AP Calculus
Hints Page
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 #1618, 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 lookup (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 ycoordinate 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(x3)/(x3)? 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 functionscheck 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
MultiSection 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)=log_{2}(x)... can I have a hint?
#10: Use change of base theorem:
log_{b}(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(a^{2}  x^{2}), how do I work with the
"a" variable and the "x" one?
#14: The "a" stands for a
constantnot a variable. Try letting it be a few different SPECIFIC examples on your calculator...like, let a=2 and you can graph
SQRT(4x^{2}),
or try a=3 and graph SQRT(9x^{2})...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 yaxis. However, if the
absolute value symbol is "outside the function" ( think of it
as y instead of f(x) ), then any negative yvalues will become
positive. In other words, any portions of the graph below the
xaxis get reflected above the xaxis.
