Inverse Variation. y varies inversely as x. REMEMBER: Direct variation y = kx where k is not equal to 0.

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1 Inverse Variation y varies inversely as x. REMEMBER: Direct variation y = kx where k is not equal to 0. Inverse variation xy = k or y = k where k is not equal to 0. x Identify whether the following functions are direct or inverse variation: x y x y 1 3 Is function y = kx or xy = k? Is function y = kx or xy = k? Function is Inverse Variation xy = 16 x = 2y Function is direct variation. y = 1 2 x Function is direct variation. y = 3x 2xy = 10 Function is indirect variation xy = 5 You try: direct or inverse variation: 1. y = -3x Direct Variation 2. xy = -40 Inverse Variation 3. xy = 1 4 Inverse Variation

2 Write an inverse variation: Assume that y varies inversely as x. If y = 18 when x = 2, write an inverse variation equation that relates x to y. xy = k (2)(18) = k 36 = k Inverse Variation Equation xy = 36 Assume that y varies inversely as x. If y = 3 when x = 12, find x when y = 4. Method 1 xy = k Step 1: Find k: Method 2 3(12) = 4(x) 36 = 4x x = 9 (12)(3) = k 36 = k Step 2: Use k to find x: 4x = 36 x = 9 You Try: 1. Assume y varies inversely as x. If y = 5 when x = 6, find x when y = 2. (5)(6) = 2x 15 = x 2. Assume y varies inversely as x. If y = 9 when x = 9, find y when x = -27. (9)(9) = -27y -3 = y

3 Rational Functions Key Terms Rational Function Excluded Values Asymptote Definitions A nonlinear function with a variable in the denominator The value of a variable that results in a denominator of zero (cannot divide by zero). A line that the graph of a function approaches but never touches. Find Excluded Values Step 1: Step 2: Step 3: Set denominator equal to zero. Solve for variable. The result of Step 2 is an excluded value. EXAMPLES: Find the excluded values for the following equations: y = 5 4x 8 y = 2 2x 4x 8 = 0 4x = 8 x = 2 x = 2 is an excluded value 2x = 0 x = 0 x = 0 is an excluded value

4 y = 3 x 2 9 x 2 9 = 0 x 2 = 9 x = ±3 excluded value: x = ±3 y = x x 2 + 5x + 6 x 2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = {-3, -2} excluded value: x = {-3, -2} Vertical Asymptote: Horizontal Asymptote: y = k. Domain Find Asymptotes y = a x h + k x = The excluded value (or the x value that makes the denominator equal to 0). All real numbers except the excluded value. Range All real numbers except y = k. EXAMPLES: y = 2 x 4 Vertical asymptote x = 0 Horizontal asymptote y = -4 y = 1 x + 1 Vertical asymptote x + 1 = 0 x = -1 Horizontal asymptote y = 0

5 Simplifying Rational Expressions Rational Expression An algebraic function whose numerator and denominator are polynomials. Simplifying Expressions Make certain numerator and denominator have a GCF of 1. Step 1: Step 2: Step 3: If possible, factor the numerator then factor the denominator. Simplify (Cross out common factors either top to bottom or diagonally NOT side to side! State excluded values. Examples: Simplify and state excluded values: Step 1: 2(r+9) (r+9)(r 1) 2r + 18 r 2 + 8r 9 Step 1: (y+10)(y 1) 2(y+10) y 2 + 9y 10 2y + 20 Step 2: 2(r+9) (r+9)(r 1) Step 2: (y+10)(y 1) 2(y+10) Step 3: 2 r 1 excluded value: r = 1 Step 3: y 1 2

6 Recognize Opposites!!! 36 x 2 5x 30 = (6 + x)(6 x) 5(x 6) = (6 + x)( 1)( 6 + x) 5(x 6) = (6 + x)(x 6) 5(x 6) = (6 + x) 5 Finding the zeros: Once simplified, set the expression equal to zero and solve. f(x) = x3 4x 2 12x x + 2 x(x 2 4x 12) (x + 2) x(x 6)(x + 2) (x + 2) x(x 6) = 0 x = 0 x 6 = 0 x = 6 x = (0, 6) Check for excluded values!

7 Multiplying / Dividing Rational Expressions Multiply Rational Expressions Step 1: Step 2: Step 3: Step 4: Simplify expressions by canceling common factors. Multiply numerators Multiply denominators Simplify if needed NOTE: Bundle an expression with a +/- in them. This is a factor, put parenthesis around the bundle to reflect as one factor. EXAMPLES: r 2 x 9t 3 3t4 r r 2 x 9t 3 3t4 r rx 3 t 1 rxt 3 y 2 3y 4 y + 5 (y 4)(y + 1) (y + 5) (y 4)(y + 1) (y + 5) (y + 1) 1 y + 5 y 2 4y (y + 5) y(y 4) (y + 5) y(y 4) 1 y (y + 1) y

8 Divide Rational Expressions Step 1: Step 2: Rewrite division problem to a multiplication problem using reciprocal. Follow multiplication steps. It s that easy!! EXAMPLES: m 3 25m 2x + 6 x 2 (x + 3) 4 15m 3 25m 12 (2x + 6) x 2 1 (x + 3) 4 15m 3 25m m m 2 2(x + 3) x 2 2 x 2 1 (x + 3)

9 y 3 y 2 10y + 16 y2 9 y 8 (y 3) (y 8) (y 2 10y + 16) (y 2 9) (y 3) (y 8)(y 2) (y 8) (y 3)(y + 3) (y 3) (y 8)(y 2) (y 8) (y 3)(y + 3) 1 (y 2) 1 (y + 3) 1 (y 2)(y + 3) 1 y 2 + y 6 q 2 + 3q (q 2 + 3q + 2) 12 (q + 2)(q + 1) 12 (q + 2)(q + 1) 12 q + 1 q (q2 + 4) (q + 1) (q2 + 4) (q + 1) (q2 + 4) (q + 1) q 3 + 2q 2 + 4q You Try! 1. k2 +5k 6 k+2 (k + 6)(k 1) (k + 2) (k + 6)(k 1) (k + 2) k2 +3k+2 (k 1) 2 (k + 1)(k + 2) (k + 1)(k 1) (k + 1)(k + 2) (k + 1)(k 1) 2. 5x 2 x 2 5x+4 10x x 1 5x 2 (x 1) (x 4)(x 1) 10x 5x 2 (x 1) (x 4)(x 1) 10x k + 6 x 2(x 4)

10 Dividing Polynomials Two methods to dividing polynomials: Divide: (2x x) 2x Can be rewritten as: 2x2 +16x 2x Method 1 Method 2 2x x 2x 2x 2 +16x 2x 2x(x+8) 2x 2x 2 2x + 16x 2x = x+8 = x + 8 EXAMPLES: (b b 4) 3b (h 2 + 9h + 18) (h + 6) (b b 4) 3b Cannot be factored (h 2 + 9h + 18) h + 6 = b2 3b + 12b 3b 4 3b = b b = (h + 6)(h + 3) (h + 6) = h + 3

11 Long Division: NOTE: When using long division, you must consider missing terms!! (y 2 + 4y + 12) (y + 3) y + 1 y + 3 y 2 + 4y (y 2 +3y) y (y + 3) 9 remainder Step 1: Divide y 2 by y (y 2 y = y) Step 2: y (y + 3) y y+3 Step 3: Subtract Step 4: Bring down next term (12) Step 5: Repeat (c 3 + 5c 6) (c 1) c 2 + c + 6 c - 1 c 3 + 0c 2 + 5c - 6 -(c 3 - c 2 ) -(c 2 + 5c) -(c 2 - c) 6c 6 -(6c 6) 0 No remainder c 2 + c + 6 C 2 term is missing must fill it!

12 Adding/Subtracting Rational Expressions When we have common denominators, just add or subtract the numerators, keep the denominator & simplify 5n n n + 3 5n + 15 n + 3 5(n + 3) n + 3 5(n + 3) n + 3 3m 5 m + 4 4m + 2 m + 4 (3m 5) (4m + 2) m + 4 m 7 m Watch for Inverse denominators 3n n 4 + 6n 4 n 3n n 4 6n n 4 3n n 4

13 Unlike Denominators: Need to find the Least Common Multiple (LCM) for the denominators. Before dealing with fractions, let s practice finding the LCM for two expressions: a 2 b 3 y and aby 2 Factors of a 2 b 3 y a a b b b y Factors of aby 2 a b y y a a b b b y y LCM = a 2 b 3 y 2 n 2 + 5n + 4 and (n + 1) 2 Factors of n 2 + 5n + 4 (n + 4)(n + 1) Factors of (n + 1) 2 (n + 1)(n + 1) (n + 4)(n + 1)(n + 1) LCM = (n + 4)(n + 1) 2 Put it together with fractions: 3t + 2 t 2 2t 3 + t + 1 t 3 3t+2 + t+1 t 2 2t 3 t 3 LCM of denominator = (t 3)(t + 1) 3t+2 + (t+1)(t+1) (t 3)(t+1) (t 3)(t+1) multiply second fraction by (t+1) because that (t+1) was the missing factor in the denominator. 3t+2 + t2 +2t+1 (t 3)(t+1) (t 3)(t+1) add the numerators, keep the denominator. t 2 +5t+3 (t 3)(t+1) final answer!

14 You Try: 1. 5n n n+3 3. n+3 n + 8n 4 4n 5n + 15 n + 3 5(n + 3) n + 3 ( 4 4 ) (n + 3 n ) + 8n 4 4n 4n n 4 4n 12n + 8 4n 4(3n + 2) 4n = 3n + 2 n lcm = 4n 2. 3m 5 4m+2 m+4 m+4 3m 5 4m 2 m + 4 m 7 m r 2 lcm = 4r2 4r r2 ( r r ) ( 5 4r ) 3r 2 r 2 ( 4 4 ) 5r 12r + 8 4r 2 7r + 8 4r x+2 + x+1 x 2 2x 3 x 3 3x + 2 (x 3)(x + 1) + x + 1 x 3 lcm = (x-3)(x+1) 3x + 2 (x + 1)(x + 1) + (x 3)(x + 1) (x 3)(x + 1) 3x + 2 (x 3)(x + 1) + x2 + 2x + 1 (x 3)(x + 1) x 2 + 5x + 3 (x 3)(x + 1) 6. d 1 d 2 3 d+5 lcm=(d-2)(d+5) (d + 5)(d 1) (d + 5)(d 2) 3(d 2) (d + 5)(d 2) d 2 + 4d 5 (d + 5)(d 2) 3d 6 (d + 5)(d 2) d 2 + d + 1 (d + 5)(d 2)

15 Mixed Expressions and Complex Fractions Mixed Expressions An expression that contains the sum of a monomial and a rational expression. For example: x+1 Mixed Expressions 3 + x + 2 x 3 Change the 3 into 3. 1 Find a common denominator (lcm). Fix by multiplying so that both fractions have the common denominator. Add or subtract. 3(x 3) 1(x 3) + x + 2 x 3 3x 9 (x 3) + x + 2 x 3 4x 7 (x 3)

16 Complex Fractions An expression that has one or more fractions in the numerator and/or denominator. For example: Simplify Complex Fractions Step 1: Change mixed numbers into improper fractions Step 2: Change mixed expressions to rational expressions Step 3: Divide by multiplying by reciprocal Step 4: Factor and simplify if possible 8x 2 v 8x 2 v 8x 2 v 8x 2 v 4x v 3 4x v 3 v3 4x v3 4x 2xv 2 2 y y y y y + 3 y (y + 3)(y 3) y y 6 5 or 2(y 3) 5

17 n 2 + 7n 18 n 2 2n + 1 n 2 81 n 1 n 2 + 7n 18 n 2 2n + 1 n 1 n 2 81 (n + 9)(n 2) (n 1)(n 1) n 1 (n + 9)(n 9) (n + 9)(n 2) (n 1)(n 1) n 1 (n + 9)(n 9) x 2 + x 1 x + 4 x 2 (x + 4) x x 1 x + 4 x 3 + 4x 2 x x 1 x + 4 x 3 + 4x 2 + x 1 x + 4 (n 2) (n 1)(n 9) (n 2) n 2 10n + 9 You Try: 1. 2 a 1 a+6 2 a 1 a+6 2. j 2 16 j 2 +10j j+8 j j 2 +10j+16 j+8 2 a a j 2 16 j j + 16 j (a + 6) = a 2a + 12 a (j + 4)(j 4)(j + 8) 15(j + 8)(j + 2) (j + 4)(j 4) 15(j + 2)

18 Rational Equations Extraneous Solution Any value that makes a denominator result in zero. Two methods to solving Rational Equations: Method 1 Method 2 Using Cross Products: 3 x + 5 = 2 x 2(x + 5) = 3x 2x + 10 = 3x 10 = x Extraneous Solution: Use Cross Products when possible Use LCM when there is addition or subtraction in the equation. Multiply ALL terms on both sides of the = by the LCM to get rid of the fractions. Use LCM: 4 y + 5y y + 1 = 5 Multiply every term by LCM y(y + 1) 4 (y)(y + 1) + 5y (y)(y + 1) = 5(y)(y + 1) y y + 1 x = 0 and x + 5 = 0 x = -5 4y y 2 = 5y 2 + 5y 4 = y Extraneous Solutions: y = 0 and y = -1 Check for extraneous solutions!

19 Be careful! Rational equations may have more than one solution. One or both of the solutions can be extraneous. YOU MUST CHECK YOUR WORK! 3x 6x 9 + x 1 x 1 = 6 LCM: x 1 Multiply all terms by this LCM 3x x 1 6x 9 (x 1) + (x 1) = 6(x 1) x 1 3x + 6x 9 = 6x 6 9x 9 = 6x 6 3x = 3 x = 1 HOWEVER 1 is an extraneous solution therefore there is NO SOLUTION! YOU TRY! Solve 2 3w = w LCM = 15w 15w( 2 3w = w ) 10 = 2w = 2w 13 = w 13 is not an extraneous solution!

Unit 9 Study Sheet Rational Expressions and Types of Equations

Algebraic Fractions: Unit 9 Study Sheet Rational Expressions and Types of Equations Simplifying Algebraic Fractions: To simplify an algebraic fraction means to reduce it to lowest terms. This is done by