1. Math 140 Quiz 3 - Summer 2004 Solution Review (Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

2. Problem 1 (38) State whether the function is a polynomial function or not. If it is, give its degree. f(x) = (5 - x 2 )/3. Note that f(x) may be rearranged in order of decreasing powers of x as: y = (-1/3) x 2 + 5/3. Which is in standard quadratic polynomial form with a 2 = -1/3, a 1 = 0, & a 0 = 5/3. So the answer is yes the function is a polynomial of degree 2.

3. Problem 2 (76) For the polynomial, list each real zero and its multiplicity. Determine whether the graph crosses or touches the x-axis at each x-intercept. f(x) = 2(x + 4)(x - 4) 3 From factor (x (x + 4) root is -4 of multiplicity (factor power) 1 and since this is an odd number the graph crosses x-axis there. From factor (x (x - 4) root is 4 of multiplicity (factor power) 3 and since this is an odd number the graph crosses x-axis there. The answer is E) None of these.

4. Problem 3 (29) Find all the intercepts for the function: f(x) = 2x 2 – 4x – 6. Note the factoring: f(x) = 2x 2 – 4x – 6 = 2(x –3)(x +1). The y-axis intercept has y-coordinate that is value of the function when x = 0. This is just f(0) = – 6. So intercept point is at (0, – 6). Any x-axis intercept has x-coordinate such that f(x) = 2(x – 3)(x + 1) = 0 => x = 3 or – 1. Thus, the x-axis intercepts are at (3, 0) and (– 1, 0).

5. Problem 4 (57) List all real values of x that must be excluded from the domain of f(x) as holes. f(x) = (9x + 1) - (20x + 7) (x + 20) (3x - 19) We could put over a common denominator and simplify to get f(x) = (7x 2 - 561x -159). _____________ (x (x + 20)(3x - 19) But the holes clearly come from x + 20 = 0 and 3x 3x - 19 = 0. That is, x-values are {-20, 19/3} or D).

6. Problem 5 (48) Find for f(x) all x-intercepts and y-intercepts (if they exist): f(x) = (x - 7)(2x + 9) (x 2 + 3x - 3) The y-axis intercept has y-coordinate that is value of f(x) when x = 0. This is f(0) = -63/(-3) = 21. So intercept point is at (0, 21). Any x-axis intercept has x-coordinate such that f(x) = 0, i.e., (x – 7)(2x + 9) = 0 => x = 7 or – 9/2. Thus, the x-axis intercepts are: (7, 0) and (– 9/2, 0).

7. Problem 6 (24) Find the coordinates of the vertex of the parabola: f(x) = x 2 + 9. The parabola shape arises from the graph of y = f(x) or y = x 2 + 9. Which in standard form, 4a(y – k) = (x – h) 2, is y - 9 = x 2. Recognize this as the translation of the simple y = x 2 parabola with vertex at (0,0) to one with vertex: ( h, k ) = (0, 9).

8. Problem 7 (52) Give the equation of the horizontal asymptote(s): f(x) = (3x 2 + 2) (3x 2 - 2) The horizontal asymptote comes from |x|  . It arises from 3x2 3x2  2  3x2 3x2. Thus, the y-asymptote is given by y = 3x 2 /(3x 2 ) = 1.

9. Problem 8 (38) Give the equation of the vertical asymptote(s): h(x) = (x - 6)(x + 2) (x 2 - 1) The vertical asymptotes come from the holes. They are arise from x2 x2 - 1 = (x (x - 1)(x + 1) = 0. Thus, the x-values and the asymptotes are given by x = 1, x =

10. Recognize this as the translation of the simple cubic y = x 3 of intercept (0,0) to one with intercept: ( h, k ) = (1, 0) and with vertical contraction (parallel to y-axis) because of factor ½, i.e., (3,8) => (3,4). This is graph A). Problem 9 (81) Graph the function and select the matching graph: f(x) = (x - 1) 3 /2.

11. Problem 10 (67) Solve the inequality, then graph its solution. Use interval notation. f(s) = s 2 - 5s - 6 < 0 Since the polynomial is in standard form proceed to Step 2 and find intervals from roots of f(s) = s2 s2 - 5s 5s - 6 = (s (s + 1)(s - 6) = 0 => s = or 6. Thus, the x-axis intercepts are: (-1, 0) and (6, 0). Step 3: The intervals: (- , -1); (-1, 6); (6,  ). Testing f(s) values: f(-2)= 8, f(0)= -6, f(7)= 8

12. Problem 10 cont’d (67) Solve the inequality, then graph its solution. Use interval notation. f(s) = s 2 - 5s - 6 < 0 Test evaluations of f(s) to get sign in intervals -2 f(-2) = 8 Above x-axis (-2, 8) 7 f(7 ) = 8 Above x-axis (7, 8) 0 f(0 ) = -6 Below x-axis (0, -6) Thus, (-1, 6) is interval when f(s) < 0. -1 0 1 2 3 4 5 6 ( )

13. Problem 11 (24) Use synthetic division. (5x 3 - 15x 2 - 26x + 24) (x - 4) Result is: 5x2 5x2 + 5x 5x - 6

14. Note f(x) = (5/3)/(x – 2/3). This is a translation of the simple inverse function, 1/x, of vertical asymptote x = 0 to a vertical asymptote: x = 2/3. and with a vertical expansion (parallel to y-axis) because of factor 5/3, i.e., (1,3) => (1,5). Problem 12 (71) Graph the function and select the matching graph: f(x) = - 5/(2 - 3x). This is graph C).

15. Problem 13 (29) Perform the indicated operation. Write result in standard form.

16. Problem 14 (33) Perform the indicated operation. Write result in standard form. (9 + 3i) - (-7 + i) (9 + 3i) - (-7 + i) = 9 - (-7) + (3 – 1) i = 16 + 2i

17. Problem 15 (33) Perform the indicated operation. Write result in standard form. i 18 i 18 = (i 2 ) 9 = (-1) 9 = (-1)(-1) 2 (-1) 2 (-1) 2 (-1) 2 =

18. Problem 16 (33) Perform the indicated operation. Write result in standard form. (4 - 7i)(5 + 2i) (4 - 7i)(5 + 2i) = 4(5) - 7(2)i 2 + [4(2) – 7(5)] i = 20 + 14 + (8 -35) i = 34 - 27i

19. Problem 17 (62) Perform the indicated operation. Write result in standard form. (8 + 2i)/(6 - 3i) (8 + 2i)(6 + 3i) = _____ (6 - 3i)(6 + 3i) 8(6) + 2(3)i 2 +[8(3)+2(6)]i = _____ 36 - 9 i 2 42 +36i = 14 +12i _ __ 45 15

20. Problem 18 (57) Use the discriminant to determine the type of solutions in the equation: 4 + 2z 2 = -3z. 2z 2 + 3z + 4 = 0 ax 2 + bx + c = 0 => a = 2, b = 3, c = 4 Since the discriminant is negative, there are two different (conjugate) imaginary solutions.

21. Problem 19 (38) Use the factor theorem to decide whether, or not the second polynomial is a factor of the first. P(x) = 6x 4 + 13x 3 - 2x 2 + x + 4, Q(x) = x + 2. Result is: P(-2) = -14 not 0. Therefore, Q(x) is not a factor of P(x). Use synthetic division to determine P(-2).

22. Problem 20 (43) Solve the equation. -8x 2 = 2x + 7 8x 2 + 2x + 7 = 0 ax 2 + bx + c = 0 => a = 8, b = 2, c = 7 This matches C) after it is simplified. Note order of signs in  does not matter here.