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Chapter 13 Quadratic Relations and Functions Corresponding Notes Date Assigned Day 1 & 2 Friday 3/9 Page 507: Monday Worksheet: HW Day 3 Consecutive Integer 2 Tuesday 3/13 Worksheet: HW Day 4 Geometric word problems Worksheet: HW Day 5 More Geometric word problem 3 Worksheet: Parabolas 5 Solving Quadratic Equations Day 3 Notes Word Problems Consecutive ntegers Day 4 Notes Geometric Word Problems Day 5 Notes More Word Problems Day 6 Notes Graphing Quadratics 3/12 Wednesday 3/14 Thursday 3/15 Assignment # 3 – 12 (all) 15 – 42 1 every 3rd ( 1st column) 4 Consecutive Integer HW 1. 7 or –3 5. 6 and 7 2. 5 or –3 6. 8 and 10 3. –10 and –8 7 4. 7 and 8 – 8 and -7 or or 7. 10. 2(x – 4)(x + 3) 3 10 Warm - up: Simplify 5 72 1. The ratio of the measures of the base and the altitude of a parallelogram is 3:4. The area of the parallelogram is 1,200 square centimeters. Find the measure of the base and altitude of the parallelogram. A = bh 1200 = (3x)(4x) 1200 =12x2 4x 12 x2 = 100 x2 – 100 = 0 3x Let 3x = base 4x = height 12 30 cm 40 cm (x + 10) (x – 10) = 0 (x + 10) = 0 (x – 10) = 0 x = -10 x = 10 2. The altitude of a triangle is 5 less than its base. The area of the triangle is 42 square inches. Find its base and altitude. x-5 x Let x = base 12 in x – 5 = altitude 7 in A = ½ bh 42 = ½ x (x – 5 ) 2(42 = ½ x (x – 5 )) 84 = x (x – 5 ) 84 = x2 - 5x 0 = x2 – 5x – 84 0 =(x – 12)(x +7) (x - 12) = 0 x = 12 (x + 7) = 0 x = -7 3. The length of a rectangle exceeds its width by 4 inches. Find the dimensions of the rectangle it its area is 96 square inches. A =wl x 96 =x(x+4) 96 =x2 + 4x 0 = x2 + 4x - 96 8 in 0 =(x – 8)(x +12) X+4 Let x = width x + 4 = length 12 in (x - 8) = 0 (x + 12) = 0 x=8 x = -12 4. If the measure of one side of a square is increased by 2 centimeters and the measure of the adjacent side is decreased by 2 centimeters, the area of the resulting rectangle is 32 square centimeters. Find the measure of one side of the square. Let x = side A =lw 32 =(x + 2)(x - 2) 32 =x2 + 2x - 2x - 4 32 =x2 - 4 6 cm -32 x-2 x+2 -32 0 =x2 – 36 0 =(x – 6)(x + 6) (x - 6) = 0 x=6 (x + 6) = 0 x = -6 5. Joe’s rectangular garden is 6 meters long and 4 meters wide. He wishes to double the area of his garden by increasing its length and width by the same amount. Find the number of meters by which each dimension must be increased. 6m A =6(4) 24 4m A =lw 48 =(x + 6)(x + 4) 48 = x2 + 6x + 4x + 24 48 = x2 + 10x + 24 -48 x+6 L= 8 m x+4 W= 6 m -48 0 = x2 + 10x - 24 0 =(x – 2)(x + 12) (x - 2) = 0 x=2 (x + 12) = 0 x = -12 March 26, 2009 Pick up 1 Homework: Worksheet # 4 Geometric Problems