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

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.

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.

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).

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 x -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).

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).

Problem 6 (24) Find the coordinates of the vertex of the parabola: f(x) = x The parabola shape arises from the graph of y = f(x) or y = x 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).

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.

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 =

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.

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

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) < ( )

Problem 11 (24) Use synthetic division. (5x x x + 24) (x - 4) Result is: 5x2 5x2 + 5x 5x - 6

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).

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

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 = i

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 =

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 = (8 -35) i = i

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 = _____ i i = i _ __ 45 15

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.

Problem 19 (38) Use the factor theorem to decide whether, or not the second polynomial is a factor of the first. P(x) = 6x x 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).

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.