Chapter 4 Quadratics 4.6 Completing the Square to Determine Max or Min Algebraically

1. Chapter 4 Quadratics4.6 Completing the Square to Determine Max or Min Algebraically

2. Humour Break

3. 4.6 Completing the Square to Determine Max or Min Recall from the Chapter 3 toolkit... y = x² + 4x + 4   y = (x + 2)(x + 2) y = ax² + bx + c y = (√x+ √4) (√x+ √4) y = (√x+ √4)² • If the square of ½ of b gives you the product of a x c, you have a perfect square • Ex. 4 ÷ 2 = 2... 2² = 4... 1 x 4 = 4 (so true)

4. 4.6 Completing the Square to Determine Max or Min Another perfect square example, where a ≠ 1... y = 16x² + 16x + 4   y = (4x + 2)(4x + 2) y = ax² + bx + c y = (√16x+ √4) (√16x+ √4) y = (√16x+ √4)² • If the square of ½ of b gives you the product of a x c, you have a perfect square • Ex. 16 ÷ 2 = 8... 8² = 64... 16 x 4 = 64 (so true)

5. 4.6 Completing the Square to Determine Max or Min • Warm-up... • Fill-in each blank with a number that would create a perfect-square trinomial • Remember, if a = 1, take ½ of “b” and square it to get the “c” that would create a perfect square... • x² + 14x + ___ • x² - 8x + ____ • x² + 3x + ____

6. 4.6 Completing the Square to Determine Max or Min • Warm-up... • Fill-in each blank with a number that would create a perfect-square trinomial • Remember, if a = 1, take ½ of “b” and square it to get the “c” that would create a perfect square... • x² + 14x + 49 • x² - 8x + 16 • x² + 3x + 2.25

7. 4.6 Completing the Square to Determine Max or Min • A quadratic relation in standard form y = ax² + bx + c can be rewritten in vertex form y = a(x – h)² + k by creating a perfect square in the equation, then factoring the square • This technique is called completing the square

8. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 1. Remove the common constant factor from both the “ax²” and the “bx” if a is ≠ 1* (Do not remove the factor from the “c”) *If a = 1, just put brackets around the 1st 2 terms Ex. 1: Complete the square for y = x² + 10x - 3

9. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 1. Remove the common constant factor from both the “ax²” and the “bx” if a is ≠ 1* (Do not remove the factor from the “c”) *If a = 1, just put brackets around the 1st 2 terms Ex. 1: Complete the square for y = x² + 10x - 3 y = 1(x² + 10x) – 3 (I showed the 1 but you don’t have to)

10. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 2. Find the constant “c” that must be added or subtracted to create a perfect square. This value =‘s the square of ½ of the coefficient of the x term “b” from step 1. Rewrite the expression by adding then subtracting this value after the x-term inside the brackets • Ex. 1... y = 1(x² + 10x +25 - 25) - 3

11. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 3. Group the three terms that form the perfect square in the brackets. Move the subtracted value outside the brackets after multiplying it by the common constant factor (if a was ≠ 1) Ex. 1... y = 1(x² + 10x +25) – 3 – 25 (here we don’t have to multiply the -25 before we move it because the common constant factor was 1)

12. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 4. Factor the perfect square and collect like terms • Ex. 1... y = 1(x² + 10x +25) – 3 – 25 ...............y = (x + 5)² - 28 The equation is now in vertex form

13. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 1. Remove the common constant factor from both the “ax²” and the “bx” if a is ≠ 1* (Do not remove the factor from the “c”) *If a = 1, just put brackets around the 1st 2 terms Ex. 2: Complete the square for y = - x² + 12x + 7

14. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 1. Remove the common constant factor from both the “ax²” and the “bx” if a is ≠ 1* (Do not remove the factor from the “c”) *If a = 1, just put brackets around the 1st 2 terms Ex. 2: Complete the square for y = - x² + 12x + 7 y = -1(x² - 12x) + 7 (Note: note the sign change)

15. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 2. Find the constant “c” that must be added or subtracted to create a perfect square. This value =‘s the square of ½ of the coefficient of the x term “b” from step 1. Rewrite the expression by adding then subtracting this value after the x-term inside the brackets • Ex. 2... y = -1(x² - 12x + 36 - 36) + 7

16. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 3. Group the three terms that form the perfect square in the brackets. Move the subtracted value outside the brackets after multiplying it by the common constant factor (if a was ≠ 1) Ex. 2... y = -1(x² - 12x + 36) + 7 + 36 (here we have to multiply the -36 by -1 before we move it and it turns into a + 36 outside the brackets)

17. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 4. Factor the perfect square and collect like terms • Ex. 2... y = -1(x² - 12x + 36) + 7 + 36 ...............y = -1(x - 6)² + 43 The equation is now in vertex form

18. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 1. Remove the common constant factor from both the “ax²” and the “bx” if a is ≠ 1* (Do not remove the factor from the “c”) *If a = 1, just put brackets around the 1st 2 terms Ex. 3: Complete the square for y = - 3x² - 24x + 1

19. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 1. Remove the common constant factor from both the “ax²” and the “bx” if a is ≠ 1* (Do not remove the factor from the “c”) *If a = 1, just put brackets around the 1st 2 terms Ex. 3: Complete the square for y = - 3x² - 24x + 1 y = - 3(x² + 8x) + 1 (Note the sign change & don’t forget to divide -24x by -3)

20. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 2. Find the constant “c” that must be added or subtracted to create a perfect square. This value =‘s the square of ½ of the coefficient of the x term “b” from step 1. Rewrite the expression by adding then subtracting this value after the x-term inside the brackets • Ex. 3... y = - 3(x² + 8x + 16 – 16) + 1

21. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 3. Group the three terms that form the perfect square in the brackets. Move the subtracted value outside the brackets after multiplying it by the common constant factor (if a was ≠ 1) Ex. 3... y = - 3(x² + 8x + 16) + 1 + 48 (Here, we have to multiply the -16 by -3 before we move it outside the brackets so we will end up with +48 outside the brackets)

22. 4.6 Completing the Square to Determine Max or Min • Steps: to complete the square of y = ax² + bx + c • 4. Factor the perfect square and collect like terms • Ex. 3... y = - 3(x + 4)² + 49 The equation is now in vertex form

23. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 1: A glassworks that makes lead-crystal bowls has a daily production cost “C” in dollars given by the relation C = 0.2b² - 10b + 650 • How many bowls should be made to minimize production cost? What is the cost when this many bowls are made?

24. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 1: C = 0.2b² - 10b + 650 C = 0.2(b² - 50b)+ 650 (Step 1) C = 0.2(b² - 50b + 625 - 625) + 650 (Step 2) C = 0.2(b² - 50b + 625) + 650 – 125 (Step 3) C = 0.2(b – 25)² + 525 (Step 4) The vertex is at (25, 525). This means that the cost is minimized when 25 bowls are made. The production cost for 25 bowls is $525

25. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 2: A gardener wants to enclose a rectangular vegetable garden with 60 concrete curb stones, each 1 m long. • (a) Write an expression for the number of curb stones used (the perimeter) and for the total area • (b) Use the perimeter expression to rewrite the area expression in terms of only one variable • (c) How should the gardener lay out the curb stones to maximize the total enclosed area?

26. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 2: Let w represent the width & l the length • The perimeter is P = 2w + 2l and the area is A = l x w 60 = 2w + 2l (substitute 60 for P) 2l = 60 – 2w (rearrange) l = 30 – w (÷ by 2)

27. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 2: Let w represent the width & l the length • (Cont’d... ) A = l x w A = (30 – w)w (Substitute 30 – w for l) A = 30w - w² (Multiply using distributive law) A = - w² + 30w (Reverse order of right side)

28. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 2: A = -w² + 30w A = -1(w² - 30w) (Step 1) A = -1(w² - 30w + 225 – 225) (Step 2) A = -1(w² - 30w + 225) + 225 (Step 3) C = -1(w - 15)² + 225 (Step 4) The vertex is at (15, 225). This means that the maximum area of the garden occurs when the width is 15m.

29. 4.6 Completing the Square to Determine Max or Min • Word Problem – Example 2: (Cont’d) The length of the garden is: l = 30 – w l = 30 – 15 l = 15m When the width and length of the garden are both 15m, the area of the garden is maximized at 225m

30. Homework • Thursday, May 19th p.390, #4, 6, 8 & 10-13 • Friday, May 20th p.392, #16, 17a, 18, 21, 23 & 23