Modeling Data with Quadratic Functions

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Modeling Data with Quadratic Functions Powered By Docstoc
					Solving Quadratic
        Functions




           Lesson 5-3
Objective
• Today, you will . . .

• solve quadratic functions by using a
  variety of methods.




                          TEKS:b2A,d1A,d3A,d3C,d3D
Some Notes on
Quadratic Functions
1. The graphs of quadratic functions are
   parabolas.
2. The solution(s) that you will be looking for are
   the x-intercepts of the parabola.
3. The x-intercepts are also called the “roots,”
   “solutions,” or the “zeros”
4. Quadratic functions can have one, two, or
   no real solutions.
5. Quadratic functions that have no real
   solutions have complex (imaginary)
   solutions.
Different Graphs of Quadratics
    Two real solutions




                                X=-6
                                X=1


         x-intercepts, roots, zeros
Different Graphs of Quadratics
    One real solutions




                             X=3

   x-intercept, root, zero
Different Graphs of Quadratics
     No real solutions




                                No
                              Solution
  No x-intercepts (roots or
           zeros)
Today you’ll find…

• The solutions to quadratic
  equations by factoring

• For example:

 GIVEN: y = Ax2 + Bx + C
 FIND: The solutions, roots, zeros,
 or x-intercepts
Solve by factoring: x2 + 9x + 20 = 0
                               1 x 20
    x2+ 9x + 20 = 0            2 x 20
                               4x5
   (x + 5)(x + 4) = 0
    x+5=0                x+4=0
    x=-5                  x=-4
   So, its two roots, solutions, zeros

         are      -5 & -4
Solve by factoring: x2 = -7x + 18

    x2   + 7x - 18 = 0                1 x 18
                                      2x9
    Hint: The sign of “B” goes with
                                      3x6
    the largest factor!
    (x - 2)(x + 9) = 0
     x-2=0          x+9=0
      x=2            x=-9
   So, its two roots, solutions, zeros

          are        2 & -9
Solve by factoring: x2 + 100 = 29x
                                 1 x 100

  x2   - 29x + 100 = 0           2 x 50
                                 4 x 25
                                 5 x 20
   (x - 4)(x - 25) = 0    10 x 10

    x-4=0          x - 25 = 0
     x=4                   x = 25
   So, its two roots, solutions, zeros

         are      4 & 25
Solve by factoring: x2 - 9 = 0
                                   1x9
         x2
          -9=0                     3x3

    (x + 3)(x - 3) = 0
    x+3=0                   x-3=0
    x=-3                     x=3
   So, its two roots, solutions, zeros are

                -3 & 3
Solve by factoring: x2 + x = 6
                                      1x6
      x 2   +x-6=0                    2x3
    Hint: The sign of “B” goes with
    the largest factor!
    (x - 2)(x + 3) = 0
     x-2=0          x+3=0
     x=2             x=-3
   So, its two roots, solutions, zeros

            are      2 & -3
Solve by factoring: x2 - 6x = - 8
                    1x8
     x=2            2x4

     x=4
 Solve by factoring: x2 - 8 = - 7x
     x=1             1x8
                     2x4

    x=-8
Solve by factoring: 2x2 + 13x + 15 = 0

 2x2 + 13x + 15 = 0             Multiply AxC                 = 30

                    Determine the factors of 30 that
                                                             1 x 30
                    give you 13
                                                             2 x 15
                                                             3 x 10

 (x + 3) (x + 10)              Write the factors

 (x + 3) (x + 10 )             Divide the #’s by A
      2        2


 (2x + 3) (x + 5)              If not divisible, send it in front of “x”,
                               if divisible then simplify.
 2x + 3 = 0            x+5=0                  Now solve both factors!
 2x = -3               x = -5                       Your solutions are
  x = -3/2                                          x = -3/2 and x = -5
Solve by factoring: 3x2 + 16x + 21 = 0

 3x2 + 16x + 21 = 0             Multiply AxC            = 63

                     Determine the factors of 63 that
                                                        1 x 63
                     give you 16
                                                        3 x 21
                                                        7x9

  (x + 7) (x + 9)              Write the factors

  (x + 7) (x + 9 )             Divide the #’s by A
       3       3


 (3x + 7) (x + 3)              If not divisible, send it in front of “x”,

                         if divisible then simplify.
3x + 7 = 0            x+3=0         Now solve both factors!
3x = -7               x = -3                      Your solutions are
 x = -7/3                                         x = -7/3 and x = -3
Solve by factoring: 2x2 – 5x = 7

 2x2 – 5x – 7 = 0       Multiply AxC                                = -14
                        Determine the factors of -14 that           1 x 14
                        give you -5
                                                                    2x7
 (x + 2) (x – 7)     Write the factors
 Remember, sign of “B” goes to the largest factor, in this case, the
 negative goes to the 7.

 (x + 2) (x – 7)         Divide the #’s by A
      2       2


 (x + 1) (2x – 7)            If not divisible, send it in front of “x”,
                              if divisible then simplify.

x+1=0               2x – 7 = 0              Now solve both factors!
x = -1              2x = 7                          Your solutions are
                    x = 7/2                        x = -1 and x = 7/2
Your turn!
Solve by factoring: 5x2 + 4x = 12
 5x2 + 4x – 12 = 0               AxC=                    -60
                                                    1 x 60 4 x 15
                                                    2 x 30   5 x 12
                                                    3 x 20   6 x 10
  (x – 6) (x + 10)    Write the factors (watch your signs!)


  (x – 6) (x + 10)    Divide the #’s by A
       5       5
  (5x – 6)(x + 2)     If not divisible, send it in front of “x”,
                      if divisible then simplify.

 5x – 6 = 0          x+2=0                Now solve both factors!
 5x = 6              x = -2
                                                    Your solutions are
 x = 6/5
                                                    x = 6/5 and x = -2
You try these by factoring . . .

1.   2x2 – 3x = 0
           x = 0 and 3/2
2.   4x2 + 5 = -9x
          x = -1 and -5/4
3.   6x2 + 55x = -9
          x = -9 and -1/6

				
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Jun Wang Jun Wang Dr
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