Math 141 Practice Test # 2: Chapter 2.3 – 2.6, 2.8, 3.1, 3.5, 3.6

Answer / Work / thinking
1. / (- 4, 0) and (3, 0) / Let y = 0 ® 0 = x2 + x – 12 ® Trinomial with a = 1, so use (x )(x ) method ® Since c is neg, difference = b = 1 and signs will be different ® Factor pair of 12 w/ diff = 1 are 3 and 4 ® (x 3)(x 4) ® Sign of b is sign of larger ® (x – 3)(x + 4) = 0 ® Set each factor = 0 ® x – 3 = 0 or x + 4 = 0 ® x = 3 or x = - 4
2. / (6, 0) and (-6, 0) / Let y = 0 ® x2 – 36 = 0 ® Solve for x2 ® x2 = 36 ® Take the square root of both sides. (Don’t forget the .) ® x = 6
3. / (10, 0) and (- 2, 0) / Let y = 0 ® x2 – 8x – 20 = 0 ® Move the constant term to the other side of the = ® x2 – 8x = 20 ® x2 – 8x + ? = 20 + ? ® Think (b / 2)2 ® (-8/2)2 ® (-4)2 = 16, so add 16 to both sides ® x2 – 8x + 16 = 20 + 16 ® Factoring the left side and simplifying the right side ® (x – 4)2 = 36 ® Take the square root of both sides ® ® x – 4 = 6 ® x – 4 = 6 or x – 4 = - 6 ® x = 10 or x = - 2
4. / (2.5, 0) and (- 1, 0) / Let y = 0 ® 2x2 – 3x – 5 = 0 ® ®®®®
5. / A. / No zeros / Discriminant b2 – 4ac < 0 ® b2 – 4ac = 02 – 4(1)(9) = 0 – 36 = - 36
B. / One zero / Discriminant = 0 ® b2 – 4ac = (-2)2 – 4(1)(1) = 4 – 4 = 0
C. / Two zeros / Discriminant > 0 ® b2 – 4ac = (-3)2 – 4(2)(1) = 9 – 8 = 1
6. / (2.5, 0) and (3, 0) / Let A = (x – 2) and substitute ® 2A2 – 3A + 1 = 0 ® Use factoring or quadratic formula ® (2A – 1)(A – 1) = 0 ® 2A – 1 = 0 or A – 1 = 0 ® A = ½ or A = 1 ® Substitute (x – 2) in for A ® x – 2 = ½ or x – 2 = 1 ® x = 2.5 or x = 3
7. / Vertex (- 11, - 121)
Minimum / Minimum because a > 0 (a = coefficient of x2). Vertex (h, k) ® h = Substitute to find k: (-11)2 + 22(-11) = 121 – 242 = - 121
8. / Vertex (- 2, - 5)
Maximum / Maximum because a < 0 (a = coefficient of x2). Vertex (h, k) ® h = . k = -5(-2)2 – 20(-2) -25 = -20 + 40 – 25 = - 5
9. / Vertex (5, 6)
Maximum / Equation is in the form y = a(x – h)2 + k where (h, k) is the vertex and a is less than zero ® opens down making vertex a maximum.
10.
10. / A. / Up / a (the coefficient of x2) is positive
B. / (5, - 25) / Vertex (h, k) ® h = . Substitute to find k:
k = (5)2 – 10(5) = 25 – 50 = - 25
C. / X = 5 / Axis of symmetry is a vertical line in the form x = h = 5.
D. / (0, 0) / Let x = 0 ® y = 02 – 10(0) = 0
E. / (0, 0) and (10, 0) / Let y = 0 ® x2 – 10x = 0 ® Factor using greatest common factor:
x(x – 10) = 0 ® Set each factor = 0 ® x = 0 or x - 10 = 0 ® x = 0 or 10
F. / (- , ) / See graph below (next page) LEFT.
G. / [-25, ) / See graph below (next page) LEFT.
H. / (5, ) / See graph below (next page) LEFT.
I. / (-, 5) / See graph below (next page) LEFT.
J. / / 11. J. /
Answer / Work / thinking
11. / A. / Up / a = 1 which is positive
B. / (3, - 16) / h = ; k = (3)2 – 6(3) – 7 = 9 – 18 – 7 = - 16
C. / x = 3 / Axis of symmetry is a vertical line in the form x = h.
D. / (0, - 7) / Let x = 0 ® y = (0)2 – 6(0) – 7 = - 7
E. / (- 1, 0) and (7, 0) / Let y = 0 ® x2 – 6x – 7 = 0 ® (x - 7)(x + 1) = 0 ® x - 7 = 0 or x + 1 = 0 ® x = 7 or x = - 1
F. / (-, ) / See graph above RIGHT.
G. / [- 16, ) / See graph above RIGHT.
H. / (3, ) / See graph above RIGHT.
I. / ( -, 3) / See graph above RIGHT.
12. / Minimum
Value of – 19 at – 2 / Minimum because a > 0 (positive) ® parabola opens up. Min / max ® Find vertex ® h = - 12 / 2 (3) = - 2. This is the where (at). Find k to find the value ® k = 3(-2)2 + 12(-2) – 7 ® - 19
13. / ( -, - 5) (2, ) / Find the zeros ® x2 + 3x – 10 = 0 ® (x + 5)(x – 2) = 0 ® x + 5 = 0 or x – 2 = 0 ® x = - 5 or x = 2. Graph the zeros. The parabola will open up because a = 1 which is positive. (Below LEFT)
13. / / 14. /
14. / / Find the zeros ® 6x2 - 5x – 6 = 0 ® (2x - 3)(3x + 2) = 0 ® 2x - 3 = 0 or 3x + 2 = 0 ® x = 3/2 or x = -2/3. Graph the zeros. The parabola will open up because a = 1 which is positive. (Above RIGHT)
15. / A. / At x = 3 or – 2 / - x2 + 4 = - x – 2 ® x2 – x – 6 = 0 ® (x – 3)(x + 2) = 0 ® x – 3 = 0 or x + 2 = 0 ® x = 3 or x = - 2
B. / [-2, ) or x - 2 / - x – 2 0 ® - x 2 ® x - 2 (When mult / div by negative, inequality turns around.)
15. / C. / (-2, 3) or – 2 < x < 3 / - x2 + 4 > - x – 2 ® - x2 + x + 6 > 0 ® Mult by – 1 ® x2 – x – 6 < 0. Find zeros (see # 21 A). Parabola opens up, so graph is below x-axis for the interval between the zeros.
D. / / - x2 + 4 > 1 ® - x2 > - 3 ® Mult by – 1 ® x2 < 3 ® x2 – 3 < 0. Find the zeros. x2 – 3 = 0 ® x2 = 3 ® Take square root of both sides ® ® x = ® a is now + (x2 – 3 < 0) so parabola opens up and the function is below the x-axis between the zeros.
Answer / Work / thinking
16. / A. / 6 seconds / Strike the ground implies d = 0 ® - 16t2 + 96t = 0 ® Factor greatest common factor ® -16t(t – 6) = 0 ® - 16t = 0 or t – 6 = 0 ® t = 0 or t = 6. t = 0 is the beginning of the throw, so the ball will strike the ground in 6 seconds.
B. / Between 2 and 4 sec / -16t2 + 96t > 128 ® -16t2 + 96t – 128 > 0 ® Mult by – 1 ® 16t2 – 96t + 128 < 0 ® Find zeros by factoring ® 16(t2 – 6t + 8) = 0 ® 16(t – 4)(t – 2) = 0 ® t – 4 = 0 or t – 2 = 0 ® t = 4 and t = 2. Looking at -16t2 + 96t – 128 > 0, the parabola opens down, so the parabola is above 0 between the zeros.
17. / A. / Revenue is 0 for prices of $0 and $3800. / -½ p2 + 1900p = 0 ® Mult by – 2 ® p2 – 3800p = 0 ® Factor GCF ® p(p – 3800) = 0 ® p = 0 or p = 3800
B. / Price range of $800 to $3000 / -½p2 + 1900p > 1,200,000 ® Mult by -2 ® p2 – 3800p < -2,400,000 ® p2 – 3800p + 2,400,000 < 0 ®Use quadratic formula ® ® p = 800 or p = 3000
C. / Unit price of $1900 / Maximum ® Vertex ®
D. / Max revenue is $1,805,000 / Substitute 1900 into R(1900) = -½(1900)2 + 1900(1900) = 1,805,000
18. / A. / 40 TVs / Minimum ® Vertex ®
B. / $400 min marginal cost / Substitute into function C(40) = 402 – 80*40 + 2000 = 400
19. / A. / 119 boxes; revenue $564 / Max ® Vertex ® ® R(119) = 9.5*119 - .04*1192 = 564
B. / P(x) = -0.04x2 + 8.25x – 250 / P(x) = R(x) – C(x) = (9.5x – 0.04x2) – (1.25x + 250) = 9.5x – 0.04x2 – 1.25x – 250 ® Combine like terms to clean up.
C. / 103 boxes; $175 in profit / Max ® Vertex ® ® P(103) = -.04*1032 + 8.25*103 – 250 = 175.39
20. / A. / X = - 7 or 4 / |2x + 3| - 5 = 6 ® |2x + 3| = 11® Means (2x + 3) is 11 units from 0, either left or right ® 2x + 3 = -11 or 2x + 3 = 11. 2x + 3 = - 11® 2x = - 14 ® x = - 7. 2x + 3 = 11 ® 2x = 8 ®x = 4
B. / No solution / |2 – x| = -1 means that the distance of (2 – x) is negative one unit from 0. Distance is always 0 or positive. This is not possible.
C. / X = 12 or - 12 / Multiply by the reciprocal of 3/4, which would be 4/3. ® |x| = 12 ® x is left or right 12 units from 0 ®x = - 12 or 12
D. / X = ½ or – ½ / 4 - |2x| = 3 ® - |2x| = - 1 ® Multiply both sides of equation by – 1 ® |2x| = 1 ® 2x is 1 unit from 0, either left or right ® 2x = -1 or 2x = 1 ® x = - ½ or x = ½
E. / X = -1 or 2 / |1 – 2x| + 6 = 9 ® |1 – 2x| = 3 ® (1 – 2x) is either 3 units left of 0 or right of 0 ® 1 – 2x = - 3 or 1 – 2x = 3 ® - 2x = - 4 or – 2x = 2 ® x = 2 or x = - 1
F. / X = 1, 3, or / |x2 – 2x| = 3 ® (x2 – 2x) is 3 units left or right of 0 ® x2 – 2x = - 3 or x2 -2x = 3 ® x2 – 2x + 3 = 0 or x2 – 2x – 3 = 0 ® x2 – 2x + 3 = 0 does not factor, so using the quadratic formula yields
x2 – 2x – 3 = 0 factors into (x – 3)(x + 1) = 0 ® x – 3 = 0 or x + 1 = 0 ® x = 3 or x = - 1

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