Quadratic Functions ppt

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    QUADRATIC FUNCTIONS

    Monika V Sikand

    Light and Life Laboratory

    Department of Physics and Engineering physics

    Stevens Institute of TechnologyHoboken, New Jersey, 07030.

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    OUTLINE

    Graphing Quadratic Functions

    Solving Quadratic Equations by Factoring

    Solving Quadratic Equations by FindingSquare Roots

    Complex Numbers

    The Quadratic Formula and the DiscriminantGraphing and Solving Quadratic Inequalities

    Modeling with Quadratic Functions

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    Graphing Quadratic Functions

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    THE GRAPH OF A QUADRATIC

    FUNCTION

    vertex

    Axis of

    symmetry

    y = x2

    y = -x2

    The parabola opens upif a>0 andopens downif a 1.

    The x-coordinate of the vertexis-b/2a.

    The axis of symmetryis the vertical

    line x = -b/2a.

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    EXAMPLE

    Graph y = 2x2-8x +6

    Solution: The coefficients for this function are

    a = 2, b = -8, c = 6.

    Since a>0, the parabola opens up.

    The x-coordinate is: x = -b/2a, x = -(-8)/2(2)

    x = 2

    The y-coordinate is: y = 2(2)2-8(2)+6

    y = -2

    Hence, the vertex is (2,-2).

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    EXAMPLE(contd.)

    Draw the vertex (2,-2)on graph.

    Draw the axis of symmetry x=-b/2a.

    Draw two points on one side of the

    axis of symmetry such as (1,0) and

    (0,6).

    Use symmetry to plot two more points

    such as (3,0), (4,6).

    Draw parabola through the plotted

    points.(2,-2)

    (1,0)

    (0,6)

    (3,0)

    (4,6)

    Axis of symmetry

    x

    y

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    VERTEX FORM OF QUADRATIC

    EQUATION

    y = a(x - h)2+ k

    The vertex is (h,k).

    The axis of symmetry is x = h.

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    GRAPHING A QUADRATIC

    FUNCTION IN VERTEX FORM

    (-3,4)

    (-7,-4)

    (-1,2)(-5,2)

    (1,-4)

    Axis of symmetry

    x

    y

    Exampley = -1/2(x + 3)2+ 4

    where a = -1/2, h = -3, k = 4. Since a

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    INTERCEPT FORM OF

    QUADRATIC EQUATION

    y = a(x - p)(x - q)

    The x intercepts are p and q.

    The axis of symmetry is halfway

    between (p,0) and (q,0).

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    GRAPHING A QUADRATIC

    FUNCTION IN INTERCEPT FORM

    (-2,0)

    (1,9)

    (4,0)

    Axis of symmetry

    x

    y

    Exampley = -(x + 2)(x - 4).

    where a = -1, p = -2, q = 4. Since

    a

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    WRITING THE QUADRATIC

    EQUATION IN STANDARD FORM(1). y = -(x + 4)(x - 9)

    = -(x2 - 9x + 4x - 36)

    = -(x2 - 5x -36)

    = -x2 + 5x + 36

    (2). y = 3(x -1)2+ 8

    = 3(x -1)(x - 1) + 8

    = 3(x2- x - x + 1) + 8= 3(x2- 2x + 1) + 8

    = 3x2- 6x + 3 + 8

    = 3x2- 6x + 11

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    SOLUTION

    Since a = -3.678 is negative, the graph of the quadratic function open

    down and the function has a maximum value. The maximum value

    occurs at:

    The corresponding value of y is:

    a. Hence, The temperature that made the greatest percent of test

    subjects comfortable was about 72.

    b. At that temperature about 92% of the subjects felt comfortable.

    x b2a

    527.32(3.678)

    72

    y 3.678(72)2

    527.3(72)18,807 92

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    REAL LIFE EXAMPLE 6

    (page 252)The Golden Gate Bridge in San Franciscohas two towers that

    rise 500 feet above the road and are connected by suspension

    cables as shown. Each cable forms a parabola with equation

    where x and y are measured in feet.

    a. What is the distance d between the towers?

    b. What is the height l above the road of a cable at its lowest

    point?

    y 1

    8940(x 2100)2 8

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    GOLDEN GATE BRIDGE

    l

    d

    y

    200ft

    500ft

    x

    Not down to scale

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    SOLUTION

    y 1

    8960(x 2100)

    2 8

    y 1

    8960 (x2

    4200x 4410000) 8

    y 1

    8960x2

    4200

    8960x (

    4410000

    8960 8)

    x b

    2a

    (4200

    8960

    )( 1

    2 18960

    )

    x 2100

    y 1

    8960(x 2100)2 8

    y 1

    8960 (2100 2100)2

    8

    y 0

    8960 8

    y 8

    Hence the vertex of the parabola is (2100,8)

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    SOLUTION(contd.)

    a. The vertex of the parabola is (2100,8), so the

    cables lowest point is 2100 feet from the left tower

    shown above. Since the heights of the two towersare the same, the symmetry of the parabola implies

    that the vertex is also 2100 feet from the right tower.

    Therefore the towers are d = 2(2100) = 4200 feet

    apart.

    b. The height l above the road of a cable at its lowest

    point is the y-coordinate of the vertex. Since the

    vertex is (2100,8), this height is l= 8 feet.

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    SOLVING QUADRATICEQUATION BY FACTORING

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    FACTORING QUADRATIC

    EXPRESSIONThe expression x2+bx+ cis a trinomial because it

    has three terms. We can use factoring to write it as

    product of two terms or binomials such as

    x2+bx+ c = (x+m)(x+n) = x2 + (m+n)x + mn

    Example:

    x2+ 8x + 15 = (x + 3)(x + 5)

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    EXAMPLE

    Problem: Factor x2- 12x - 28

    Solution:

    x2- 12x - 28 = (x+m)(x+n) where mn = -28 and m+n = -12

    Factors

    of -28

    -1,28 1,-28 -2,14 2,-14 -4,7 4,-7

    Sum of

    factors

    27 -27 12 -12 3 -3

    The table shows that m = 2 and n = -14.

    So, x2- 12x - 28 = (x + 2)(x - 14)

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    SPECIAL FACTORING

    PATTERNS1. Difference of two squares: a2- b2= (a+b)(a-b)

    Example: x2

    - 9 = (x+3)(x-3)

    2. Perfect square Trinomial: a2+ 2ab + b2= (a+b)2

    Example: x2+ 12x +36 = (x+6)2

    3. Perfect square Trinomial: a2- 2ab + b2= (a-b)2

    Example: x2-8x +16 = (x-4)2

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    FACTORING MONOMIALS FIRST

    Monomialis an expression that has only one term.

    Factor the quadratic expression:

    a. 5x2- 20 = 5(x2-4)

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

    b. 6p2 + 15p + 9 = 3(2p2+ 5p + 3)

    = 3(2p + 3)(p + 1)

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    SOLVING QUADRATIC

    EQUATIONS

    x2 3x 18 0

    (x 6)(x 3) 0

    x 6 0

    x 6

    or

    x 3 0,

    x 3

    Solve: Solve:2t2 17t 45 3t 5

    2t2 20t 50 0

    t2 10t 25 0

    (t 5)2 0

    t 5 0

    t 5

    The solution is 5.

    The solutions are -6 and 3.

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    REAL LIFE EXAMPLE

    You have made a rectangular stained glass window

    that is 2 feet by 4 feet. You have 7 square feet of clear

    glass to create a border of uniform width around the

    window. What should the width of the border be?

    x xx x

    x

    x xx

    2

    44+2x

    2+2x

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    ZERO PRODUCT PROPERTY

    Let A and B be real numbers or algebraic

    expressions. If AB=0, then either A = 0 or B = 0

    Solve(a) x2+ 3x -18 = 0

    (x + 6)(x - 3) = 0.Hence either x + 6 = 0 or x - 3 = 0.

    The solutions are x = -6 or x = 3

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    ZERO PRODUCT PROPERTY

    Let A and B be real numbers or algebraic

    expressions. If AB=0, then either A = 0 or B = 0

    Solve(a) 2t2- 17t + 45 = 3t - 5

    2t2- 20t + 50 = 0

    t2- 10t + 25 = 0

    (t - 5)2

    = 0t - 5 = 0

    t = 5

    Hence the solutions is 5.

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    FINDING ZEROES OF QUADRATIC

    FUNCTIONS

    Find zeros of y = x2-x -6

    Solution:

    y = x2 - 3x + 2x - (32)

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

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

    Hence the zeros of the function are -2 and 3.

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    SOLVING QUADRATICEQUATIONS BY FINDING

    SQUARE ROOTS

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    SQUARE ROOT

    A number r is a square rootof a number s if r2= s.

    A positive number s has square roots denoted by

    Example:

    Since 32= 9 and (-3)2= 9. The square roots of 9 are:

    sor s

    9 3

    9 3.

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    PROPERTIES OF SQUARE ROOTS

    (a>0, b>0)

    PRODUCT PROPERTY:

    QUOTIENT PROPERTY:

    ab a. b

    a

    b

    a

    b

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    SIMPLIFY

    (a). 24

    4 . 6

    2 6

    (b). 7

    2

    7

    2

    2

    2

    14

    2

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    Solving a quadratic equation

    2x2 117

    2x2

    16x2 8

    x 8

    x2 2

    Hence the solutions are:

    2 2 and 2 2

    1

    3

    (x 5)2 7

    (x 5)2 21

    (x 5) 21

    x 5 21

    5 21 5 21

    Hence the solutions are:

    and

    Solve: Solve:

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    REAL LIFE EXAMPLE

    A stunt man working on the set of a movie is to fall

    out of a window 100 feet above the ground. For the

    stunt mans safety, an air cushion 26 feet wide by

    30 feet long by 9 feet high is positioned on theground below the window.

    a. For how many seconds will the stunt man fall before

    he reaches the cushion?b. A movie camera operating at a speed of 24 frames

    per second records the stunt mans fall. How many

    frames of film show the stunt man falling?

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    SOLUTION

    a. The stunt mansinitial height is ho= 100 feet, so his height as a

    function of time can be modeled by function h = -16t2 + 100.

    Since the height of the cushion is 9 feet above the ground, the

    the time taken by the stunt man to reach the cushion is:h = -16t2 + 100

    9 = -16t2 + 100

    -91 = -16t2

    91/16 = t2

    or t 2.4. Thus, it takes about 2.4 seconds for the stuntman to reach the cushion.

    b. The number of frames of filmthat show the stunt man falling is

    given by the product (2.4sec)(24 frames/sec), or about 57

    frames.

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    COMPLEX NUMBERS

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    COMPLEX NUMBER

    A complex number written in standard form is a

    number a+bi where a and b are real numbers.

    The number a is the real part of the complex

    number and number bi is the imaginary part.

    If b0, then a+bi is an imaginary number. If a=0 and

    b0, then a+bi is a pure imaginary number.

    A complex plane has a horizontal axis called the

    real axisand a vertical axis called the imaginary axis.

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    THE SQUARE ROOT OF A

    NEGATIVE NUMBER

    1. If r is a positive real number, then

    whereiis the imaginary unitdefined as

    2. By property (1), it follows that

    r i r

    (i r)2 r

    i 1

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    SOLVING A QUADRATIC

    EQUATION

    3x2 10 26

    Solution

    3x2 10 26

    3x2 36

    x2

    12

    x 12

    x i 12

    x 2i 3

    Hence, the solutions are

    and 2i 3

    2i 3

    Solve:

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    PLOTTING COMPLEX NUMBERS

    Plot 2-3i in the complex

    plane.

    To plot 2-3i , start at theorigin, move 2 units to the

    right and then move 3 units

    down.

    0

    2-3i

    x

    y

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    ADDING AND SUBTRACTING

    COMPLEX NUMBERS

    Sum of complex numbers:

    (a+bi) + (c+di) = (a+c) + i(b+d)

    Example:

    (4-i) + (3+2i) = (4+3) + i(-1+2)

    = 7 + i

    Difference of complex numbers:

    (a+bi) - (c+di) = (a-c) + i(b-d)

    Example:

    (7-5i) - (1-5i) = (7-1) + i(-5+5)

    = 6 + 0i

    = 6

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    MULTIPLYING THE COMPLEX

    NUMBERS

    Write the expression as a complex number in standard form.

    a. 5i(-2+i)= -10i + 5i2

    = -10i + 5(-1)

    = -5-10i

    b. (7-4i)(-1+2i)= -7 + 14i + 4i - 8i2

    = -7 + 18i - 8(-1)= 1 + 18i

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    DIVIDING COMPLEX NUMBERS

    Write the quotient in standard form.

    Solution:Multiply the numerator and denominator by the complex

    conjugate of the denominator.

    5 3i

    1 2i

    5 3i

    1 2i 1 2i

    1 2i 5 10i 3i 6i2

    1 2i 2i 4i2

    113i

    5

    1

    5

    13

    5i

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    ABSOLUTE VALUES OF COMPLEX

    NUMBER

    Find the absolute value of each complex number.

    (a)3 4i 32 42 25 5

    (b)2i 0 (2i) 02

    (2)2

    2

    (c)1 5i (1)2 52 265.10

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    THE QUADRATIC FORMULA

    Let a, b, and c be real numbers such that a0. The

    solutions of the quadratic equation ax2+ bx +c = 0

    are:

    x b b

    2 4ac

    2a

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    SOLVING QUADRATIC EQUATION

    WITH TWO REAL SOLUTIONS

    Solve

    2x2 x 5

    Solution:

    2x2x 5 0

    xb b

    2 4ac

    2a

    x 1 12

    4(2)(5)2(2)

    x1 41

    4

    x 1 41

    4 1.35

    and

    x 1 41

    4 1.85

    The solutions are:

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    SOLVING QUADRATIC EQUATION

    WITH ONE REAL SOLUTIONS

    Solve

    x2 x 5x 9

    Solution:x2 x 5x 9

    x2 6x 9 0

    x6 (6)2 4(1)(9)

    2(1)

    x6 0

    2

    x3

    Hence, the solution is 3.

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    SOLVING QUADRATIC EQUATION

    WITH TWO IMAGINARY SOLUTIONS

    Solve

    x2

    2x 2

    Solution

    x2 2x 2

    x2

    2x 2 0

    x2 22 4(1)(2)

    2(1)

    x

    2 4

    2

    x2 2i

    2

    x1 i

    The solutions are:

    1+i and 1-i

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    DISCRIMINANT

    In the quadratic formula, the expression b2-4ac

    under the radical sign is called the discriminant

    of the associated equation ax2

    + bx + c = 0.

    x

    b b2 4ac

    2a

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    EXAMPLE: TWO REAL SOLUTIONS

    Solve

    x2 6x 8 0

    Discriminant:

    b2 4ac (6)2 4(1)(8) 4

    Solutions:

    xb b2 4ac

    2a

    x 6 (6)2 4(1)(8)2

    x6 4

    2

    6 2

    2314, 2

    Hence there are two realsolutions: 4,2

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    EXAMPLE: ONE REAL SOLUTION

    Solve

    x2 6x 9 0

    Discriminant:

    b2 4ac (6)2 4(1)(9) 0

    Solutions:

    xb b2 4ac

    2a

    x 6 (6)2

    4(1)(9)2

    x6 0

    2

    6

    23

    Hence, there is one real solution: 3

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    EXAMPLE: TWO IMAGINARY

    SOLUTIONS

    Find the discriminant of the quadratic equation and give the number

    and type of solutions of the equation.

    Solve

    x2

    6x10 0

    Discriminant:

    b2

    4ac (6)2

    4(1)(10) 4

    Solutions:

    x b b

    2 4ac

    2a

    x 6 (6)

    2 4(1)(10)

    2

    x 6 4

    262 1

    2 3 i

    Hence there are twoimaginary solutions:

    3+i and 3-i

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    SOLUTION(contd.)

    h 16t2 vot ho

    5 16t2 45t 6

    0 16t2

    45t1

    t45 2089

    32

    t 0.022

    ort 2.8

    Rejecting the negative solution, the

    baton is in the air for about 2.8seconds.

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    GRAPHING AND SOLVINGQUADRATIC INEQUALITIES

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    GRAPHING A QUADRATIC

    INEQUALITY IN TWO VARIABLES

    y ax2+bx +c y ax2+bx +c

    Draw parabola with equation y = ax2+bx +c . Makethe parabola dashed for inequalities with < or >and

    solid for inequalities with or .

    Choose a point (x,y) inside the parabola and check

    whether the point is a solution of the inequality.

    If a point (x,y) is a solution, shade the region inside

    the parabola. If it is not the solution, shade the region

    outside the parabola.

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    EXAMPLE

    0(1,0) x

    y

    4

    1

    Graph y > x2-2x -3

    Solution:

    Graph y = x2-2x -3. Since theinequality is >, make parabola dashed.

    Test the point (1,0) inside the

    parabola. y > x2

    -2x -3

    0 > 12-2(1)-3

    0 > -4So, (1,0) is a solution of the inequality.

    Shade the region inside the parabola.

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    REAL LIFE EXAMPLE

    You are building a wooden bookcase. You want

    to choose a thickness d(in inches) for the

    shelves so that each is strong enough to support

    60 pounds of books without breaking. A shelf cansafely support a weight of W ( in pounds)

    provided that

    W 300d2

    a. Graph the given inequality

    b. If you make each shelf 0.75 inch thick, can it

    support a weight of 60 pounds?

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    SOLUTION

    y

    0

    50

    100

    150

    200

    250300

    W

    d0.5 1.0 1.5

    a. Graph W = 300d2 for non negative values of

    d. Since the inequality symbol is make the

    parabola solid. Test a point inside the

    parabola, such as (0.5, 240).

    W 300d2

    240 300(0.5)2240 75

    Since the chosen point is not a solution,

    shade the region outside (below) the

    parabola.

    b. The point (0.75,60) lies in the shaded regionof the graph from part (a), so (0.75,60) is a

    solution of the given inequality.

    c. Therefore, a shelf that is 0.75 inch thick can

    support a weight of 60 pounds.

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    GRAPHING A SYSTEM OF

    QUADRATIC INEQUALITY

    x

    yGraph the system of quadratic inequalities.

    yx2-4 Inequality 1

    y

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    QUADRATIC INEQUALITY IN

    ONE VARIABLE

    1. To solve ax2+ bx + c < 0 (or ax2 + bx + c 0),

    graph y = ax2+ bx + c and identify the x values

    for which the graph lies below (or on and

    below) the x-axis.

    2. To solve ax2+ bx + c > 0 (or ax2 + bx + c 0),

    graph y = ax2+ bx + c and identify the x values

    for which the graph lies above(or on andbelow)the x-axis.

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    EXAMPLE

    x

    y

    3

    1

    1 5

    Solve x2- 6x + 5 < 0

    Solution:

    1. The solution consist of the x-values for

    which the graph of y = x2 - 6x + 5 lies

    below the x-axis.2. Find the graphs x-intercepts by letting y

    = 0

    0 = x2 - 6x + 5

    0 = (x-1)(x-5)

    x = 1 or x = 5

    Sketch a parabola that opens up and has1 and 5 as x-intercepts.

    3. The graph lies below the x-axis between

    x = 1 and x = 5.

    4. The solution of the given inequality is

    1

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    EXAMPLE

    Solve 2x2 + 3x -3 0

    Solution:

    1. The solution consist of the x-values for which the graph ofy = 2x2 + 3x -3 lies on and above the x-axis.

    2. Find the graphs x-intercepts by letting y = 0

    0 = 2x2 + 3x -3

    x3 32 4(2)(3)

    2(2)

    x3 33

    4

    x 0.69or

    x 2.19

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    EXAMPLE(contd.)

    -2.19 4 x

    y

    0.69

    1

    1. Sketch a parabola that opens up

    and has 0.69 and -2.19 as x-

    intercepts.

    2. The graph lies on and above the x-

    axisto the left of ( and including) x

    = -2.19 and to the right of ( and

    including) x = 0.69.

    3. The solution of the given inequality

    is approximately x -2.19 or x

    0.69.

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    SOLVING A QUADRATIC

    INEQUALITY ALGEBRAICALLY

    Solve:

    x2+ 2x 8

    Solution:

    First replace the inequality symbol with equal sign.x2+ 2x = 8

    x2+ 2x - 8 = 0

    (x+4)(x-2) = 0

    x = - 4 or x = 2

    The numbers -4 and 2 are the critical x-valuesof the inequality x2+ 2x

    8. Plot -4 and 2 on a number line.

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    SOLUTION (contd.)

    -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

    These critical x-values partition the number line into

    three intervals.

    Test x = -5, (-5)2+ 2(-5) = 15 8

    Test x = 0, (0)2 +2(0) = 0 8Test x = -3, (3)2+ 2(3) = 15 8

    Hence the solution is -4 x 2.

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    REAL LIFE EXAMPLE

    For a driver aged x years, a study found that the drivers reaction time V(x)

    (in milliseconds) to a visual stimulus such as traffic can be modeled by:

    V(x) = 0.005x2 - 0.23x + 22, 16 x 70

    At what ages does a drivers reaction time tend to be greater than 25

    milliseconds?

    Solution:

    The values of x for which

    V(x) > 25

    0.005x2- 0.23x +22 > 25

    0.005x2- 0.23x - 3 > 0The solution consists of the x- values for which the graph lies above the x-

    axis.

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    SOLUTION(contd.)

    The graphs x intercept is found by letting y =0 and using the quadratic

    formula to solve for x.

    0.005x2- 0.23x - 3 = 0

    x 0.23 (0.23)2 4 0.005 (3)

    2 0.005

    x0.23 .0529 .06

    .01

    x0.23 .1129

    .01

    x

    0.23 0.34

    .01

    x 57approx.

    or

    x 11approx

    Rejecting the negative value, thegraphs x-intercept is about 57. The

    graph of 0.005x2 - 0.23x - 3 = 0 lies in

    the domain 16 x 70. The graph lies

    above the x-axis when 57 < x 70.

    Hence the drivers over 57 years old

    tend to have reaction times greater than

    25 milliseconds.

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    MODELING WITHQUADRATIC FUNCTIONS

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    QUADRATIC FUNCTION IN

    VERTEX FORM

    (2,-3)

    (4,1)

    x

    y

    1

    1

    Write the quadratic function for the parabola

    shown.

    Solution:

    The vertex shown is (h,k) = (2,-3)Using the vertex form of the quadratic function.

    y = a(x-h)2+ k

    y = a(x-2)2- 3

    Use the other given point (4,1) to find a.

    1 = a(4-2)2- 3

    1 = 4a - 3

    4 = 4a1 = a

    Hence the quadratic function for the parabola is

    y = (x-2)2-3

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    QUADRATIC FUNCTION IN

    INTERCEPT FORM

    (-1,2)

    (-2)

    (3)

    x

    y

    1

    1

    Write the quadratic function for the parabola

    shown.

    Solution:

    The x intercepts shown are p = -2, q = 3Using the intercept form of the quadratic

    function.

    y = a(x-p)(x-q)

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

    Use the other given point (-1,2) to find a.

    2 = a(-1+2)(-1-3)

    2 = -4a-1/2 = a

    Hence the quadratic function for the parabola

    is y = -1/2(x+2)(x-3)

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    END