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Math141 – Practice Test # 3 Sections 3.6, 3.2, 3.3, 4.1, 4.2 Answer 1. 1/3, - 2 ANSWER KEY Work Use synthetic division to find the other polynomial factor. 2 3 1 12 4 3 6 10 4 5 2 0 Read the bottom row: The last zero is the remainder; the other numbers are the coefficients of the other polynomial factor: 3x2 + 5x – 2. Set this factor = 0 3x2 + 5x – 2 = 0 Factor or use the quadratic formula (3x – 1)(x + 2) = 0 3x – 1 = 0 or x + 2 = 0 3x = 1 x = 1/3 or x = -2 2. -2, 1, 2 Since the degree (highest exponent of x) is 3, there are 3 zeros. There are 2 signs changes in f(x) indicating 2 or 0 positive zeros. f(- x) = -x3 – x2 + 4x + 4 One sign change indicates 1 negative zero. Possible rational zeros (p / q) come from factors of 4 1, 2, 4. Using the Remainder Theorem: f(1) = 0, factors of 1 indicating that 1 is a zero. Using synthetic division gives x2 – 4 = 0 x2 = 4 Use square root property x = 2, x = - 2. [Synthetic division: 1 1 1 4 1 3. 4. 4 4 1 0 0 4 0 A. X=2 X=-2 Y=3 No oblique asymptotes Vertical asymptotes are found by setting the denominator of the reduced fraction = 0. x2 – 4 = 0 Using square root property x2 = 4 x 2 4 x = 2, x = - 2. To find the horizontal or oblique asymptotes: compare the degree of the numerator N with the degree of the denominator D. N = 2, D = 2 N = D Horizontal asymptote at y = a/b or lead coefficient of numerator / lead coefficient of denominator. Y = 3 /1 = 3. There are no oblique asymptotes if there are horizontal asymptotes. B. X=-3 X=3 Y=0 No oblique asymptotes. Vertical: x2 – 9 = 0 (x + 3)(x – 3) = 0 x = -3, x = 3. Horizontal / Oblique asymptotes: N = 1, D = 2, so N < D Horizontal at y = 0 (the x-axis). There are no oblique asymptotes if there are horizontal asymptotes. C. X=3 No horizontal asymptotes Y = 3x + 10 (oblique) Vertical: x – 3 = 0 x = 3. Horizontal / Oblique: N = 2, D = 1 N = D + 1 Oblique. Using long division 3x + 10 2 X – 3 | 3x + x – 2 3x2 – 9x Change signs to subtract 10x – 2 10x – 30 Rational function (fraction) that does not reduce. Y-intercept: Let x = 0 y = (2*0 + 3) / (0 – 1) y = 3 / - 1 = - 3 (0, -3) X-intercept: Let y = 0. Fractions = 0 when the numerator = 0 2x + 3 = 0 2x = - 3 x = -3 / or – 1.5 (- 1.5, 0) Vertical asymptote: denominator = 0 x – 1 = 0 x = 1 (dashed line) Horizontal / Oblique: N = 1, D = 1 N = D Horizontal at y = a / b = 2 / 1 = 2 (No oblique asymptotes when there are horizontal ones.) Rational function: Domain: all real numbers, x - 2, 5. Fraction reduces to R(x) = ( x 2) = 1 . 5. ( x 5)( x 2) x5 Y –intercept: Let x = 0 y = 1 / (0 – 5) = 1 / -5 or (0, -1/5) X-intercept(s): Let y = 0 which means the numerator = 0 No x in numerator no x-intercepts. Vertical asymptote: denominator = 0 x – 5 = 0 x = 5 (line) Horizontal / Oblique: N = 0, D = 1 N < D horizontal at y = 0 (x-axis) GAP at x = - 2 (from the factor that divided out). A. 25 Rational function that does not factor and reduce. y-intercept: Let x = 0 y = (02 + 4*0 + 16) / (0 – 3) 16 / - 3 (0, -5 1/3 ) x-intercept(s): Let y = 0 numerator = 0 x2 + 4x + 16 = 0 doesn’t factor use quadratic formula no x-intercepts Vertical: reduced denominator = 0 x – 3 = 0 x = 3 Horizontal / Oblique: N = 2, D = 1 N = D + 1 oblique use long division x + 7 Y = x + 7 is the oblique asymptote. x – 3 | x2 + 4x + 16 x2 – 3x Change signs to subtract 7x + 16 7x – 21 37 F(g(3)) = f(32) = f(9) = 3(9) – 2 = 27 – 2 = 25 B. 64 G(f(-2)) = g(3*-2 – 2) = g(-6 – 2) = g(-8) = (-8)2 = 64 C. 3x2 – 2 D. E. (6x – 2)2 or 36x2 – 24x + 4 No A. 209 209 function rule of f 1 function rule of f -1 (inverse) 209 B. 39 39 function rule of f -1 12 function rule of f 39 9. Yes Passes horizontal line test. 10. C Only one to pass the horizontal line test. x2 4 5 Switch x and y and solve for y. x 5 y 4 Square both sides of the equation. x2 = 5y + 4 x2 – 4 = 5y Divide by 5 on both sides to get answer. 6. 7. 8. 11. 12. 13. y y3 x9 2 -1, multiplicity of 2, 3, i, and – i F(g(x)) = f(x2) = 3(x2) – 2 = 3x2 -2 G(f(2x)) = g(3*2x – 2) = g(6x – 2) = (6x – 2)2 To be inverses, f(g(x)) = g(f(x)) = x. f(g(x)) = 3(x2) – 2 x Switch x and y and solve for y. x = 2y3 – 9 x + 9 = 2y3 Divide by 2 on both sides, then take the cube root. Change the problem to g(x) = x5 – x4 – 4x3 – 4x2 – 5x – 3. These answers will then be correct. Degree (highest exponent) = 5 5 solutions (roots or zeros). Look at the sign changes in f(x) to find there is 1 positive zero. Look at f(-x) = x5 – x4 + 4x3 - 4x2 + 5x – 3 to find there are 4 or 2 or 0 negative zeros. Possible rational zeros p q factors of cons tan t term 3 1 or 3 factors of leading coefficient 1 Since f(1) = - 16, 1 is not a zero. Since f(3) = 0, 3 is a zero. Since f(-1) = 0, - 1 is a zero. Since f(-3) = -240, -3 is not a zero. So far, we have one positive zero (3) and one negative zero (-1). Use synthetic division for x = 3: 3 1 1 4 4 5 3 1 3 6 6 6 3 2 2 2 1 0 Then use synthetic division for x = - 1: 1 1 2 2 2 1 1 1 1 1 1 1 1 1 0 We now have a polynomial factor of x3 + x2 +x + 1 Since we have one negative zero, we know that we have at least one more. -3 didn’t work, so it must be – 1. Substitute – 1 into the polynomial factor (-1)3 + (-1)2 + (-1) – 1 = 0 to see that this is true. There are now 3 zeros: 3, -1, and – 1. Use synthetic division to find a new polynomial factor: 1 1 1 1 1 1 0 1 1 0 1 0 The new polynomial factor is x2 + 1. Since this is a quadratic, use the square root property to solve: x2 = - 1 x 2 1 x = i. 14. A. T 4800 S k is the constant of proportionality. “Indirectly” translates as T k S . Substitute 80 for T and 60 for S, multiply by 60 on both sides of the =, to find k = 4800. B. 15. 120 min About 14.7 ohms Substitute 40 for S in the formula developed in 15A. kL . Substitute 10 for R, 50 r2 Multiply by 0.000036 0.0036 = 50k Translating “directly” and “inversely” gives R for L and 0.006 for r 10 50k 0.000036 k= 0.0000072 (calculator will say 7.2 E – 6 meaning 7.2 X 10 – 6. This is scientific notation move the decimal point left 6 places because the exponent is – 6.) Substitute the second set of information into the formula R 0.0000072 100 2 (0.007)