Chapter 4 Quadratics4.6 Completing the Square to

Determine Max or Min Algebraically

Humour Break

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)

4.6 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)

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 + ____

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

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

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 termsEx. 1: Complete the square for y = x² + 10x - 3

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 termsEx. 1: Complete the square for y = x² + 10x - 3 y = 1(x² + 10x) – 3 (I showed the 1 but you don’t have to)

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

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)

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 - 28The equation is now in vertex form

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 termsEx. 2: Complete the square for y = - x² + 12x + 7

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 termsEx. 2: Complete the square for y = - x² + 12x + 7 y = -1(x² - 12x) + 7 (Note: note the sign change)

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

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)

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 + 43The equation is now in vertex form

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 termsEx. 3: Complete the square for y = - 3x² - 24x + 1

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 termsEx. 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)

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

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)

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 + 49The equation is now in vertex form

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?

4.6 Completing the Square to Determine Max or Min

• Word Problem – Example 1:C = 0.2b² - 10b + 650C = 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

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?

4.6 Completing the Square to Determine Max or Min

• Word Problem – Example 2:Let w represent the width & l the length(a) 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)

4.6 Completing the Square to Determine Max or Min

• Word Problem – Example 2:Let w represent the width & l the length(a) (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)

4.6 Completing the Square to Determine Max or Min

• Word Problem – Example 2:A = -w² + 30wA = -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.

4.6 Completing the Square to Determine Max or Min

• Word Problem – Example 2: (Cont’d)The length of the garden is:l = 30 – wl = 30 – 15l = 15mWhen the width and length of the garden are

both 15m, the area of the garden is maximized at 225m

Homework

• Thursday, May 19th

p.390, #4, 6, 8 & 10-13• Friday, May 20th

p.392, #16, 17a, 18, 21, 23 & 23