Math 143 Final Review Spring 2007

perpendicular to the given line 4x – 5y = 6 goes through (-2, 3) 1. Given line Line in question perpendicular to the given line 4x – 5y = 6 goes through (-2, 3) -5y = -4x + 6 -5 4 m = Equation: 4 5 6 5 y = x – -5 y – 3 x + 2 = 4 4 5 m = 4y – 12 = -5x – 10 4y = -5x + 2 -5 4 1 2 Slope-intercept form y = x + 5x + 4y = 2 Standard form

2. Given f(x). Draw g(x) = -f(-x + 1) + 3 1. Hor shifts 2. Flips 3. Vert shifts 3 -3 3 Shift one unit left Flip over the y-axis Flip over the x-axis Shift three units up -3

x – 2 x + 3 f(x) = x2 – 2x g(x) = 3x + 1 h(x) = 3. (f o g)(x) = (3x + 1)2 – 2(3x + 1) = 9x2 + 6x + 1 – 6x – 2 = 9x2 – 1 4. g 2 2 6 = 3 + 1 = + 1 x – 1 x – 1 x – 1 x – 1 x – 1 6 = + x – 1 x + 5 x – 1 =

x – 2 x + 3 f(x) = x2 – 2x g(x) = 3x + 1 h(x) = 2 2 3x 2 5. = = g(x) – 1 (3x + 1) – 1 f(x + h) – f(x) h [(x + h)2 – 2(x + h)] – [x2 – 2x] 6. = h x2 + 2xh + h2 – 2x – 2h – x2 + 2x = h 2xh + h2 – 2h h(2x + h – 2) = = h h = 2x + h – 2

x – 2 x + 3 f(x) = x2 – 2x g(x) = 3x + 1 h(x) = 7. h-1(x) = ? Inverse Original function y – 2 y + 3 x – 2 x + 3 x = h(x) = xy + 3x = y – 2 x – 2 x + 3 y = xy – y = -3x – 2 y(x – 1) = -3x – 2 -3x - 2 x – 1 y = -3x - 2 x – 1 h-1(x) =

x – 2 x + 3 f(x) = x2 – 2x g(x) = 3x + 1 h(x) = (4) – 2 (4) + 3 8. h(4) = 2 7 =

It passes the vertical line test 9. Is f(x) a function? f(x) 3 Yes Why or why not? -3 3 It passes the vertical line test -3 Does f(x) have an inverse function? No Why or why not? It does not pass the horizontal line test

10. a. What is the domain of f(x) ? f(x) 3 (-4, 5] -3 3 b. What is the range of f(x) ? -3 [0, 3] c. What is f(5) ? f(5) = 3

11. a. For what value(s) of x does f(x) = 2 ? f(x) 3 When x = 3 or x = -3 -3 3 -3 b. Identify any x-intercepts (0, 0) c. Identify any y-intercepts (0, 0)

12. a. For what values of x is the graph of f(x) increasing ? f(x) 3 From x = - 4 to x = -3 -3 3 From x = 0 to x = 5 -3 b. For what values of x is the graph of f(x) decreasing ? From x = -3 to x = 0 c. For what values of x is the graph of f(x) constant ? None

13. Is the degree of f(x) even or odd? How do you know? -3 3 The degree is odd -3 The endpoint behavior is different on the right and left. 14. What can be said about the leading coefficient of f(x)? Why? -9 The leading coefficient is positive. The right side of the graph rises.

15. What is the minimum degree of f(x)? How do you know? 3 15. What is the minimum degree of f(x)? How do you know? -3 3 Minimum degree = 5 -3 The graph makes four turns. 16. Counting multiplicities, what is the minimum number of real zeros of f(x) -9 The minimum number of real zeros is 5. At least 2 zeros at x = -5, at least two zeros at x = -1, and at least 1 zero at x = 3.

-5 even multiplicity -1 even multiplicity 3 odd multiplicity f(x) 3 17. State each zero of f(x) and whether its multiplicity is even or odd. -3 3 -3 Xeros at x = -5 even multiplicity -1 even multiplicity 3 odd multiplicity 18. Write one possible equation for f(x) -9 f(x) = (x + 5)2(x+ 1)2(x – 3)

19. 2x2 + 4 , x > 3 Evaluate each of the following f(x) = Ö7 – x , x £ 3 a. f(4) b. f(3) c. f(- 9) = 2(4)2 + 4 = 36 = = 2 Ö7 – 3 = = 4 Ö7 + 9

20. List all the possible rational zeros of: 20. List all the possible rational zeros of: f(x) = 3x4 – 13x3 + 22x2 – 18x + 4 factors of p: ±4, ± 2, ± 1 factors of q: ±3, ± 1 possible rational zeros: ±4, ± 2, ± 4/3, ±1, ± 2/3, ± 1/3

The other two zeros must be the solutions of 3x2 – 6x + 6 = 0 21. Find all the zeros of: f(x) = 3x4 – 13x3 + 22x2 – 18x + 4 By seeing the graph of the function or by plugging the values from problem 20 into the function you can determine that there are zeros at 2 and at 1/3 The other two zeros must be the solutions of 3x2 – 6x + 6 = 0 1/3 3 -13 22 -18 4 1 -4 6 -4 2 3 -12 18 -12 0 x2 – 2x + 2 = 0 6 -1 12 3 -6 6 0 2 ± Ö4 – 4(1)(2) x = 2 2 ± Ö - 4 = = 1 ± i 2 Zeros: 1/3, 2, 1+ i, 1 - i

22.Graph the system: x2 + y2 £ 9 and y > x + 1 Border: x2 + y2 = 9 Test (0, 0) 0 + 0 £ 9 True Border: y = x + 1 Test (0, 0) 0 > 0 + 1 False

23. Match each equation to its type (x – 3)2 + (y + 1)2 = 16 A. Linear B. Quadratic C. Higher order polynomial D. Rational E. Circle F. Ellipse H. Hyperbola J. Parabola A y = 2x – 3 x2 + 3x + 2 D y = x2 – 9 H (x – 3)2 – 2(y + 1)2 = 9 C y = 4x4 – 3x2 + 5

23. Complete the square and draw the graph 9x2 + 16y2 – 18x + 64y – 71 = 0 9(x2 – 2x + __) + 16(y2 + 4y + __) = 71 + __ + __ 1 4 9 64 9(x – 1)2 + 16(y + 2)2 = 144 (x – 1)2 (y + 2)2 + = 1 16 9

23. Complete the square and draw the graph b. 16x2 –y2 + 64x – 2y + 67 = 0 16(x2 + 4x + __) – 1 (y2 + 2y + __) = -67 + __ + __ 4 1 64 -1 16(x + 2)2 – 1(y + 1)2 = -4 (y + 1)2 (x + 2)2 – = 1 4 9

25. Evaluate each of the following. 0! = 1 1 1! = 5! = 120

nCr 26. Evaluate: 8 3 8! (8)(7)(6) 5! = = 3! 5! 3! 5! = 56 8 3 8! (8)(7)(6) 5! = = 3! 5! 3! 5! = 56 On the calculator: 8 3 8 nCr 3 = 56 =

27. Find the third term of (x – 3)9 9 2 (x)7 (-3)2 3rd term = = 36(x7)(9) = 324x7

Both answers work in the original problem 1.5 Solve Ö2x + 3 – Öx – 2 = 2 2 2 Ö2x + 3 = 2 – Öx – 2 2x + 3 = 4 – 4Öx – 2 + (x – 2) 2 2 x + 1 = -4Öx – 2 x2 + 2x + 1 = 16(x – 2) x2 – 14x + 33 = 0 (x – 3)(x – 11) = 0 Both answers work in the original problem x = 3 or x = 11

1.6 Solve: Öx – 3 Öx – 10 = 0 x1/2 – 3x1/4 – 10 = 0 Let: a = x1/4 The form of this equation looks like a quadratic equation. x1/2 – 3x1/4 – 10 = 0 Let: a = x1/4 a2 – 3a – 10 = 0 (a – 5)(a + 2) = 0 a = 5 or a = -2 so x1/4 = 5 or x1/4 = -2 (x1/4)4 = (5)4 (x1/4)4 = (-2)4 x = 625 x = 16 16 does not work in the original equation, but 625 does work. x = 625

Find the key values of x by solving |5x – 2| = 13 1.8a. Solve: |5x – 2| > 13 Find the key values of x by solving |5x – 2| = 13 5x – 2 = 13 or 5x – 2 = -13 5x = 15 5x = -11 x = 3 x = -11/5 Now test numbers from the intervals created by these key values. Use the original problem to test. T F T 3 -11/5 Solution: x < -11/5 or x > 3

Find the key values of x by solving 4x2 + 7x = -3 1.8b. Solve: 4x2 + 7x < -3 Find the key values of x by solving 4x2 + 7x = -3 4x2 + 7x + 3 = 0 (4x + 3)(x + 1) = 0 x = -3/4 or x = -1 Now test numbers from the intervals created by these key values. Use the original problem to test. F T F -3/4 -1 Solution: -1 < x < -3/4

2.7 a. f(x) = (1/2)x3 – 4 Inverse Original function x = (1/2)y3 – 4 y = (1/2)x3 - 4 y3 = 2x + 8 3 y = Ö2x + 8 3 f-1(x) = Ö2x + 8

3.4. f(x) = 6x3 + 25x2 – 24x + 5 one of the zeros is -5 - 5 6 25 -24 5 factors of p: ±5, ± 1 factors of q: ±6, ± 3, ±2, ± 1 possible rational zeros: ±5, ± 5/2, ± 5/3, ±1, ± 5/6, ± 1/2, ± 1/3, ± 1/6 one of the zeros is -5 The other two zeros are the solutions to - 5 6 25 -24 5 -30 25 - 5 6x2 – 5x + 1 = 0 6 -5 1 0 (3x – 1)(2x – 1) = 0 x = 1/3 x = 1/2 zeros: - 5, 1/3, 1/2

The solutions for x2 – 2x + 2 = 0 are 3.5 Find all the roots of x4 – 4x3 + 16x2 – 24x + 20 = 0 given that 1 – 3i is a root. Since 1 – 3i is a root, 1 + 3i is also a root so (x – 1 + 3i) and (x – 1 – 3i) are factors of the equation (x – 1 – 3i)(x – 1 + 3i) = x2 – 2x + 10 See problem 21 x2 – 2x – 2 The solutions for x2 – 2x + 2 = 0 are x2 – 2x + 10 x4 – 4x3 + 16x2 – 24x + 20 - + - x4 – 2x3 + 10x2 – 2x3 + 6x2 – 24x + 20 x = 1 ± i + – 2x3 + 4x2 – 20x - + 2x2 – 4x + 20 2x2 – 4x + 20 Roots: 1 – 3i, 1 + 3i, 1 – i, 1 + i

3.6 Graph the function: f(x) = x + 2 Vertical asymptote: x = -2 Horizontal asymptote: y = -3 x-intercept: (0,0) y-intercept: (0,0) x y -4 -3 -1 1 -6 -9 3 -1

4.2 f(x) = logb x is shown on the graph a. When x = 5, logb 5 = 1 f-1(x) so b = 5 5 -5 5 c. For f-1(x) Domain: all real numbers -5 Range: y > 0

log 140.3 4.3 log3 140.3 = » 4.5 log 3 4.4 a. Solve: 22x – 1 + 3 = 35 22x – 1 = 32 ln 22x – 1 = ln 32 (2x – 1) ln 2 = ln 32 2x – 1 = 5 2x = 6 x = 3

x = -4 does not work in the original equation. 4.4 b. Solve: log3 (x – 5) + log3 (x + 3) = 2 log3 (x2 – 2x – 15) = 2 x2 – 2x – 15 = 32 x2 – 2x – 24 = 0 (x – 6)(x + 4) = 0 x = 6 or x = -4 x = -4 does not work in the original equation. Solution: x = 6

A blood alcohol level of 0.112 corresponds to an accident risk of 25% 4.5 R = 6e12.77x If R = 25% accident risk 25 = 6e12.77x 25/6 = e12.77x ln (25/6) = ln e12.77x ln (25/6) = 12.77x x = 0.112 A blood alcohol level of 0.112 corresponds to an accident risk of 25%

7.1 Write the standard form of the equation of an ellipse given foci: (0,-3) and (0,3) vertices: (0,-4) and (0,4) Center: (0, 0) V The equation must be in the form: F x2 y2 a2 b2 + = 1 b = distance from the center to a vertex = 4 F c = distance from the center to a focus point c = 3 V c2 = b2 – a2 9 = 16 – a2 x2 y2 7 16 Equation: + = 1 a2 = 7

The parabola is vertical with an equation in the form: 7.3 Find the equation of a parabola with vertex at (2, -3) and focus at (2, -5) The parabola is vertical with an equation in the form: (x – h)2 = 4p(y – k) (h, k ) = (2, -3) V p = -2 F Equation: (x – 2)2 = 8(y + 3)