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Transformations was quite a laborious unit. I don’t know how else to summarize my struggles. Reciprocal functions was a problem for me, but I think I know how to properly do them now. I just have to eliminate making minor mistakes. Nevertheless, I think I’m prepared for the pretest and test. I’ll put as much as I can below to hasten those that are reading in this. Let us all go good and rock these upcoming tests ! Good luck !

This will all be unordered because I’m too lazy to put them in a specific arrangement and with disorder, uhmm.. I had an explanation, but I forgot it after five seconds. Oh well.

• sine is equivalent to cosine shifted right by ¼ of the period
  Ex1. Period=2п ; sin(x) = cos(x - pi/2)
  Ex2. Period=16 ; sin(x) = cos(x - 4)
  
  
• cosine is equivalent to sine shifted left by ¼ of the period
  Ex1. Period=2π ; cos(x) = sin(x + pi/2)
  Ex2. Period=16 ; cos(x) = sin(x + 4)
  
  
• the values of parameter C above are the smallest absolute value of shifts assuming the parameter A remains constant
  There are an infinite number of shifts where sin(x) = cos(x) and cos(x) = sin(x).
  Ex1. Period=16 ; cos(x) = sin(x + 20)
  
  
• general equation for sin function is:
  f(x) = AsinB (x-C) + D
  
  
• general equation for cos function is:
  f(x) = AcosB (x-C) + D
  
  
• amplitude = | A |
  Ex1. f(x) = -4sin(x + 3) ; amplitude = 4
  
  
• B is not the period, but it determines the period
  f(x) = -4sin5(x + 3) ; period = 2π / B = 2π / 5
  
  Note that the equation consists of 2π because the period for sin(x) and cos(x) is 2π. The period for tan(x) is π, so period = π / B.
  
  
• What is the period and values of parameter B and C if f(x) = -4sin[ (5x + 7) / 2 ]?
  X must be isolated to find the period. We can factor 5/2 from [ (5x + 7) / 2 ]. This makes the equation into f(x) = -4sin[ 5/2(x + 7/5) ].
  
  Parameter B is 5/2. Parameter C is -7/5. The period = 4π/5.
  If you are wondering how Parameter C was obtained, here is one way to do it. You factor 5/2 from 5x/2 to isolate x. Then you factor 5/2 from 7/2. Here are two ways to do that.
  
  1)Something multiplied by 5/2 must equal 7/2 because of 5/2(x - C). So 5/2 * C = -7/2. Isolating C gives -7/5.
  2)Multiply -7/2 by the reciprocal of 5/2.
  
  
• Parameter C shifts the graph horizontally.
  Parameter B compresses/stretches the graph horizontally. The x values are multiplied by a factor of 1/B.
  
  Ex1. f(x) = -4sin[ 5/2(x + 7/5) ] ; Parameter B is 5/2, so all X values are multiplied by 1/(5/2), which equals 2/5.
  Parameter A stretches/compresses the graph vertically. If |A| is greater than 1, then the graph is stretched. If |A| is less than 1, but greater than 0, the graph is compressed.
  Parameter D shifts the graph vertically.
  
  
• Stretches before translations.
  
  
• Even functions are symmetrical about the y-axis.
  A function is even if f(-x) = f(x).
  Ex1. f(x) = x^2 ; If we plug in -x, then f(-x) = (-x)^2 = x^2
  f(-x) = f(x), therefore the function is even.
  
  
• Odd functions are symmetrical about the x-axis.
  If you rotate the graph 180 degrees and the graph appears to have not moved, then the graph is odd.
  A function is odd if f(-x) = -f(x) *multiplying y by -1
  
  Ex1. f(x) = x^3 - x ; If we plug in -x, then f(-x) = (-x)^3 - (-x) = -x^3 + 3
  f(-x) = -f(x), therefore the function is odd. Every term in the function must consist of x where the exponents are all odd if a function were to be odd.
  
  
• -f(x) produces a reflection over the x-axis
  y values are multiplied by a negative value
  
  
• f(-x) produces a reflection over the y-axis
  x values are multiplied by a negative value
  
  
• switch X and Y of function when finding its inverse
  
  Ex1. f(x) = x^5 - 2
  y = x^5 - 2
  x = y^5 - 2
  x + 2 = y^5
  y = root 5 of (x + 2)
  
  
• Invariant points are points on graphs that don't change
  1 and -1 are invariant points if the graph touches those points.
  
  
• If a graph touches zero, the reciprocal part of that graph goes on forever without reaching the asymptote.
  If a graph goes on forever without touching an asymptote, the reciprocal part of that graph touches zero.
  Analyze tan(x) and cot(x) and you will definitely notice this. Where tan(x) touches zero will be where where vertical asymptotes are for cot(x)
  However, graphs can go on forever without there being an asymptote. If that is the case, the reciprocal part of that graph approaches zero forever without actually touching it.
  
  

Below are a few of the answers to the questions from the most recent slideshow provided by Mr. K. I apologize for the monumental size of the images below. Hover over them too :P I didn't really provide explanations too. The vertical asymptote for the funny looking graph below is -2.5 too.

Make sure your bookmarks are here too !!
Circular Functions and Transformations
Tags must be Transformations, CircularFunctions, and PC40SW09

I also posted my answers to the Trigonometric modeling problems. I don't know if I'm right, so comment along !
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