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Chapter 13 Quadratic Relations and Functions

Corresponding Notes. Date Assigned. Assignment. #. Day 1 & 2. Solving Quadratic Equations. Friday 3/9. Page 507: 3 – 12 (all) 15 – 42 every 3 rd ( 1 st column). 1. Day 3 Notes Word Problems Consecutive ntegers. Monday 3/12. Worksheet: HW Day 3 Consecutive Integer. 2.

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Chapter 13 Quadratic Relations and Functions

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  1. Corresponding Notes Date Assigned Assignment # Day 1 & 2 Solving Quadratic Equations Friday 3/9 Page 507: 3 – 12 (all) 15 – 42 every 3rd ( 1st column) 1 Day 3 Notes Word Problems Consecutive ntegers Monday 3/12 Worksheet: HW Day 3 Consecutive Integer 2 Day 4 Notes Geometric Word Problems Tuesday 3/13 Worksheet: HW Day 4 Geometric word problems 3 Day 5 Notes More Word Problems Wednesday 3/14 Worksheet: HW Day 5 More Geometric word problem 4 Day 6 Notes Graphing Quadratics Thursday 3/15 Worksheet: Parabolas 5 Chapter 13 Quadratic Relations and Functions

  2. 7 or –3 5 or –3 3. 7 4. 7 and 8 or – 8 and -7 5. 6 and 7 6. 8 and 10 or –10 and –8 2(x – 4)(x + 3) 10. Consecutive Integer HW

  3. Warm - up: Simplify

  4. Theratioof 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 12 x2 = 100 3x x2 – 100 = 0 Let 3x = base 4x = height 30 cm 40 cm (x + 10) (x – 10) = 0 (x + 10) = 0 (x – 10) = 0 x = -10 x = 10

  5. 2. The altitudeof a triangle is 5 less than its base. The areaof the triangle is 42 square inches. Find its base and altitude. A = ½ bh 42 = ½ x (x – 5 ) x - 5 2(42 = ½ x (x – 5 )) 84 = x (x – 5 ) x 84 = x2 - 5x 0 = x2 – 5x – 84 12 in 7 in Let x = base x – 5 = altitude 0 =(x – 12)(x +7) (x - 12) = 0 (x + 7) = 0 x = 12 x = -7

  6. 3. The lengthof a rectangle exceeds its width by 4 inches. Find the dimensions of the rectangle it its area is 96 square inches. A =wl 96 =x(x+4) x 96 =x2 + 4x X + 4 0 = x2 + 4x - 96 8 in 12 in Let x = width x + 4 = length 0 =(x – 8)(x +12) (x - 8) = 0 (x + 12) = 0 x = 8 x = -12

  7. 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 6 cm 32 =(x + 2)(x - 2) 32 =x2 + 2x - 2x - 4 32 =x2 - 4 -32 -32 x - 2 0 =x2 – 36 0 =(x – 6)(x + 6) x + 2 (x - 6) = 0 (x + 6) = 0 x = 6 x = -6

  8. 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. A =lw 6 m 48 =(x + 6)(x + 4) A =6(4) 24 4 m 48 = x2 + 6x + 4x + 24 48 = x2 + 10x + 24 -48 -48 0 = x2 + 10x - 24 x + 6 0 =(x – 2)(x + 12) L= 8 m W= 6 m (x - 2) = 0 (x + 12) = 0 x + 4 x = 2 x = -12

  9. March 26, 2009 Pick up 1 Homework: Worksheet # 4 Geometric Problems

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