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Name Class 6-1 Date Reteaching Roots and Radical Expressions For any real numbers a and b and any positive integer n, if a raised to the nth power equals b, then a is an nth root of b. Use the radical sign to write a root. The following expressions are equivalent: index power an 5 b g radicand n !b ! b5a radical sign Problem What are the real-number roots of each radical expression? 3 a. !343 b. 4 1 Å625 3 Because (7)3 5 343, 7 is a third (cube) root of 343. 3 Therefore, !343 5 7. (Notice that (27)3 5 2343, so 27 is not a cube root of 343.) 1 1 1 Because Q 15 R 4 5 625 and Q 215 R 4 5 625 , both 15 and 215 are real-number fourth roots of 625 . c. !20.064 Because (20.4)3 5 20.064,20.4 is a cube root of 20.064 and is, in fact, the only one. 3 So, !20.064 5 20.4. d. !225 Because (5)2 5 (25)2 5 25, neither 5 nor 25 are second (square) roots of 225. There are no real-number square roots of 225. Exercises Find the real-number roots of each radical expression. 3 1. !169 213, 13 4. 7. 3 4 2. !729 9 218 212 5. 4 225 no real sq root 8. !0.1296 20.6, 0.6 Å Å 4 10. !20.0001 no real 4th root 4 3. !0.0016 20.2, 0.2 Å121 2 2 211 , 11 4 11. 5 1 1 Å 243 3 6. 3 125 5 Å 216 6 3 9. !20.343 20.7 12. 3 8 Å 125 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 9 2 5 Name Class 6-1 Date Reteaching (continued) Roots and Radical Expressions n You cannot assume that "an 5 a. For example, "(26)2 5 !36 5 6, not 26. This leads to the following property for any real number a: n If n is odd "an 5 a If n is even "an 5 u a u n Problem What is the simplified form of each radical expression? 3 a. " 1000x3y9 3 3 " 1000x3y9 5 " 103x3(y3)3 Write each factor as a cube. 3 5" (10xy3)3 Write as the cube of a product. 5 10xy3 Simplify. 8 256g b. 4 4 16 Åh k 4 256g8 Å h4k16 5 4 44(g2)4 Å h4(k4)4 Write each factor as a power of 4. 2 4 4g 5 4 a 4b Å hk 5 4g2 u hu k4 Write as the fourth power of a quotient. Simplify. The absolute value symbols are needed to ensure the root is positive when h is negative. Note that 4g2 and k4 are never negative. Exercises Simplify each radical expression. Use absolute value symbols when needed. 3 13. "36x2 6»x… 16. "x20 "y8 14. " 216y3 6y (x 1 3)3 17. 3 Å (x 2 4)6 x10 y4 27z3 19. 3 Å (z 1 12)6 (y 2 4)8 22. 4 Å (z 1 9)4 3z (z 1 12)2 (y 2 4)2 »z 1 9… 15. x 1 3 (x 2 4)2 4 20. " 2401x12 7»x3… 6 6 a b 23. 3 3 Å c a 2b 2 c 1 Å 100x2 5 18. " x10y15z5 x2y3z 1331 21. 3 Å x3 11 x 3 24. " 2x3y6 2xy2 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 10 1 10»x… Name Class 6-2 Date Reteaching Multiplying and Dividing Radical Expressions You can simplify a radical if the radicand has a factor that is a perfect nth power and n is the index of the radical. For example: n n !xynz 5 y !xz Problem What is the simplest form of each product? 3 3 a. ! 12 ? ! 10 n n n Use !a ? !b 5 !ab. 3 3 3 ! 12 ? ! 10 5 ! 12 ? 10 3 5 "22 ? 3 ? 2 ? 5 Write as a product of factors. 3 3 5" 2 ?3?5 Find perfect third powers. 3 3 5 "23 ? "3 ? 5 n Use !ab 5 !a ? !b. 3 5 2! 15 Use "an 5 a to simplify. n n n b. "7xy3 ? "21xy2 n n n Use !a ? !b 5 !ab. "7xy3 ? "21xy2 5 "7xy3 ? 21xy2 5 "7xy2y ? 3 ? 7xy2 Write as a product of factors. 5 "72x2(y2)2 ? 3y Find perfect second powers. 5 7xy2"3y Use !an 5 a to simplify. n Exercises Simplify each product. 3 3 1. !15x ? !35x 5x"21 2. " 50y2 ? " 20y 10y 4. 5"7x3y ? "28y2 5. 2" 9x5y2 ? " 2x2y5 70xy"xy 3 3 3 2x2y2 !18xy 3 6. !3 Q !12 2 !21 R 6 2 3"7 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 19 3 3 3. " 36x2y5 ? " 26x2y 26xy2 !x Name Class Date Reteaching (continued) 6-2 Multiplying and Dividing Radical Expressions Rationalizing the denominator means that you are rewriting the expression so that no radicals appear in the denominator and there are no fractions inside the radical. Problem What is the simplest form of !9y !2x ? Rationalize the denominator and simplify. Assume that all variables are positive. !9y !2x 5 9y Å 2x Rewrite as a square root of a fraction. 5 9y ? 2x Å 2x ? 2x Make the denominator a perfect square. 5 18xy Å 4x2 Simplify. 5 !18xy Write the denominator as a product of perfect squares. "22 ? x2 5 "18xy 2x Simplify the denominator. 5 "32 ? 2 ? x ? y 2x Simplify the numerator. 5 3"2xy 2x Use !an 5 a to simplify. n Exercises Rationalize the denominator of each expression. Assume that all variables are positive. 3 "6ab2 8. 3 "2a4b !5 "5x 7. x !x 3 11. 4"k9 3 16"k5 3 k "k 4 12. 4 4 "9y " 9x3y 9. 4 x "x 3 "3b a 4 "10 13. 4 "z2 3x5 x2"15xy Ä 5y 5y 4 "10z2 z 10. 3 19a2b " 19ac2 14. 3 4 c2 Ä abc Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 20 "10xy3 "30xy 6 "12y2 Name Class Date Reteaching 6-3 Binomial Radical Expressions Two radical expressions are like radicals if they have the same index and the same radicand. Compare radical expressions to the terms in a polynomial expression. Like terms: 4x3 11x3 Unlike terms: 4y3 11x3 Like radicals: 4!6 3 3 11 ! 6 Unlike radicals: 4!5 3 3 11 ! 6 The power and the variable are the same 4y2 Either the power or the variable are not the same. The index and the radicand are the same 2 4 "6 Either the index or the radicand are not the same. When adding or subtracting radical expressions, simplify each radical so that you can find like radicals. Problem What is the sum? !63 1 !28 !63 1 !28 5 !9 ? 7 1 !4 ? 7 Factor each radicand. 5 "3 2 ? 7 1 "2 2 ? 7 Find perfect squares. 5 "3 2 "7 1 "2 2 "7 n n n Use !ab 5 !a ? !b. 5 3 !7 1 2 !7 Use "an 5 a to simplify. 5 5 !7 Add like radicals. n The sum is 5!7. Exercises Simplify. 1. !150 2 !24 3 !6 3 3 3 4. 5 !2 2 !54 2 !2 3 3 3 2. !135 1 !40 5 !5 3. 6 !3 2 !75 !3 3 3 3 5. 2!48 1 !147 2 !27 0 6. 8 !3x 2 !24x 1 !192x 3 10 !3x Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 29 Name Class Date Reteaching (continued) 6-3 Binomial Radical Expressions • Conjugates, such as !a 1 !b and !a 2 !b, differ only in the sign of the second term. If a and b are rational numbers, then the product of conjugates produce a rational number: 2 2 Q !a 1 !b R Q !a 2 !b R 5 Q !a R 2 Q !b R 5 a 2 b. • You can use the conjugate of a radical denominator to rationalize the denominator. Problem What is the product? Q 2 !7 2 !5 R Q 2 !7 1 !5 R Q 2 !7 2 !5 R Q 2 !7 1 !5 R These are conjugates. 2 5 Q 2 !7 R 2 Q !5 R 2 Use the difference of squares formula. 5 28 2 5 5 23 Simplify. Problem How can you write the expression with a rationalized denominator? 4 !2 1 1 !3 4 !2 1 1 !3 5 4 !2 1 2 !3 ? 1 1 !3 1 2 !3 Use the conjugate of 1 1 !3 to rationalize the denominator. 5 4 !2 2 4 !6 123 Multiply. 5 A4 !2 2 4 !6B 4 !2 2 4 !6 5 2 22 2 Simplify. 5 24 !2 1 4 !6 5 22 !2 1 2 !6 2 Exercises Simplify. Rationalize all denominators. 7. A3 1 !6B A3 2 !6B 3 10. 2 2 !7 2 1 !7 211 1 4"7 3 8. 2 !3 1 1 "3 1 1 2 5 2 !3 11. A2 !8 2 6B A !8 2 4B 40 2 28 !2 9. Q 4 !6 2 1 R Q !6 1 4 R 12. !5 2 1 !3 2 !5 2 !15 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 30 20 1 15 !6 Name Class Date Reteaching 6-4 Rational Exponents You can simplify a number with a rational exponent by converting the expression to a radical expression: 1 1 2 92 5 !9 53 n xn 5 !x, for n . 0 1 3 83 5 ! 852 You can simplify the product of numbers with rational exponents m and n by raising the number to the sum of the exponents using the rule am ? an 5 am1n Problem What is the simplified form of each expression? 1 1 1 1 a. 364 ? 364 1 1 364 ? 364 5 364 1 4 Use am ? an 5 am1n . 1 5 362 Add. 2 5 !36 Use xn 5 !x. 56 Simplify. 1 n 3 2 b. Write Q 6x3 R Q 2x4 R in simplified form. 3 2 Q 6x3 R Q 2x4 R 3 2 5 6 ? 2 ? x3 ? x4 2 Commutative and Associative Properties of Multiplication 3 5 6 ? 2 ? x3 1 4 Use xm ? xn 5 xm1n. 17 5 12x12 Simplify. Exercises Simplify each expression. Assume that all variables are positive. 1 2 1 2 1 1 7 2. Q 2y4 R Q 3y3 R 6y 12 1. 53 ? 53 5 13 4. 2y3 y5 2y 15 1 1 1 1 5. 54 ? 54 !5 1 2 6. Q 23x 6 R Q 7x 6 R 221 !x Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 39 1 3. (211)3 ? (211)3 ? (211)3 211 Name Class Date Reteaching (continued) 6-4 Rational Exponents To write an expression with rational exponents in simplest form, simplify all exponents and write every exponent as a positive number using the following rules for a 2 0 and rational numbers m and n: 1 a2n 5 n a 1 a2m 5 am (am)n 5 amn (ab)m 5 ambm Problem 2 What is A8x9y23 B 23 in simplest form? 2 (8x9y23)23 2 5 A23 x9 y23 B 23 2 2 Factor any numerical coefficients. 2 5 A23 B 23 Ax9 B 23 Ay23 B 23 Use the property (ab)m 5 ambm. 5 222x26y2 Multiply exponents, using the property (am)n 5 amn. y2 5 2 6 2 x y2 5 6 4x Write every exponent as a positive number. Simplify. Exercises Write each expression in simplest form. Assume that all variables are positive. 1 1 7. A16x2 y8 B 22 4xy4 1 10. A25x26 y2 B 2 5y x3 1 8. Az23 B 9 1 1 4 9. Q 2x4 R 1 z3 16z4 12. a b 25x8 2 11. A8a23 b9 B 3 4b6 a2 1 x2 5 2 1 13. a 21 b x 5y 5 y 3 16. A9z10 B 2 27z15 16x 1 22 1 2 14. A27m9 n23 B 23 1 17. (2243)25 21 3 n2 9m6 1 32r2 4 15. a 4 b 2r 2 2s s 2 x5 10 18. a 1 b y2 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 40 5x4 4z2 x4 y5 Name Class Date Reteaching 6-5 Solving Square Root and Other Radical Equations Equations containing radicals can be solved by isolating the radical on one side of the equation, and then raising both sides to the same power that would undo the radical. Problem What is the solution of the radical equation? 2!2x 1 2 2 2 5 10 2 !2x 1 2 2 2 5 10 2 !2x 1 2 5 12 Add 2 to each side. !2x 1 2 5 6 Divide each side by 2. (!2x 1 2)2 5 62 Square each side to undo the radical. 2x 1 2 5 36 Simplify. 2x 5 34 Subtract 2 from each side. x 5 17 Divide each side by 2. Check the solution in the original equation. Check 2 !2x 1 2 2 2 5 10 2 !2(17) 1 2 2 2 0 10 2 !36 2 2 0 10 Write the original equation. Replace x by 17. Simplify. 12 2 2 0 10 10 5 10 The solution is 17. Exercises Solve. Check your solutions. 1 2. 3 !2x 5 12 8 1. x 2 5 13 169 1 4. (3x 1 4)2 2 1 5 4 7 1 7. (x 1 2) 2 2 5 5 0 23 1 3. !3x 1 5 5 11 12 5. (6 2 x) 2 1 2 5 5 23 6. !3x 1 13 5 4 1 8. !3 2 2x 2 2 5 3 211 3 9. !5x 1 2 2 3 5 0 5 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 49 Name Class 6-5 Date Reteaching (continued) Solving Square Root and Other Radical Equations An extraneous solution may satisfy equations in your work, but it does not make the original equation true. Always check possible solutions in the original equation. Problem What is the solution? Check your results. !17 2 x 2 3 5 x !17 2 x 2 3 5 x !17 2 x 5 x 1 3 Add 3 to each side to get the radical alone on one side of the equal sign. A !17 2 xB 2 5 (x 1 3)2 Square each side. 17 2 x 5 x2 1 6x 1 9 0 5 x2 1 7x 2 8 Rewrite in standard form. 0 5 (x 2 1)(x 1 8) Factor. x 2 1 5 0 or x 1 8 5 0 x 5 1 or Set each factor equal to 0 using the Zero Product Property. x 5 28 Check !17 2 x 2 3 5 x !17 2 1 2 3 0 1 !16 2 3 0 1 151 !17 2 x 2 3 5 x !17 2 (28) 2 3 0 28 !25 2 3 0 28 2 2 28 The only solution is 1. Exercises Solve. Check for extraneous solutions. 10. !5x 1 1 5 !4x 1 3 2 11. !x2 1 3 5 x 1 1 no 12. !3x 5 !x 1 6 3 solution 13. x 5 !x 1 7 1 5 9 14. x 2 3 !x 2 4 5 0 16 15. !x 1 2 5 x 2 4 7 16. !2x 2 10 5 x 2 5 5, 7 17. !3x 2 6 5 2 2 x 2 18. !x 2 1 1 7 5 x 10 19. !5x 1 1 5 !3x 1 15 7 20. !x 1 9 5 x 1 7 25 21. x 2 !x 1 2 5 40 47 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 50 Name 6-6 Class Date Reteaching Function Operations When you combine functions using addition, subtraction, multiplication, or division, the domain of the resulting function has to include the domains of both of the original functions. Problem Let f (x) 5 x2 2 4 and g(x) 5 !x. What is the solution of each function operation? What is the domain of the result? a. ( f 1 g)(x) 5 f (x) 1 g(x) 5 (x2 2 4) 1 (!x) 5 x2 1 !x 2 4 b. ( f 2 g)(x) 5 f (x) 2 g(x) 5 (x2 2 4) 2 (!x) 5 x2 2 !x 2 4 c. (g 2 f )(x) 5 g(x) 2 f (x) 5 (!x) 2 (x2 2 4) 5 2x2 1 !x 1 4 d. ( f ? g)(x) 5 f (x) ? g(x) 5 (x2 2 4)(!x) 5 x2 !x 2 4 !x The domain of f is all real numbers. The domain of g is all x $ 0. For parts a2d, there are no additional restrictions on the values for x, so the domain for each of these is x $ 0. f f (x) (x2 2 4) !x x2 2 4 e. g (x) 5 5 5 x g(x) !x As before, the domain is x $ 0. But, because the denominator cannot be zero, eliminate any values of x for which g(x) 5 0. The only value for which !x 5 0 is x 5 0. Therefore, f the domain of g is x . 0. f. g g(x) !x (x) 5 5 2 f f(x) x 24 Similarly, begin with x $ 0 and eliminate any values of x that make the denominator g f(x) zero: x2 2 4 5 0 when x 5 22 and x 5 2. Therefore, the domain of f is x $ 0 combined with x 2 22 and x 2 2. In other words, the domain is x $ 0 and x 2 2, or all nonnegative numbers except 2. Exercises Let f (x) 5 4x 2 3 and g(x) 5 x2 1 2. Perform each function operation and then find the domain of the result. 1. ( f 1 g)(x) 2. ( f 2 g)(x) x2 1 4x 2 1; all real numbers 4. ( f · g)(x) 4x3 2 3x2 1 8x 2 6; all real numbers 2x2 1 4x 2 5; all real numbers f 5. g (x) 4x 2 3 ; x2 1 2 all real numbers 3. (g 2 f )(x) x2 2 4x 1 5; all real numbers g 6. (x) f x2 1 2 4x 2 3 ; Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 59 x u 34 Name Class 6-6 Date Reteaching (continued) Function Operations • One way to combine two functions is by forming a composite. • A composite is written (g + f ) or g( f (x)). The two different functions are g and f. • Evaluate the inner function f (x) first. • Use this value, the first output, as the input for the second function, g(x). Problem What is the value of the expression g( f (2)) given the inner function, f (x) 5 3x 2 5 and the outer function, g(x) 5 x2 1 2? 1st input x ⫽ 2 f(x) 1st output, 1, becomes 2nd input 3x ⫺ 5 3(2) ⫺ 5 g(x) x2 2nd output ⫹ 2 3 12 ⫹ 2 6 ⫺ 5 1 ⫹ 2 1 1st output 3 2nd output Exercises Evaluate the expression g( f (5)) using the same functions for g and f as in the Example. Fill in blanks 7–14 on the chart. Use one color highlighter to highlight the first input. Use a second color to highlight the first output and the second input. Use a third color to highlight the second output, which is the answer. 1st input x ⫽ 5 f(x) 1st output, 3x ⫺ 5 10. 5) ⫺ 5 7. 3(— 10 , ——— becomes 2nd input g(x) x2 2nd output ⫹2 14. 102 ——— 2 11. 10 — ⫹2 8. 15 — ⫺ 5 12. 100 —⫹2 10 9. ——— 102 13. ——— Given f (x) 5 x2 1 4x and g(x) 5 2x 1 3, evaluate each expression. 15. f (g(2)) 77 16. g( f (2.5)) 35.5 17. g( f (25)) 13 18. f (g(25)) 21 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 60 Name Class Date Reteaching 6-7 Inverse Relations and Functions • Inverse operations “undo” each other. Addition and subtraction are inverse operations. So are multiplication and division. The inverse of cubing a number is taking its cube root. • If two functions are inverses, they consist of inverse operations performed in the opposite order. Problem What is the inverse of the relation described by f (x) 5 x 1 1? f (x) 5 x 1 1 y5x11 Rewrite the equation using y, if necessary. x5y11 Interchange x and y. x215y Solve for y. y5x21 The resulting function is the inverse of the original function. So, f 21 (x) 5 x 2 1. Exercises Find the inverse of each function. 2. y 5 3x3 1 2 1. y 5 4x 2 5 f 21 5 x 1 5 4 f 21 5 4. y 5 0.5x 1 2 f 21 5 2x 2 4 x 7. f (x) 5 5 f 21(x) 5 5x 3. y 5 (x 1 1)3 3 2 "x 2 3 3 f 21 5 "x 2 1 5. f (x) 5 x 1 3 f 21(x) 5 x 2 3 6. f (x) 5 2(x 2 2) 8. f (x) 5 4x 1 2 9. y 5 x f 21(x) 4 f 21(x) 5 x 1 2 x22 4 5 x21 10. y 5 x 2 3 11. y 5 2 f 21 5 x 1 3 f 21 5 2x 1 1 2 13. f (x) 5 !x 1 2 14. f (x) 5 3 x 2 1 f 21(x) 5 32(x 1 1) f21 (x) 5 x2 2 2 for x L 22 16. f (x) 5 2(x 2 5)2 f 21(x) 556 "x2 17. y 5 !x 1 4 f 21 5 (x 2 4)2 f 21 5 x 12. y 5 x3 2 8 3 f 21 5 "x 1 8 x13 5 f 21(x) 5 5x 2 3 15. f (x) 5 18. y 5 8x 1 1 for x L 0 1 f 21 5 x 2 8 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 69 Name 6-7 Class Date Reteaching (continued) Inverse Relations and Functions Examine the graphs of f (x) 5 !x 2 2 and its inverse, f 21(x) 5 x2 1 2, at the right. 6 5 4 3 2 1 Notice that the range of f and the domain of f 21 are the same: the set of all real numbers x $ 0. Similarly, the domain of f and the range of f 21 are the same: the set of all real numbers x $ 2. y f ⫺1 f x O 1 2 3 4 5 6 This inverse relationship is true for all relations whenever both f and f 21 are defined. Problem What are the domain and range of the inverse of the function f (x) 5 !3 2 x? f is defined for 3 2 x $ 0 or x # 3. Therefore, the domain of f and the range of f 21 is the set of all x # 3. The range of f is the set of all x $ 0. So, the domain of f 21 is the set of all x $ 0. f 6 5 4 3 2 1 23 22 21 O Exercises y x 1 2 3 Name the domain and range of the inverse of the function. 1 19. y 5 2x 2 1 20. y 5 2 2 x The domain and the range is domain: x u 2; the set of all real numbers. range: y u 0 21. y 5 !x 1 5 domain: x L 0; range: y L 25 22. y 5 !2x 1 8 domain: x L 8; range: y K 0 23. y 5 3 !x 1 2 domain: x L 2; range: y L 0 24. y 5 (x 2 6)2 domain: x L 0; range: all real numbers 25. y 5 x2 2 6 1 26. y 5 x 1 4 27. y 5 domain: x L 26; range: all real numbers domain: x u 0; range: y u 24 1 (x 1 4)2 domain: x S 0; range: y u 24 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 70 Name Class 6-8 Date Reteaching Graphing Radical Functions The graph of y 5 a!x 2 h 1 k is a translation h units horizontally and k units vertically of y 5 a!x. The value of a determines a vertical stretch or compression of y 5 !x. Problem 6 What is the graph of y 5 2 !x 2 5 1 3? y 4 y 5 2 !x 2 5 1 3 2 x a52 O h55 k53 2 4 6 Translate the graph of y 5 2 !x right five units and up three units. The graph of y 5 2 !x looks like the graph of y 5 !x with a vertical stretch by a factor of 2. Exercises Graph each function. 1. y 5 !x 2 4 1 1 4 y 2. y 5 !x 2 4 2 O 2 3. y 5 !x 1 1 5. y 5 2 !x 2 1 7. y 5 2!x 1 1 6. y 5 22 !x 1 3 1 4 y 2 2 2 y 8. y 5 !x 1 3 2 4 x y ⫺4 ⫺2 O ⫺2 4 ⫺4 ⫺4 6 y 10. y 5 2!x 2 2 4 2 O x ⫺4 ⫺2 O O ⫺2 9. y 5 3 !x 1 2 6 y 4 x ⫺2 O ⫺2 2 4 x ⫺4 ⫺2 O ⫺2 2 2 y 4. y 5 2!x 1 2 2 3 x ⫺2 O ⫺2 2 ⫺4 y 2 x ⫺2 O ⫺2 x 6 4 y x Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 79 x Name 6-8 Class Date Reteaching (continued) Graphing Radical Functions Graphs can be used to find solutions of equations containing radical expressions. Problem What is the minimum braking distance of a bicycle with a speed of 22 mph? You can find the minimum braking distance d, in feet, of a bicycle travelling s miles per hour using the equation s 5 5.5 !d 1 0.002 . We want to find the value of d when s 5 22. In other words, solve the equation 5.5 !d 1 0.002 5 22. Graph Y1=5.5√(X+0.002) and Y2=22. Try different values until you find an appropriate window. Then use the intersect feature to find the coordinates of the point of intersection. Intersection x ⫽ 15.998 Y ⫽ 22 The minimum braking distance will be about 16 ft. Exercises Solve the equation by graphing. Round the answer to the nearest hundredth, if necessary. If there is no solution, explain why. 11. !3x 1 1 5 5 8 12. !4x 1 1 5 9 20 13. !2 2 5x 5 4 22.8 14. !3x 1 5 5 7 14.67 15. !7x 1 2 5 11 17 16. !2x 2 1 5 !1 2 2x 0.5 17. !x 2 2 5 !2 2 3x no solution; x 5 1 is extraneous 18. 7!x 2 3 5 2!2x 1 1 3.68 19. !2x 2 5 5 !4 2 x 3 20. !2x 1 7 5 3!5x 1 2 20.26 Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 80