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

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)


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)


• 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 + ____


• 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


• 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


• 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




*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)


• 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


• 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)


• 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




*If a = 1, just put brackets around the 1st 2 termsEx. 2: Complete the square for y = - x² + 12x + 7




*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)




• Ex. 2... y = -1(x² - 12x + 36 - 36) + 7


• 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)



terms• Ex. 2... y = -1(x² - 12x + 36) + 7 + 36...............y = -1(x - 6)² + 43The equation is now in vertex form




*If a = 1, just put brackets around the 1st 2 termsEx. 3: Complete the square for y = - 3x² - 24x + 1




*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)




• Ex. 3... y = - 3(x² + 8x + 16 – 16) + 1


• 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)



terms• Ex. 3... y = - 3(x + 4)² + 49The equation is now in vertex form


• 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?


• 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


• 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?


• 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)


• 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)


• 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.


• 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

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

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