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By: Emma Stevens Teaching Transformations of Functions using Music.

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Presentation on theme: "By: Emma Stevens Teaching Transformations of Functions using Music."— Presentation transcript:

1 By: Emma Stevens Teaching Transformations of Functions using Music

2 Objectives  Students will be able to analyze nonlinear relationships to explain how the change in one variable results in the change of another.  They will explain how that change affects the graphs and then apply the changes to a melody of notes.

3 Previous knowledge  Students should have an understanding about a function f(x).  They should understand that a function inputs x values (the domain or the pre-image) and then outputs f(x) values or y values (the range or image).

4 Keys of the Piano

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6 x f(x) xf(x) xf(x)

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8 Things to Keep in Mind The function repeats its self; so that f(42)= f(0) =0, f(43)= f(1)= 0, f(44)=f(2)=4 and so on. The x-values are mod 42 Also this does not take in consideration of flats and sharps in new keys so the song with the transformation may not sound right. i.e. the notes are in (mod 7) – The y-values So 0 = 7 = C, 1 = 8 = D, 2 = 9 = E, 3 = 10 = F…. The beat is the same throughout the song: i.e. each note is the same count (it is a note a beat)-meaning it doesn’t take in account for rhythm

9  g(x) = f(1/2x)  This transformation inputs x values, then multiply each x value by 1/2 and the output is f(1/2x). when x=0, g(0) = f(1/2*0) = f(0) =0 when x=2, g(2) = f(1/2*2) = f(1) =0 when x=4, g(4) = f(1/2*4) = f(2) = 4 And so on…  But some values of x will not have a corresponding y value(because the x values and f(x) values are both integers): when x=1, g(1) = f(1/2*1)= f(1/2) when x=3, g(3) = f(1/2*3) = f(3/2) And so on… 

10 x f(x) 1/2xg(x) xf(x)1/2xg(x) xf(x)1/2xg(x)

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12  We will call this transformation h(x) = f(2x)  This transformation inputs the x values, multiplies each x value by 2 and outputs f(2x) values. when x=0, h(0)=f(2*0)= f(0)= 0 when x=1, h(1)=f(2*1)= f(2)= 4 when x=2, h(2)=f(2*2)= f(4)= 5 And so on…  Notice that f(1), f(3), f(5) and so on don’t exist in this transformation; therefore only half of the notes are in the function 

13 x f(x) 2xh(x) x f(x) 2xh(x) = = = = = = =121 x f(x) 2xh(x) = = = = = = = = = = = = = =401

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15  j(x) = f(x+4)  This transformation inputs the x values, adds 4 to each x value and outputs f(x+4) values. when x=1, j(1)= f(1+4)= f(5)= 5 when x=2, j(2)=f(2+4)= f(6)= 5 when x=3, j(3)=f(3+4)= f(7)= 4 And so on…  Notice that the starting note of the function is actually the fifth note of the original function and f(1), f(2), f(3), and f(4) have been taken off the front of the song and attached to back of the song. 

16 x f(x) x+4j(x) x f(x) x+4j(x) x f(x) x+4j(x) = = = =34

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18  k(x) = f(x) + 2  This transformation inputs the x values, adds 2 to each f(x) value and outputs f(x)+2 values. when x=1, k(1)=f(1)+2= 0+2=2 when x=2, k(2)=f(2)+2= 0+2=2 when x=3, k(3)=f(3)+2=4+2 =6 And so on…  Notice that the starting note of the function is two notes higher than the original function 

19 xf(x) k(x) xf(x) k(x) x f(x) k(x)

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21  r(x) = 3f(x)  The transformation that inputs x values and outputs 3f(x) values; Therefore: When x=1, r(1)=3f(1)= 3(0)=0 When x=2, r(2)=3f(2)= 3(0)=0 When x=3, r(3)=3f(3)= 3(4)=12 And so on…  This song stretches the original notes further apart, so that they have a greater distance between them. 

22 xf(x) r(x) xf(x) r(x) x f(x) 4(x)

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24  s(x) =(1/2) f(x)  This transformation inputs x values, then multiply each f(x) value by 1/2 and the output is (1/2)f(x). when x=0, s(0) = (1/2)f(0) = (1/2)(0) = 0 when x=2, s(1) = (1/2)f(2) = (1/2)0 = 0 when x=4, s(2) = (1/2)f(2) = (1/2)4 = 2 And so on…  But some values of x will not have a corresponding y value(because the x values and f(x) values are both integers): when x=4, s(4) = (1/2)f(4)= (1/2)5 = 2.5 when x=11, s(11) = (1/2)f(11) = (1/2)1=0.5 And so on… 

25 xf(x) s(x) xf(x) s(x) x f(x) s(x)

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27 This transformation inputs x values and outputs f(-x) values Therefore: When x= 1, m(1)=f(-1)=f(41) = 0 When x= 2, m(2)=f(-2))=f(40)= 1 When x= 3, m(3)=f(-3)= f(39)= 1 And so on…  m(x) = f(-x)  The we will actually play the last note first  i.e. the song is backwards

28 x f(x) -xm(x) 0 0 0= = = = = = = = = = = = = =29 0 x f(x) -xm(x) = = = = = = = = = = = = = =15 4 x f(x) -xm(x) = = = = = = = = = = = = = =1 0

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31  p(x) = - f(x)  The transformation that inputs x values and outputs -f(x) values; Therefore: When x=1, p(1)=-f(1)= 0 When x=2, p(2)=-f(2)= 0 When x=3, p(3)=-f(3)= - 4 And so on…  This song has completely different notes than the original tune and the notes are lower. 

32 xf(x) p(x) xf(x) p(x) x f(x) p(x)

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34 Putting it All Together!

35 Summarization of Transformations

36 Special Thanks to Betty Clifford for advising me though this project.


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