Poly_tut_sol1 - Polynomials Polynomials tutorial solutions...

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Polynomials Polynomials tutorial solutions 1 a) Fill out the spreadsheet cells as follows A B C D E F 1 2 Start Value -5 3 Increment 0.1 1 4 5 X Y 6 =C2 =2*A6*A6*A6+3*A6*A6- 12*A6+32 7 =A6+C$3 Copy Down 8 Copy Down 9 To construct the graph, follow the instructions below. i. Highlight x and y values ii. Select chart icon iii. Select XY (scatter plot) iv. Select chart sub-type v. Select finish vi. Experiment with modifying the appearance of the chart 1 y = 2x^3+3x^2-12x+32 -100 -50 0 50 100 150 -6 -4 -2 0 2 4 x y
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Polynomials b) Proceed as part a) Note that the curve is not defined for x < 1 2. a) Let y 2x 3 2x 3 y 3 y x 2 = - = + + = b) ( 29 2x 3 Let y x 4 x 4 y 2x 3 xy 2x 4y 3 x(y 2) (4y 3) 4y 3 x y 2 - = + + = - - =- - - =- + - =- - x = – 2 y 3 y 4 - + c) 2 2 Let y x 1 x y 1 x y 1 = + = - = ± - But a inverse function should not be multi-valued. Therefore define our inverse function as being x y 1 = + - Also the value is not defined for y <1 (values of complex) x y 1 for y 1 = - 2 y = (x-1)^(2/3)-1 -1.5 -1 -0.5 0 0.5 1 1.5 2 0 1 2 3 4 5 6 x y
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Polynomials 3. a) The function y = ax + b goes through the points x = 0, y = 3 and x = 2, y = -1 3 b and 1 2a b ∴ = - = + Substitute for b = 3 gives – 1 = 2a + 3 2a 3 2a 4 a 2 -= + =- =- Therefore linear function given as y = 3 - 2x b) The function y = ax + b goes through the points x = -1, y = 2 and x = 3, y = 4 2 a b and 4 3a b _____________ sub 2 4a 1 a 2 ∴ =-+ = + - =- ∴ = subtraction – 2 = –4a a = 2 1 Substitute for a = 1/2 gives 2 = – 2 1 + b 1 2 b 2 5 b 2 = + = Therefore linear function given as 1 5 y x 2 2 = + or 2y x 5 = + 3
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Polynomials 4. a) The function 2 y ax bx c = + + goes through the points x = 1 y = 2, x = 2 y = 3 and x = 4, y = 5 Therefore 2 a b c (i) 3 4a b c (ii) 5 16a 4b c (iii) = + + =+ + = ++ 3 = 4a + 2b + c (ii) 2 a b c (i) 3 4a b c (ii) 5 16a 4b c (iii) = + + =+ + = ++ subtract (i) from (ii) 1 = 3a + b (iv) subtract (ii) from (iii) 2 = 12a + 2b (v) (iv) 2 2 6a 2b (v) 2 12a 2b × = + = + subtract (iv) × 2 from (v) 0 = 6a a = 0 substitute for a = 0 in (iv) gives 1 = b substitute for a = 0 and b = 1 in (i) gives 2 = 0 + 1 + c c = 1 Therefore y = x + 1 and the points are colinear.
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Poly_tut_sol1 - Polynomials Polynomials tutorial solutions...

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