hornblower Posted March 31, 2015 Share Posted March 31, 2015 If a student were struggling with questions like this, can you recommend an amazing resource to make it clearer? Can anyone point us in the right direction? An awesome textbook? A study guide? A magical being to cram this stuff into the student's head? I don't know enough about what this stuff is even called to find appropriate resources. Everything seems to be either too simple or much harder & not quite explaining the questions the student is dealing with. Current textbook is absolutely worthless. Instructor means well but is baffled that people don't understand him.At this point unfortunately it's less a matter of working towards true deep understanding and more a matter of 'just teach me the steps so I can pass the darned exam'. Not ideal, I know but it is what it is... So something that breaks the questions down into: "do this, then do this, look out for this .... " I used to have semester end cram guides like this for my science courses but I'm not finding the right sort of thing for this....Some sample questions are: 1. consider the polynomial function p(x) = 3x^3 - 4x^2 - 7x + 6 a) state the possible rational zeros of p(x) b) Determine all of the real zeroes of p(x) & show work. 2. A rectangle with length x and width y has a perimeter of 40 cm. Determine length of the diagonal of the rectangle as a function of x and state the domain of the function. 3. determine the equation of the polynomial function shown below. Assume it is a polynomial function of minimum degree. sketch attached 4. write 4 ln x-2 ln(x+2) -1/2 ln z - ln y as a single logarithm 5. determine the inverse of f(x) = log base3 (2-x) + 1 6. sketch y=1 + log base2 (x+4) 7. state range of y=4-5e^(x-2) 8. solve for x log base2 x + log base2 (x-6) = 4 11. sketch angle 5pi/3 in standard position the determine the exact value of sin 5pi/3 , cos 5pi/3 and tan 5pi/312. find all values of angle theta such that cos theta = - 1/2 and 0 ≤ theta <2pi pls & thank you! Quote Link to comment Share on other sites More sharing options...
kiana Posted March 31, 2015 Share Posted March 31, 2015 Schaum's outline of Precalculus. 50 bajillion (ok, really 700+) worked examples to show you the steps. Cheap. Quite possibly available in your local library if you want it immediately. 1 Quote Link to comment Share on other sites More sharing options...
hornblower Posted March 31, 2015 Author Share Posted March 31, 2015 awesome! Library has it, I have hold on it & I should be able to get it tonight or tomorrow a.m. thank you kiana! Quote Link to comment Share on other sites More sharing options...
JanetC Posted March 31, 2015 Share Posted March 31, 2015 patrickjmt.com 1 Quote Link to comment Share on other sites More sharing options...
Arcadia Posted March 31, 2015 Share Posted March 31, 2015 I second Schaum's for exam prep. If the student finds Schuams too hard to start with, Barron's EZ guides are a lot easier. Also SparkNotes Pre-Calculus (SparkChart) http://www.amazon.com/Barrons-E-Z-Precalculus-Lawrence-Leff/dp/0764144650 http://www.amazon.com/Pre-Calculus-SparkCharts-SparkNotes/dp/1586636227 If you are near a Barnes & Noble bookstore, hit the exam prep shelves. Lots of choices there. The student would have a high chance of finding an exam prep book that fits him/her. ETA: The Schaum's easy outlines series won't useful for us. My older read through the Barron's EZ series up to their calculus book from our library. He is my bookworm though. 1 Quote Link to comment Share on other sites More sharing options...
MarkT Posted March 31, 2015 Share Posted March 31, 2015 see If your library has "Pre-Calculus Problem Solver" by REA I used the "Calculus Problem Solver" back in college before the WWW existed! Today I would probably just look it up on the Web. 1 Quote Link to comment Share on other sites More sharing options...
hornblower Posted March 31, 2015 Author Share Posted March 31, 2015 Thanks! We have a mathways.com account too and that's been helpful in many ways but more was needed. Also found this site which looks good http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm Got the Schaum's, put the REA book on hold... I see a glimmer of hope. :) Her final is on April 21 so there's still lots of time for panicking ;) thx everyone. Quote Link to comment Share on other sites More sharing options...
mathwonk Posted March 31, 2015 Share Posted March 31, 2015 this may not help, but i couldn't resist a few comments. In regard to question #3, the statement about "minimum degree" does not make precise sense, although one sort of knows what is meant. A more precise statement would be to assume all roots are real, and that no root has multiplicity greater than 3. I.e. you cannot "see" complex roots nor a root of multiplicity more than 3 with the naked eye, so they seem to mean the polynomial has the lowest degree for any graph that "looks roughly like this". so in that picture the first root is a double root, since the graph is tangent but does not cross the x axis,a nd the second root is a triple root since the graph is tangent and does cross, and the third root (left to right) is a simple root. as to question #1, the most basic rule of algebra is the "root/factor" theorem, i.e. r is a root if and only if (x-r) is a factor. so if x = a/b is a root, (in lowest terms) then (x-a/b) is a factor, and since all coefficients are integers, this implies that (bx-a) is a factor. but that forces b to be a factor of the lead coefficent 3, and a to be a factor of the constant coefficient 6, so we know b = 1,-1,3, or -3, and a = 1,-1,2,-2,3,-3,6, or -6. that's a lot of possibilities and you just have to try them all. well i guess you only have to try a = 1,2,3,6, if you try all the negative possibilities for b. that's still 16 possibilities. oh i guess 1/1 = 3/3 and 2 = 6/3 so some of them are duplicates. luckily an integer works. i.e. b =1 in this case works, and a = 2. so x = 2 is a root. then after finding one root, you divide out to reduce the degree of the polynomial i.e. you get (x-2)q(x) = your original cubic, where q is a quadratic. then you factor or use the quadratic formula to solve q(x) = 0. now you use root/factor backwards to finish off question #3. i.e. if you have a double root of a, and a triple root of b and a simple root of c, the polynomial should be a constant multiple of (x-a)^2.(x-b)^3.(x-c). oh, here is another flaw in the statement, since it is pretty much impossible to see the value of that constant mutiple too, so i have just taken it to be 1. (am i right here? (no, see below) does someone know more about this?) and just saying "minimum degree" does not help here, since multiplying by a non zero constant does not raise degree. Aha! we CAN find that constant since taking x=0 gives us the constant term of the polynomial, and we can presumably see that as the height where the graph crosses the y axis (i can't see those numbers on that thumbnail). so we can set that value equal to constant.a^2.b^3.c, and solve for the constant. so in a nutshell, the x intercepts give you the roots, (and the tangencies give the multiplicities) which give you the factors, and the y intercept gives you the constant term. in general, a graph that crosses the x axis has a root of odd multiplicity and one which does not cross it, has a root of even multiplicity. A graph that meets the x axis tangentially has a root of multiplicity greater than one, and a graph that crosses transversely (non tangentially) has a simple root (multiplicity one). You cannot really see the difference between a root of mult = 2, or mult = 4 say, or mult = 6, so when they say "minimum degree, they just mean only consider multiplicities 1,2 and 3. I.e. all we really know about this graph is the first root has even mult ≥ 2, and the second has odd mult ≥ 3, and well maybe that the third one has mult = 1. (But even there, how do we know that if we zoomed in a billion times the graph might not look slightly tangent?) that's a lot of logarithm problems. the main fact to remember is that log(XY) = log(X) + log(Y), and log(X/Y) = log(X) - log(Y), with any base. oh yes, and if the base is b, then b^(log(x)) = x, indeed that is the meaning of the term "log(x), with base b". in general one cannot find angles with given cosine, nor the cosine of a given angle, unless they occur in a very simple right triangle, i.e. 30-60-90, or 45-45-90. so be sure to practice on these triangles, e.g. in a 30-60-90, the hypotenuse is twice one leg, so problems with cos(t) = 1/2, or -1/2, must involve this triangle. of course you have to know that 60 degrees = pi/3 radians, i.e. 2pi radians = 360 degrees. good luck. 1 Quote Link to comment Share on other sites More sharing options...
hornblower Posted March 31, 2015 Author Share Posted March 31, 2015 thx mathwonk! I think that will be very helpful. Quote Link to comment Share on other sites More sharing options...
mathwonk Posted March 31, 2015 Share Posted March 31, 2015 you are welcome. by the way, some time ago i believe i wrote short guides to both algebra and trigonometry in these forums, but i don't know how to search for old content. Quote Link to comment Share on other sites More sharing options...
wapiti Posted March 31, 2015 Share Posted March 31, 2015 some time ago i believe i wrote short guides to both algebra and trigonometry in these forums, but i don't know how to search for old content. http://forums.welltrainedmind.com/topic/404665-trigonometry-help/?p=4078129 ? or http://forums.welltrainedmind.com/topic/457791-tell-me-about-doing-some-beginning-calculus-with-just-an-algebra-background/?do=findComment&comment=4847890 Quote Link to comment Share on other sites More sharing options...
mathwonk Posted March 31, 2015 Share Posted March 31, 2015 thank you! Quote Link to comment Share on other sites More sharing options...
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