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6.1 maxima and minima solutions

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									                                      6.1 Maxima and Minima
We have already learned that quadratic equations can be written in different forms.

Standard Form: y  ax 2  bx  c
               y  a  x  h  k
                              2
Vertex Form:
Factored Form: y = a(x – r)(x – s)

Last unit we learned how to convert an equation from standard form to factored form. In factored
form, we can easily find the x-intercepts of the equation. What about equations that do not have x-
intercepts? We can convert these equations into vertex form. To do this, we can apply the
technique of completing the square.

For example, the equation y = 2x2 + 12x + 11 can be converted to vertex form as y = 2(x + 3)2 - 7.

To complete the square we must remember the perfect square trinomial:
             (a + b)2 = a2 + 2ab + b2 and         (a – b)2 = a2 – 2ab + b2

Find the values of c so that each expression is a perfect square trinomial:
   a) x2 – 20x + c                 b) x2 + 8x + c             c) x2 – 14x + c

   - 20 = 2ab                        8 = 2ab                    -14 = 2ab
   - 20 = 2(1)b                      8 = 2(1)b                  -14 = 2(1)b
   -10 = b                           4=b                         -7 = b

   c = b2                            c = b2                     c = b2
   c = (-10)2                        c = 42                     c = (-7)2
   c = 100                           c = 16                     c = 49

Use completing the square to factor: y = 3x2 – 12x + 11

   1. Group the terms containing x
      y = (3x2 – 12x) + 11

   2. Common factor the coefficients of the first 2 terms (this is the ‘a’ value)
      y = 3(x2 – 4x) + 11

   3. Complete the square inside the brackets. This is done by adding and subtracting a number
      which makes the binomial in the brackets a perfect square.
      y = 3[x2 – 4x + (-2)2 – (-2)2] + 11

   4. Take (-2)2 out of the brackets by multiplying it by 3(the ‘a’ value).
      y = 3(x2 – 4x + 4) – 3(4) + 11

   5. Factor the perfect square trinomial (PST) and simplify.
      y = 3(x – 2)2 – 12 + 11
      y = 3(x – 2)2 – 1

See examples 2 – 4 on pages 267 – 269 in the textbook.
Use completing the square to sketch the graphs of the following parabolas.
   a) y = x2 + 2x + 7                                b) y = -3x2 + 18x - 28




Direction of opening:

Vertex:

Equation of AOS:

Min/Max Value:




Practice: pg. 270 – 271 # 2bdf, 3bg, 5, 6a, 8ac, 10bd, 12, 15

								
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