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 1 16 2 8 4 4 x y 1 3 Is function y = kx or xy = k? Is function y = kx or xy = k? 2 6 3 9 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
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
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
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
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
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!
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
Divide Rational Expressions Step 1: Step 2: Rewrite division problem to a multiplication problem using reciprocal. Follow multiplication steps. It s that easy!! EXAMPLES: 4 12 15m 3 25m 2x + 6 x 2 (x + 3) 4 15m 3 25m 12 (2x + 6) x 2 1 (x + 3) 4 15m 3 25m 12 1 3m 2 5 3 5 9m 2 2(x + 3) x 2 2 x 2 1 (x + 3)
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 + 2 12 (q 2 + 3q + 2) 12 (q + 2)(q + 1) 12 (q + 2)(q + 1) 12 q + 1 q 2 + 4 (q2 + 4) (q + 1) (q2 + 4) (q + 1) (q2 + 4) (q + 1) q 3 + 2q 2 + 4q + 8 12 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)
Dividing Polynomials Two methods to dividing polynomials: Divide: (2x 2 + 16x) 2x Can be rewritten as: 2x2 +16x 2x Method 1 Method 2 2x 2 + 16x 2x 2x 2 +16x 2x 2x(x+8) 2x 2x 2 2x + 16x 2x = x+8 = x + 8 EXAMPLES: (b 2 + 12b 4) 3b (h 2 + 9h + 18) (h + 6) (b 2 + 12b 4) 3b Cannot be factored (h 2 + 9h + 18) h + 6 = b2 3b + 12b 3b 4 3b = b 3 + 4 4 3b = (h + 6)(h + 3) (h + 6) = h + 3
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 + 12 -(y 2 +3y) y + 12 -(y + 3) 9 remainder Step 1: Divide y 2 by y (y 2 y = y) Step 2: y (y + 3) y + 1 + 9 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!
Adding/Subtracting Rational Expressions When we have common denominators, just add or subtract the numerators, keep the denominator & simplify 5n n + 3 + 15 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 + 4 5 Watch for Inverse denominators 3n n 4 + 6n 4 n 3n n 4 6n n 4 3n n 4
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!
You Try: 1. 5n n+3 + 15 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 + 12 + 8n 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 + 4 4. 5 3r 2 lcm = 4r2 4r r2 ( r r ) ( 5 4r ) 3r 2 r 2 ( 4 4 ) 5r 12r + 8 4r 2 7r + 8 4r 2 5. 3x+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)
Mixed Expressions and Complex Fractions Mixed Expressions An expression that contains the sum of a monomial and a rational expression. For example: 2 + 4 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)
Complex Fractions An expression that has one or more fractions in the numerator and/or denominator. For example: 3 1 2 4 3 4 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 + 3 5 y 2 9 2 y + 3 5 y 2 9 2 y + 3 y2 9 5 2 (y + 3)(y 3) y + 3 5 2y 6 5 or 2(y 3) 5
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 + 4 + x 1 x + 4 x 3 + 4x 2 x + 4 + 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+16 15 j+8 j2 16 15 j 2 +10j+16 j+8 2 a a + 6 1 j 2 16 j 2 + 10j + 16 j + 8 15 2(a + 6) = a 2a + 12 a (j + 4)(j 4)(j + 8) 15(j + 8)(j + 2) (j + 4)(j 4) 15(j + 2)
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 + 4 + 5y 2 = 5y 2 + 5y 4 = y Extraneous Solutions: y = 0 and y = -1 Check for extraneous solutions!
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 = 2 15 + 12 5w LCM = 15w 15w( 2 3w = 2 15 + 12 5w ) 10 = 2w + 36 26 = 2w 13 = w 13 is not an extraneous solution!