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Lesson 8.3 , For use with pages 565-571

4 x + 5. x – 1. Graph. y =. 1. 2. – . x. ANSWER. 2. Find the vertical and horizontal asymptotes of y = . ANSWER. vertical: x = 1; horizontal: y = 4. Lesson 8.3 , For use with pages 565-571. 8.4 Rationalize Expressions. x 2 – 2 x – 15. x 2 – 2 x – 15.

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Lesson 8.3 , For use with pages 565-571

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  1. 4x + 5 x – 1 Graph y = 1. 2 – . x ANSWER 2. Find the vertical and horizontal asymptotes ofy = . ANSWER vertical:x = 1; horizontal:y = 4 Lesson 8.3, For use with pages 565-571

  2. 8.4 Rationalize Expressions

  3. x2 – 2x – 15 x2 – 2x – 15 Simplify : x2– 9 x2– 9 (x +3)(x –5) (x +3)(x –5) = (x +3)(x –3) (x +3)(x –3) = x – 5 x – 5 = x – 3 x – 3 ANSWER EXAMPLE 1 Simplify a rational expression SOLUTION Factor numerator and denominator. Divide out common factor. Simplified form

  4. Packaging A company makes a tin to hold flavored popcorn. The tin is a rectangular prism with a square base. The company is designing a new tin with the same base and twice the height of the old tin. EXAMPLE 2 Solve a multi-step problem • Find the surface area and volume of each tin. • Calculate the ratio of surface area to volume for each tin. • What do the ratios tell you about the efficiencies of the two tins?

  5. S 2s2 + 4sh S 2s2 + 8sh = = V s2h V 2s2h 2s(s + 4h) s(22 + 4h) = = 2s(sh) s(sh) 2s + 4h s + 4h = = sh sh EXAMPLE 2 Solve a multi-step problem SOLUTION New tin Old tin S = 2s2 + 4sh S = 2s2 + 4s(2h) STEP 1 Find surface area, S. = 2s2 + 8sh V = s2h V = s2(2h) Find volume, V. = 2s2h STEP 2 Write ratio ofStoV. Divide out common factor. Simplified form

  6. STEP 3 Because the left side of the inequality has a greater numerator than the right side and both have the same (positive) denominator. The ratio of surface area to volume is greater for the old tin than for the new tin. So, the old tin is less efficient than the new tin. s + 4h 2s + 4h > sh sh EXAMPLE 2 Solve a multi-step problem

  7. 2(x + 1) 2(x + 1) 1. (x + 1)(x + 3) (x + 1)(x + 3) 2(x +1) = (x +1)(x + 3) 2 2 = x + 3 x + 3 ANSWER for Examples 1 and 2 GUIDED PRACTICE Simplify the expression, if possible. SOLUTION Divide out common factor. Simplified form

  8. 40x + 20 40x + 20 10x + 30 10x + 30 20(2x +1) = 10(x + 3) 20(2x +1) = 10(x + 3) 2(2x +1) 2(2x +1) = x + 3 x + 3 ANSWER for Examples 1 and 2 GUIDED PRACTICE 2. SOLUTION Factor numerator and denominator. Divide out common factor. Simplified form

  9. 4 4 4 x(x + 2) x(x + 2) x(x + 2) ANSWER for Examples 1 and 2 GUIDED PRACTICE 3. SOLUTION Simplified form

  10. x + 4 x + 4 x 2 –16 x 2 –16 (x + 4) = (x + 4)(x – 4) (x + 4) = (x + 4)(x – 4) 1 1 = x – 4 x – 4 ANSWER for Examples 1 and 2 GUIDED PRACTICE 4. SOLUTION Factor numerator and denominator. Divide out common factor. Simplified form

  11. (x – 3)(x + 1) = (x – 3)(x + 2) x2 – 2x – 3 x2 – 2x – 3 (x – 3)(x + 1) = x2 – x – 6 x2 – x – 6 (x – 3)(x + 2) x + 1 x + 1 = x + 2 x + 2 ANSWER for Examples 1 and 2 GUIDED PRACTICE 5. SOLUTION Factor numerator and denominator. Divide out common factor. Simplified form

  12. 2x(x + 5) = (3x + 1)(x + 5) 2x2 + 10x 2x2 + 10x 2x(x + 5) = 3x2 + 16x + 5 3x2 + 16x + 5 2x 2x (3x + 1)(x + 5) 3x + 1 3x + 1 = ANSWER for Examples 1 and 2 GUIDED PRACTICE 6. SOLUTION Factor numerator and denominator. Divide out common factor. Simplified form

  13. for Examples 1 and 2 GUIDED PRACTICE 7. What If? In Example 2, suppose the new popcorn tin is the same height as the old tin but has a base with sides twice as long. What is the ratio of surface area to volume for this tin? SOLUTION New tin Old tin S = 2s2 + 4sh S = 2 (2s)2 + 4(2s)h STEP 1 Find surface area, S. = 8s2 + 8sh V = s2h V = (2s)2h Find volume, V. = 4s2h

  14. 4s(2s + 2h) = 4s(sh) 2s + 4h 2s + 4h 2s + 4h = = = = = sh sh sh ANSWER 2s2 + 4sh 8s2 + 8sh = = s2h 2s2h s(2s + 4h) = s(sh) S S V V for Examples 1 and 2 GUIDED PRACTICE STEP 2 Write ratio ofStoV. Divide out common factor. Simplified form

  15. 56x7y4 7x4y3 8x 3y 8xy3 = 4y 2x y2 8 7 x x6y3y = 7x6y = 8 x y3 ANSWER The correct answer is B. EXAMPLE 3 Standardized Test Practice SOLUTION Multiply numerators and denominators. Factor and divide out common factors. Simplified form

  16. 3x –3x2 x2 + x – 20 x2 + x – 20 x2 + 4x – 5 3x 3x 3x –3x2 3x(1– x) x2 + 4x – 5 (x –1)(x +5) (x + 5)(x – 4) = 3x 3x(1– x)(x + 5)(x – 4) = (x –1)(x + 5)(3x) 3x(–1)(x – 1)(x + 5)(x – 4) = (x – 1)(x + 5)(3x) 3x(–1)(x – 1)(x + 5)(x – 4) = (x – 1)(x + 5)(3x) EXAMPLE 4 Multiply rational expressions Multiply: SOLUTION Factor numerators and denominators. Multiply numerators and denominators. Rewrite 1– xas (– 1)(x – 1). Divide out common factors.

  17. ANSWER –x + 4 EXAMPLE 4 Multiply rational expressions = (–1)(x – 4) Simplify. = –x + 4 Multiply.

  18. x + 2 x + 2 x + 2 Multiply: (x2 + 3x + 9) (x2 + 3x + 9) x3 – 27 x3 – 27 x3 – 27 x2 + 3x + 9 = 1 (x + 2)(x2 + 3x + 9) (x + 2)(x2 + 3x + 9) = = (x – 3)(x2 + 3x + 9) (x – 3)(x2 + 3x + 9) x + 2 x + 2 = x – 3 x – 3 ANSWER EXAMPLE 5 Multiply a rational expression by a polynomial SOLUTION Write polynomial as a rational expression. Factor denominator. Divide out common factors. Simplified form

  19. 6xy2 3x5 y2 8. 8xy 9x3y 18x6y4 6xy2 3x5 y2 = 72x4y2 2xy 9x3y 18 x4y2x2y2 = 18 4 x4 y2 x2y2 = 4 for Examples 3, 4 and 5 GUIDED PRACTICE Multiply the expressions. Simplify the result. SOLUTION Multiply numerators and denominators. Factor and divide out common factors. Simplified form

  20. 2x(x –5) (x –5)(x +5) 2x2 – 10x x + 3 x2– 25 2x2 x + 3 = (x) 2x 2x(x –5) (x + 3) = (x –5)(x + 5)2x (x) 2x(x –5) (x + 3) = (x –5)(x + 5)2x (x) x + 3 = x(x + 5) for Examples 3, 4 and 5 GUIDED PRACTICE 2x2 – 10x x + 3 9. x2– 25 2x2 SOLUTION Factor numerators and denominators. Multiply numerators and denominators. Divide out common factors. Simplified form

  21. x + 5 x2+x + 1 x3– 1 x + 5 x2+x + 1 = (x – 1) (x2+x + 1) 1 (x + 5) (x2+x + 1) = (x – 1) (x2+x + 1) (x + 5) (x2+x + 1) = (x – 1) (x2+x + 1) x + 5 = x – 1 for Examples 3, 4 and 5 GUIDED PRACTICE x + 5 10. x2+x + 1 x3– 1 SOLUTION Factor denominators. Multiply numerators and denominators. Divide out common factors. Simplified form

  22. Divide : 7x 7x x2 – 6x x2 – 6x 2x – 10 2x – 10 x2 – 11x + 30 x2 – 11x + 30 7x x2 – 11x + 30 = 2x – 10 x2 – 6x (x – 5)(x – 6) 7x = 2(x – 5) x(x – 6) 7x(x – 5)(x – 6) = 2(x – 5)(x)(x – 6) = 7 7 ANSWER 2 2 EXAMPLE 6 Divide rational expressions SOLUTION Multiply by reciprocal. Factor. Divide out common factors. Simplified form

  23. 6x2 + x – 15 Divide : (3x2 + 5x) 4x2 6x2 + x – 15 (3x2 + 5x) 4x2 6x2 + x – 15 1 = 3x2 + 5x 4x2 (3x + 5)(2x – 3) 1 = x(3x + 5) 4x2 (3x + 5)(2x – 3) = 4x2(x)(3x + 5) 2x – 3 2x – 3 = 4x3 4x3 ANSWER EXAMPLE 7 Divide a rational expression by a polynomial SOLUTION Multiply by reciprocal. Factor. Divide out common factors. Simplified form

  24. 11. x2 – 2x x2 – 2x 4x 4x 5x – 20 5x – 20 x2 – 6x + 8 x2 – 6x + 8 x2 – 6x + 8 4x = 5x – 20 x2 – 2x 4(x)(x – 4)(x – 2) = 5(x – 4)(x)(x – 2) 4(x)(x – 4)(x – 2) = 5(x – 4)(x)(x – 2) 4 = 5 for Examples 6 and 7 GUIDED PRACTICE Divide the expressions. Simplify the result. SOLUTION Multiply by reciprocal. Factor. Divide out common factors. Simplified form

  25. 2x2 + 3x – 5 12. (2x2 + 5x) 6x 2x2 + 3x – 5 (2x2 + 5x) 6x 2x2 + 3x – 5 1 = 6x (2x2 + 5x) (2x + 5)(x – 1) = 6x(x)(2 x + 5) (2x + 5)(x – 1) = 6x(x)(2 x + 5) x – 1 = 6x2 for Examples 6 and 7 GUIDED PRACTICE SOLUTION Multiply by reciprocal. Factor. Divide out common factors. Simplified form

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