Monthly Schedule

(Algebra II, Periods E and F)

W 9/7/05	First day of class.
Th 9/8/05	HW due: §1-1 #1, 2, 4, 6, 8; §1-2 all. Follow the HW guidelines as you do this assignment. Quiz on the alphabet. Daily HW reports will begin later. Don’t worry about them for the moment.
F 9/9/05	HW due: Finish the assignment that was made previously, if you have not already done so. Today I will collect the fallacious proof that pi equals 5. Your job is to fill in the handout, giving a reason for each step. Quiz on the alphabet (postponed from yesterday). Possible Quiz on the HW guidelines. Extra credit: Do the assignment due today in the MODD class.
M 9/12/05	HW due: Read §1-3; write §1-3 #1-37 eoo. (The abbreviation “eoo” can be found on the abbreviations page.) This is a brief review of Algebra I.
T 9/13/05	HW due: Read §1-4; write §1-4 #1-33 eoo.
W 9/14/05	HW due: Write §1-5 #1-33 eoo, 36. Note that this assignment was originally not stated correctly. The last problem in the set is #36, not #37.
Th 9/15/05	HW due: Write §1-6 #1-29 eoo.
F 9/16/05	HW due: Read §1-7; write §1-7 #1, 2, 13, 14. In class: Review. Here are the answers to the question we did not have time to get to in class: 14.(a) reflexive (b) identity for addition (c) mult. prop. of equality (d) identity for addition (e) factoring (or distributive property in reverse) (f) subtraction prop. of equality (or you could say “Cancellation Property of Equality for Addition” as on the top of p. 37)
M 9/19/05	Test on Chapter 1. Under new math department policies, you will be required to take a different test as a make-up test if you miss the test today. The make-up test will be offered at 7:00 a.m. on Tuesday, 9/20/2005, and is not guaranteed to be of equivalent difficulty.
T 9/20/05	HW due: Read §§2-1 and 2-2; write §2-2 #13-16 all.
W 9/21/05	HW due: Read §2-3; write §2-3 #1, 2, 6, and one other completely made-up example from your own imagination.
Th 9/22/05	HW due: Write §2-4 #2-10 even, 11-26 all. Your sketches for #11-26 are permitted to be sloppy. Reading §2-4 is not required. However, if you do read the section, you must promise me that you will never use the so-called “vertical line test” (middle of p. 66) unless you can explain why it works.
F 9/23/05	HW due: Read remaining portions of §§3-1 and 3-2 (reading notes optional tonight); write §3-2 #1-20 all. You must practice sketching these quickly, less than 1 minute each. (Do not plot neatly on graph paper as stated.) If you hate graphing and understand the concept, you may pick any 2 even-numbered problems plus #22. However, if you do that, your answer to #22 must be in complete sentences and must be correct.
M 9/26/05	HW due: p. 70 #R1-R3 all, p. 106 #R1-R3 all. In class: Review. Answers to the review questions for your study and reference: p. 70 R1.(a) Rewrite as y = –x + 2. Plot as a segment (not a full line) from (–1, 3) to (3, –1). The segment slopes downward as you travel from left to right, and the y-intercept is at (0, 2). The graph represents a function with R = [–1, 3]. It is perhaps interesting that the range is identical to the domain here. (b) Rewrite as y = –x^2 + 2 and start plotting points. The graph consists of exactly 5 ordered pairs, namely the points (–2, –2), (–1, 1), (0, 2), (1, 1), and (2, –2). Do not connect the points with a smooth curve, since the domain is only 5 points. The graph represents a function with R = {–2, 1, 2}. (c) This is the same problem as (b), except with the roles of x and y interchanged. The graph consists of exactly 5 ordered pairs, namely the points (–2, –2), (1, –1), (2, 0), (1, 1), and (–2, 2). Do not connect the points with a smooth curve, since the domain is only 3 points. (Note that except for x = 2, each value in the domain produces two values in the range.) The graph represents a relation that is not a function with R = {–2, –1, 0, 1, 2}. (d) Rewrite as |y| = –x + 2, and then use your knowledge of absolute value to rewrite as follows: y = –x + 2 or y = –(–x + 2) The first equation is a line segment with slope –1 and y-intercept at (0, 2). The second can be written as y = x – 2, which is a line segment with slope 1 and y-intercept at (0, –2). Why segments and not full lines? That is because the domain is only the interval [–2, 2]. Therefore, you get two segments meeting to look like a giant “>” symbol: one from the point (–2, 4) to the point (2, 0), and the other from the point (–2, –4) to the point (2, 0). The graph represents a relation that is not a function with R = [–4, 4]. R2.(a) Yes. (b) No. (c) Yes. (d) No. (e) No. (f) Yes. R3.(a) Let x-axis be time t (sec.), and let y-axis be speed v (mph). D = [0, 10], R = [0, 5]. Curve is linear with negative slope at first, then tapers to a less negative slope so as to make an overall “L” shape. (b) Let x-axis be time t (min.), and let y-axis be temperature T (degrees Fahrenheit). D = [0, 20], R = [68, 375]. If the oven is thermostatically controlled for 375° F., then the curve may be shaped like an “S”: increasing gradually at first, then more steeply, then gradually until ending at the point (20, 375). Numerous other answers are possible. (c) If the location is far northern Alaska in the middle of the summer, then the curve would resemble the illustration in #R2(a). Let x-axis be time t in days, where 0 represents 12:00 midnight on July 1, 2005. Let y-axis be the sun’s angle of elevation q, measured in degrees. D = [–1.5, 1.3], R = [30, 50]. If the location is elsewhere in the world, then the curve may need to be shifted downward. Note that most locations, where the sun is visible for only part of the day, would have a curve that looks more like Figure 13-6i on p. 750, except that the numbers on the x and y axes would need to be adjusted, and all intervals of x values for which the curve lies below the x-axis should be deleted from the domain. That is because “the angle at which you have to look up to see the sun” is undefined at night. (d) Let x-axis be time t (hrs.) after 12:00 midnight on a typical day. Let y-axis be # of cars C. The graph is not a curve of the type seen in the other problems, but instead a step function that shows some unpredictable upward and downward blips. D = [0, 24], R = {0, 1, 2, 3, . . . , 94, 95}. Note that while the domain is an interval, the range consists of whole numbers only. p. 106 R1. Let y = f (x) = 3x – 7 (a) f (–1) = 3(–1) – 7 = –10; f (2) = 3(2) – 7 = –1; f (5) = 3(5) – 7 = 8 (b) Presumably you can do this. Note that the slope from (–1, –10) to (2, –1) is 3, as is the slope from (2, –1) to (5, 8), since both segments involve going to the right 3 units and up 9 units. As you learned in geometry, the points are collinear because the slopes match. If you cannot remember how to compute slope (Dy/Dx), then you need to ask for help as soon as possible. R2.(a) Graph has slope of 2/5 and y-intercept at (0, –3). (b) Graph has slope of –7/3 and y-intercept at (0, 7). The x-intercept at (3, 0) makes plotting easier. (c) Graph is a vertical line passing through the point (3, 0). (d) Graph is a horizontal line passing through the point (0, –4). R3.(a) Point-slope form (from geometry class). (b) The point is (4, –3). Remember how to do this: You look at the value that is being subtracted from x and the value that is being subtracted from y. What value is being subtracted from x? Answer: 4. What value is being subtracted from y? Answer: –3. Therefore the known point through which the line must pass is the point (4, –3). (c) Graph has slope of –5/2 and passes through the point (4, –3). That is enough information to sketch the graph. If you look ahead to part (d), you can also plot the y-intercept at (0, 7), but that is optional. (d) Distribute to get y + 3 = –5x/2 + 10, which simplifies to y = –5x/2 + 7. This clearly shows a slope of –5/2 and a y-intercept at (0, 7). (e) Multiply the equation by the common denominator (2) to get 2y = –5x + 14, which can be transformed by a small bit of algebra into 5x + 2y = 14.
T 9/27/05	Test on Chapter 2 plus §§3-1 and 3-2. There will be approximately 5 minutes for questions before we begin the test.
W 9/28/05	HW due: §3-3 #1-14 all.
Th 9/29/05	HW due: §3-4 #17, 9-21 mo3, §3-5 #2, 4, 5 (choose any 2).
F 9/30/05	HW due: §3-5 #17, 19. Today’s assignment, as well as the assignment due yesterday, will be collected. (Actually, I forgot to collect the assignment in F period. F period students should plan on having this or another double assignment collected next week.)

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Last updated: 06 Oct 2005