AP Calculus
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Questions about the Individual Worksheets

Worksheet III: Mixed Review Problems

I just need a quick refresher course in finding roots of equations.  On the third individual worksheet, on the first question, are you just looking for the factored answers to the equation?  I don't understand the problem fully. 

For the general review sheet, problem #1, you do the
same type of factoring, but you have to go the extra step and solve for the zeros (aka roots, solutions) of the polynomial equation.  To help you on that one, you factor by grouping (2/2 grouping), and then determine what makes each factor equal zero.  (The final answers should be 2i, - 2i, and -1/3 since it said to solve for all roots.)

On #10, I forget the formulas for finding the period and amplitude and couldn't find them in any textbooks...could you refresh my memory of them?  And on #16-18, I completely forget what I'm doing with those.  Can you help me get started on them?

On #10 for any sine or cosine function of the form y = A sin(Bx+C)+D, you can find the following:
Amplitude = |A|
Period = (2 PI)/B
Phase Shift = - C/B
Vertical Shift = D
For #16, you have to get sine by itself and then look-up (or hopefully know) what angles on the unit circle produce that coordinate value.  After solving for sinx = +/- SQRT(2)/2, what are the corresponding angles with those possible y-coordinate points?  #17, You should equal the equation to zero and then factor the trinomial.  #18, You should use the Sine Double Angle Identity, equal the equation to zero, and then factor what's common to solve. 

I was just wondering if you could help start me off on 13, 14, and 15 of the mixed review because I have no idea how to do any of them.  Also, is the answer to number 2 of the mixed review: sq rt(x-3)/(x-3)?  Can you explain the difference between arccos and secant?

#13 - You are given that sin(theta) = -4/5 and the inequality tells you the angle lies in Quadrant III (so that you know the sign that should be on the various trig functions there).  You need to use the double angle identities to evaluate the three trig functions--check the back of your unit circle sheet for the identities to use.
#14 - To evaluate inverse trig functions, I usually rephrase the problem:  For what angle is cosine = -1/2?  Look on your circle sheet if you don't remember the specific values.
#15 - Rephrase: For what angle is cosecant equal to the value in the problem?
#2 - No, your answer is not correct.  To find the inverse of a function, swap the X and Y and solve this new equation for the Y variable.  The resulting expression will be your inverse function.
The difference between Arccos and Secant: 
Arccos is the same thing as Inverse Cosine and it is used to cancel out the cosine function when directly applied to each other.  Arccos(cos X) = X  or cos(Arccos X) = X
Evaluating inverse trig functions means you are finding the angle on the unit circle that has the given coordinate. 
Secant is the multiplicative reciprocal of cosine, or 1/cosine.  It cancels out the cosine function only when they are multiplied together:  secX * cosX = 1

Questions about the Multi-Section Graphing Packet

Section I      Section II      Section III      Section IV


Section I Questions

In Section I, since my calculator can't do logs of base 2, I'm having trouble with #10 which is f(x)=log2(x)... can I have a hint? 

#10:  Use change of base theorem: logb(x)=log(x)/log(b) to graph it.  You will type in y=log(x)/log(2) for that particular problem.

Also, with #14, which is f(x)=SQRT(a2 - x2), how do I work with the "a" variable and the "x" one?

#14:  The "a" stands for a constant--not a variable. Try letting it be a few different SPECIFIC examples on your calculator...like, let a=2 and you can graph SQRT(4-x2),  or try a=3 and graph SQRT(9-x2)...then generalize the pattern you find to work for any constant, a.

In the Section I, question 15, what are you asking?  I didn't know if that was supposed to be a matrix or something else.

# 15: [x] is the greatest integer function, otherwise known as the step function, because it looks like a set of stairs...your calculator will graph it...graph  y =int(x)  where the int function is probably under your "MATH" menu someplace.  Also, you should be in DOT mode so that the individual "steps" on the graph are not connected to each other.


Section II Questions

In the Section II, I forgot how to draw a reflection of all those things.  Do you just negate the function?  It's the vertical line test that tells if the reflection is a function, right?

Section II:  No, you do not just negate the function... you have to draw the line y=x (diagonal through quadrants 1 and 3) and reflect the image across this line.  And, no, the vertical line test does not tell whether the reflection is a function...it tells you whether the ORIGINAL graph is a function.   There was another test we used in conjunction with the VLT to tell whether the reflected inverse was also a function.

In the Section II, on the characteristic that assures the reflection is a function, is it when it is raised to a positive/negative odd power?

For the last question, you are being way too specific in your rule. I am looking for something about "line tests"....see if that starts you in the right direction.


Section III Questions

In Section III, do you want us to just use the negated function.  Like, plugging in a negative x for x and then graphing that?

Section III:  Each part is slightly different from the others...the only part that you actually plug in a negative x is problem 1, part b...the next part you make the entire function negative (as opposed to plugging in a negative), and so on.


Section IV Questions

In Section IV, I am not understanding how the absolute value symbols affect the graphs.  Can you explain this concept a little more?

The key to understanding these transformations is in knowing where the absolute value symbol is applied:  If the absolute value symbol is "inside the function" (just around the x), then it makes the negative x's look like the positive x's.  In other words, the left half of your graph will be a reflection of whatever is on the right half and your graph will be symmetric about the y-axis.  However, if the absolute value symbol is "outside the function" ( think of it as |y| instead of |f(x)| ), then any negative y-values will become positive.  In other words, any portions of the graph below the x-axis get reflected above the x-axis.



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