
                                        2.   B
C. about 2                              3.   C
                                        4.   D

D. about 0.1           A   B    C   D
                       Use Exponential Functions to
                       Solve Problems
DEPRECIATION Some people say that the value of a
new car decreases as soon as it is driven off the
dealer’s lot. The function V = 25,000 ● 0.82t models
the depreciation of the value of a new car that
originally cost $25,000. V represents the value of the
car and t represents the time in years from the time
the car was purchased. Graph the function. What
values of V and t are meaningful in the function?
Use a graphing calculator to graph the function.
                       Use Exponential Functions to
                       Solve Problems
Answer:




Only the values of 0 ≤ V ≤ 25,000 and t ≥ 0 are
meaningful in the context of the problem.
                       Use Exponential Functions to
                       Solve Problems
B. What is the value of the car after one year?
V = 25,000 ● 0.82t        Original equation
V = 25,000 ● 0.821        t=1
V = 20,500                Use a calculator.




Answer: After one year, the car's value is about
        $20,500.
                       Use Exponential Functions to
                       Solve Problems
C. What is the value of the car after five years?
V = 25,000 ● 0.82t        Original equation
V = 25,000 ● 0.825        t=5
V = 9268.50               Use a calculator.




Answer: After five years, the car’s value is about
        $9,270.
A. Depreciation The function V = 22,000 ● 0.82t models
the depreciation of the value of a new car that originally
cost $22,000. V represents the value of the car and t
represents the time in years from the time the car was
purchased. Graph the function.

A.                   B.                               0%
                                             1.           A
                                             2.           B
                                             3.           C
C.                   D.                      4.           D

                                                  A   B    C   D
B. What is the value of the car after one year?

A. $21,000


B. $23,600                                    0%




                                        1.        A
C. $18,040                              2.        B
                                        3.        C
                                        4.        D
D. $20,000                                A   B    C   D
C. What is the value of the car after three years?

A. $12,130


B. $25,120                                    0%




                                         1.       A
C. $10,000                               2.       B
                                         3.       C
                                         4.       D
D. $15,000                                A   B    C   D
              Identify Exponential Behavior

Determine whether the set of data displays
exponential behavior. Explain why or why not.




Method 1 Look for a Pattern
The domain values are at regular intervals of 10.
Look for a common factor among the range values.
10           25       62.5      156.25


     × 2.5    × 2.5     × 2.5
           Identify Exponential Behavior

Answer: Since the domain values are at regular
        intervals and the range values have a common
        factor, the data are probably exponential. The
        equation for the data may involve (2.5)x.
Method 2      Graph the Data
Answer: The graph shows a
        rapidly increasing value
        of y as x increases. This
        is a characteristic of
        exponential behavior.
Determine whether the set of data displays
exponential behavior.




A. no
                               0%
B. yes

C. cannot be
   determined                                1.   A
                                             2.   B
                           A    B   C
                                             3.   C
Five-Minute Check (over Lesson 9-5)
Main Ideas and Vocabulary
California Standards
Key Concept: General Equation for Exponential Growth
Example 1: Real-World Example: Exponential Growth
Example 2: Real-World Example: Compound Interest
Key Concept: General Equation for Exponential Decay
Example 3: Real-World Example: Exponential Decay
• Solve problems involving exponential growth.
• Solve problems involving exponential decay.



• exponential growth
• compound interest
• exponential decay
Preparation for Algebra II Standard 12.0 Students
know the laws of fractional exponents, understand
exponential functions, and use these functions in
problems involving exponential growth and decay. (Key)
                       Exponential Growth

A. POPULATION In 2005 the town of Flat Creek had a
population of about 280,000 and a growth rate of
0.85% per year. Write an equation to represent the
population of Flat Creek since 2005.
The rate 0.85% can be written has 0.0085.
y = C(1 + r)t            General equation for
                         exponential growth
y = 280,000(1 + 0.0085)t C = 280,000 and r = 0.0085
y = 280,000(1.0085)t     Simplify.
Answer: An equation to represent the population of Flat
        Creek is y = 280,000(1.0085)t, where y is the
        population and t is the number of years since
        2005.
                        Exponential Growth

B. POPULATION In 2005 the town of Flat Creek had a
population of about 280,000 and a growth rate of
0.85% per year. According to the equation, what will
be the population of Flat Creek in the year 2015?
In 2015, t will equal 2015 – 2005 or 10.
y = 280,000(1.0085)t       Equation for population of Flat
                           Creek
y = 280,000(1.0085)10      t = 10
y ≈ 304,731              Use a calculator.
Answer: In 2015, there will be about 304,731 people in
          Flat Creek.
A. POPULATION In 2000, Scioto School District had a
student population of about 4500 students, and a
growth rate of about 0.15% per year. Write an
equation to represent the student population of the
Scioto School District since the year 2000.
A. y = 4500(1.0015)

B. y = 4500(1.0015)t
                                              A.    A
                      t
C. y = 4500(0.0015)                           B.    B
                                         0%   0%   0%   0%
                                              C.    C
D. y = (1.0015)t


                                     A




                                              B




                                                   C




                                                        D
                                              D.    D
B. POPULATION In 2000, Scioto School District had a
student population of about 4500 students, and a
growth rate of about 0.15% per year. According to the
equation, what will be the student population of the
Scioto School District in the year 2006?
A. about 9000 students

B. about 4600 students
                                               A.    A
C. about 4540 students                         B.    B
                                          0%   0%   0%   0%
                                               C.    C
D. about 4700 students


                                      A




                                               B




                                                    C




                                                         D
                                               D.    D
                     Compound Interest

COLLEGE When Jing May was born, her
grandparents invested $1000 in a fixed rate savings
account at a rate of 7% compounded annually. The
money will go to Jing May when she turns 18 to help
with her college expenses. What amount of money
will Jing May receive from the investment?

                        Compound interest equation


                        P = 1000, r = 7% or 0.07,
                        n = 1, and t = 18
                     Compound Interest

A = 1000(1.07)18        Compound interest equation
A = 3379.93             Simplify.




Answer: She will receive about $3380.
COMPOUND INTEREST When Lucy was 10 years
old, her father invested $2500 in a fixed rate savings
account at a rate of 8% compounded semiannually.
When Lucy turns 18, the money will help to buy her a
car. What amount of money will Lucy receive from
the investment?
A. about $4682                                 0%
                                      1.       A
B. about $5000                        2.       B
                                      3.       C
C. about $4600                        4.       D
                                           A   B    C   D



D. about $4500
                          Exponential Decay

A. CHARITY During an economic recession, a
charitable organization found that its donations
dropped by 1.1% per year. Before the recession, its
donations were $390,000. Write an equation to
represent the charity’s donations since the beginning
of the recession.
y = C(1 + r)t                 General equation for
                              exponential growth
y = 390,000(1 – 0.011)t         C = 390,000 and r = 1.1%
                                or 0.011
y = 390,000(0.989)t             Simplify.
Answer: y = 390,000(0.989)t
                      Exponential Decay

B. CHARITY During an economic recession, a
charitable organization found that its donations
dropped by 1.1% per year. Before the recession, its
donations were $390,000. Estimate the amount of the
donations 5 years after the start of the recession.

y = 390,000(0.989)t         General equation for
                            exponential growth
y = 390,000(0.989)5         t=5
y = 369,016.74
Answer: The amount of donations should be about
        $369,017.
A. CHARITY A charitable organization found that the
value of its clothing donations dropped by 2.5% per
year. Before this downturn in donations, the
organization received clothing valued at $24,000.
Write an equation to represent the value of the
charity’s clothing donations since the beginning of
the downturn.
                                              1.   A
              t                  0%
A. y = (0.975)                                2.   B
                                              3.   C
B. y = 24,000(0.975)t
                                              4.   D
C. y = 24,000(1.975)t

D. y = 24,000(0.975)         A   B    C   D
B. CHARITY A charitable organization found that the
value of its clothing donations dropped by 2.5% per
year. Before this downturn in donations, the
organization received clothing valued at $24,000.
Estimate the value of the clothing donations 3 years
after the start of the downturn.
A. about $23,000                             1.   A
                                0%
                                             2.   B
B. about $21,000                             3.   C
                                             4.   D
C. about $22,245

D. about $24,000
                            A   B    C   D
Five-Minute Checks

Image Bank

Math Tools




Animation Menu

Exploring Quadratic Functions
Lesson 9-1 (over Chapter 8)
Lesson 9-2 (over Lesson 9-1)
Lesson 9-3 (over Lesson 9-2)
Lesson 9-4 (over Lesson 9-3)
Lesson 9-5 (over Lesson 9-4)
Lesson 9-6 (over Lesson 9-5)
To use the images that are on the
following three slides in your own
presentation:
1. Exit this presentation.
2. Open a chapter presentation using a
   full installation of Microsoft® PowerPoint®
   in editing mode and scroll to the Image
   Bank slides.
3. Select an image, copy it, and paste it
   into your presentation.
9-1 Families of Quadratic Functions
9-2 Solving Quadratic Equations By Graphing
9-3 Completing the Square
                            (over Chapter 8)


Factor the polynomial a2 – 5a + 9, if possible. If the
polynomial cannot be factored using integers, write
prime.

A. (a + 3)(a – 3)

B. (a + 3)(a + 3)
                                                   A. A
C. (a – 3)(a – 3)                                  B. 0% B
                                         0%   0%             0%
                                                   C. C

                                     A




                                              B




                                                     C




                                                             D
D. prime                                           D. D
                            (over Chapter 8)


Factor the polynomial 6z2 – z – 1, if possible. If the
polynomial cannot be factored using integers, write
prime.

A. (2z – 1)(3z – 1)
                                                0%


B. (2z – 1)(3z + 1)                    1.           A
                                       2.           B
C. (2z + 1)(3z – 1)                    3.           C
                                       4.           D
                                            A   B    C   D

D. prime
                          (over Chapter 8)


Solve the equation 5x2 = 125.

A. {–5, 5}

                                            0%
B. {–25, 25}
                                           1.        A
                                           2.        B
C.
                                           3.        C
                                           4.        D
D.
                                       A    B    C   D
                           (over Chapter 8)


Solve the equation 2x2 + 11x – 21 = 0.

A.


B.


C.                                                A. A
                                        0%   0%
                                                  B. 0% B   0%
D.                                                C. C

                                    A




                                             B




                                                    C




                                                            D
                                                  D. D
                           (over Chapter 8)


A certain basketball player’s hang time can be
described by 4t2 = 1, where t is time in seconds.
How long is this player’s hang time?

A. 4 seconds
                                               0%


B. 2 seconds                          1.           A
                                      2.           B
C.                                    3.           C
                                      4.           D
                                           A   B    C   D

D.
                           (over Chapter 8)



The area of a rectangle is given by 6y2 + 5y – 6 and
the width is given by 3y – 2. What is the length?

A. 2y + 3
                                             0%



                                            1.        A
B. y + 3
                                            2.        B
                                            3.        C
C. 2y – 3                                   4.        D

                                        A    B    C   D


D. y – 3
                         (over Lesson 9-1)


Which of the following options shows a graph and
the solution of y = x2 + 2x – 1?
A.              B.




C.              D.                            A.     A
                                              B.     B
                                        0%   0%    0%    0%
                                              C.     C


                                    A




                                             B




                                                   C




                                                         D
                                              D.     D
                            (over Lesson 9-1)


For y = –x2 + 2, identify the equation of axis of
symmetry, and the coordinates of the vertex of the
graph of the equation. Identify the vertex as a
maximum or a minimum.

A.
                                                0%


                                       1.       A
B.                                     2.       B
                                       3.       C
C. x = 0; (0, 2); maximum              4.       D
                                            A   B    C   D



D. x = 0; (0, 2); minimum
                              (over Lesson 9-1)


For y = x2 – 5x, identify the equation of the axis of
symmetry, and the coordinates of the vertex of the
graph of the equation. Identify the vertex as a
maximum or a minimum.
                                                0%
A. x = –2.5; (–2.5, 18.75);
   maximum                                     1.        A
B. x = –2.5; (–2.5, 18.75);                    2.        B
   minimum                                     3.        C
C. x = 2.5; (2.5, –6.25);                      4.        D
   maximum
                                           A    B    C   D

D. x = 2.5; (2.5, –6.25);
   minimum
                            (over Lesson 9-1)


What is the maximum height of a rocket fired
straight up if the height in feet is described by
h = –16t2 + 64t + 1, where t is time in seconds?

A. 66 ft

B. 65 ft
                                                 A.     A
C. 64 ft                                         B.     B
                                           0%   0%    0%    0%
                                                 C.     C


                                       A




                                                B




                                                      C




                                                            D
D. 63 ft                                         D.     D
                          (over Lesson 9-1)


The graph of which of the following equations has
a vertex at (2, –1)?

A. y = x2 – 4x + 2

                                              0%

B. y = 2x2 + 4x + 2
                                     1.           A
                                     2.           B
C. y = x2 – 4x + 3                   3.           C
                                     4.           D
                                          A   B    C   D

D. y = x – 4x – 2
        2
                         (over Lesson 9-2)


Which of the following options shows a graph and
the solution of m2 – 2m – 3 = 0?

A.              B.




C.              D.                             A.     A
                                               B.     B
                                        0%   0%     0%    0%
                                               C.     C


                                    A




                                             B




                                                    C




                                                          D
                                               D.     D
                          (over Lesson 9-2)


Which choice shows the graph of w2 + 5w – 1 = 0,
and its roots or the consecutive integers between
which the roots lie?
A.              B.

                                              0%
                                     1.   A
                                     2.   B
C.              D.                   3.   C
                                     4.   D
                                          A   B    C   D
                          (over Lesson 9-2)


Use a quadratic equation to find two numbers
whose difference is 3 and whose product is 10.

A.
                                            0%


B. –5, 2 or 5, –2                          1.        A
                                           2.        B
C. –5, –2 or 5, 2                          3.        C
                                           4.        D

D.                                     A    B    C   D
                           (over Lesson 9-2)


Which of the following are the roots of
x2 + 6x + 8 = 0?

A. 2 and 4


B. 8 and 1


C. –8 and –1                                      A. A
                                        0%   0%
                                                  B. 0% B   0%
                                                  C. C
D. –4 and –2
                                    A




                                             B




                                                    C




                                                            D
                                                  D. D
                           (over Lesson 9-3)


Solve the equation x2 + 8x + 16 = 16.


A. –8, 0


B. 8, 0


C. 2, 16                                          A. A
                                        0%   0%
                                                  B. 0% B   0%
                                                  C. C
D. –2, 16
                                    A




                                             B




                                                    C




                                                            D
                                                  D. D
                            (over Lesson 9-3)


Solve the equation x2 – 6x – 2 = 5.


A. 1, –7

                                                0%
B. –1, –7
                                       1.           A
                                       2.           B
C. –1, 7                               3.           C
                                       4.           D
D. 1, 7                                     A   B    C   D
                            (over Lesson 9-3)


Find the value of c that makes z2 + z + c a perfect
square.

A.
                                              0%


B.                                           1.        A
                                             2.        B
                                             3.        C
C.
                                             4.        D

                                         A    B    C   D
D.
                          (over Lesson 9-3)


The area of a square can be tripled by increasing
the length and width by 10 inches. What is the
original length of the square? Round to the nearest
tenth of an inch.

A. 2.24 in.

B. 3.7 in.
                                               A.     A
C. 10 in.                               0%   0%
                                               B.
                                                    0%
                                                      B
                                                          0%
                                               C.     C


                                    A




                                             B




                                                    C




                                                          D
D. 13.7 in.                                    D.     D
                           (over Lesson 9-3)


Use the roots of x2 – 14x + 49 = 81 to determine
which statement is true.

A. The sum of the roots is
   greater than the product of
   the roots.
                                               0%
B. The product of the roots is
                                      1.       A
   greater than the sum of the
   roots.                             2.       B
C. The product of the roots is        3.       C
   positive.                          4.       D
                                           A   B    C   D
D. The sum of the roots is
   negative.
                          (over Lesson 9-4)


Solve the equation x2 – 6x – 7 = 0 by using the
Quadratic Formula. Round to the nearest tenth if
necessary.

A. 1, 7

B. –1, 7

                                                 A. A
C. –7, 1
                                       0%   0%
                                                 B. 0% B   0%
                                                 C. C
D. Ø
                                   A




                                            B




                                                   C




                                                           D
                                                 D. D
                          (over Lesson 9-4)


Solve the equation y2 – 11y = –30 by using the
Quadratic Formula. Round to the nearest tenth if
necessary.

A. 5, 6
                                              0%

B. –5, 6                             1.           A
                                     2.           B
C. –5, –6                            3.           C
                                     4.           D
                                          A   B    C   D
D. Ø
                           (over Lesson 9-4)


Solve the equation 5z2 + 16z + 3 = 0 by using the
Quadratic Formula. Round to the nearest tenth if
necessary.

A.                                           0%



                                            1.        A
B.
                                            2.        B
                                            3.        C
C.                                          4.        D

                                        A    B    C   D
D. Ø
                          (over Lesson 9-4)


Solve the equation 4n2 = –19n – 25 by using the
Quadratic Formula. Round to the nearest tenth if
necessary.

A.

B.

                                                 A. A
C.
                                       0%   0%
                                                 B. 0% B   0%
                                                 C. C
D. Ø
                                   A




                                            B




                                                   C




                                                           D
                                                 D. D
                             (over Lesson 9-4)


Without graphing, determine the number of
x-intercepts of the graph of f(x) = 3x2 + x + 7.

A. 0

                                                 0%
B. 1
                                        1.           A
C. 2                                    2.           B
                                        3.           C
                                        4.           D
D. 3                                         A   B    C   D
                           (over Lesson 9-4)



The work shown is the first step to finding the roots

of which equation?

A. 3x2 – 2x + 4 = 0
                                                 0%

                                        1.           A
B. 4x – 3x – 4 = 0
      2
                                        2.           B
                                        3.           C
C. 2x2 – 3x + 4 = 0
                                        4.           D
D. 2x2 + 3x – 4 = 0                          A   B    C   D
                          (over Lesson 9-5)


Which choice shows the graph of y = 4x, and also
states the y-intercept and the approximate value of
40.6? Use a calculator to confirm the value.
A.               B.




C.               D.                             A.     A
                                         0%   0%B.   0%B   0%
                                                C.     C


                                     A




                                              B




                                                     C




                                                           D
                                                D.     D
                            (over Lesson 9-5)


Which choice states and explains
whether or not the data in the given
table display exponential behavior?
A. Yes, the domain values are at a
    regular intervals and the range
    values have a common difference.
B. Yes, the domain values are at regular
    intervals and the range values have a       1.           A
    common ratio of 3.
                                                2.       0%
                                                             B
C. No, the domain values are at regular
                                                3.           C
    intervals and the range values have a
    common ratio of 3.                          4.           D
D. No, the domain values are at regular              A   B    C   D




    intervals and the range values have a
    common difference.
                              (over Lesson 9-5)


A tournament begins with 64 teams. After each round, only
of the teams remain. Write an exponential function for the
number of teams remaining after x rounds of play. How many
teams remain after 3 rounds?

A.            8 teams
                                            1.    A
                                                      0%
B.            16 teams                      2.    B
                                            3.    C
C.             8 teams                      4.    D

                                                  A   B    C   D


D.             16 teams
                            (over Lesson 9-5)



What is the y-intercept of the graph of y = 2x – 5?
A. –5


B. –4


C. 3
                                                   A. A
                                         0%   0%
                                                   B. 0% B   0%
D. 11                                              C. C

                                     A




                                              B




                                                     C




                                                             D
                                                   D. D
This slide is intentionally blank.

								
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