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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) 0 0 1 0 2 4 3 4 4 5 5 5 6 4 7 3 8 3 9 2 10 2 11 1 12 1 13 0 xf(x) 14 4 15 4 16 3 17 3 18 2 19 2 20 1 21 4 22 4 23 3 24 3 25 2 26 2 27 1 xf(x) 28 0 29 0 30 4 31 4 32 5 33 5 34 4 35 3 36 3 37 2 38 2 39 1 40 1 41 0

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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) 0 0 00 1 00.5 2 4 10 3 4 1.5 4 5 24 5 5 2.5 6 4 34 7 3 3.5 8 3 45 9 2 4.5 10 2 55 11 1 5.5 12 1 64 13 0 6.5 xf(x)1/2xg(x) 14 473 15 47.5 16 383 17 38.5 18 292 19 29.5 20 1102 21 410.5 22 4111 23 311.5 24 3121 25 212.5 26 2130 27 113.5 xf(x)1/2xg(x) 28 0144 29 014.5 30 4154 31 415.5 32 5163 33 516.5 34 4173 35 317.5 36 3182 37 218.5 38 2192 39 119.5 40 1201 41 020.5

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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) 0 0 00 1 024 2 4 45 3 4 64 4 5 83 5 5 102 6 4 121 7 3 144 8 3 163 9 2 182 10 2 201 11 1 224 12 1 243 13 0 262 x f(x) 2xh(x) 14 4280 15 4304 16 3325 17 3344 18 2363 19 2382 20 1401 21 442=00 22 444=24 23 346=45 24 348=64 25 250=83 26 252=102 27 154=121 x f(x) 2xh(x) 28 056=144 29 058=163 30 460=182 31 462=201 32 564=224 33 566=243 34 468=262 35 370=280 36 372=304 37 274=325 38 276=344 39 178=363 40 180=382 41 082=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) 0 0 4 5 1 05 5 2 4 6 4 3 4 7 3 4 5 8 3 5 5 9 2 6 4 10 2 7 3 11 1 8 3 12 1 9 2 13 0 10 2 144 11 1 154 12 1 163 13 0 173 x f(x) x+4j(x) 14 4182 15 4192 16 3201 17 3214 18 2224 19 2233 20 1243 21 4252 22 4262 23 3271 24 3280 25 2290 26 2304 27 1314 x f(x) x+4j(x) 28 0325 29 0335 30 4344 31 4353 32 5363 33 5372 34 4382 35 3391 36 3401 37 2410 38 242=00 39 143=10 40 144=24 41 045=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) 0 0 2 1 02 2 4 6 3 4 6 4 5 7 5 5 7 6 4 6 7 3 5 8 3 5 9 2 4 10 2 4 11 1 3 12 1 3 13 0 2 xf(x) k(x) 14 46 15 46 16 35 17 35 18 24 19 24 20 13 21 46 22 46 23 35 24 35 25 24 26 24 27 13 x f(x) k(x) 28 02 29 02 30 46 31 46 32 57 33 57 34 46 35 35 36 35 37 24 38 24 39 13 40 13 41 02

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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) 0 00 1 00 2 412 3 4 4 515 5 5 6 412 7 39 8 39 9 26 10 26 11 13 12 13 13 00 xf(x) r(x) 14 412 15 412 16 39 17 39 18 26 19 26 20 13 21 412 22 412 23 39 24 39 25 26 26 26 27 13 x f(x) 4(x) 28 00 29 00 30 412 31 412 32 515 33 515 34 412 35 39 36 39 37 26 38 26 39 13 40 13 41 00

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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) 0 00 1 00 2 4 2 3 4 2 4 5 5 5 6 4 2 7 3 8 3 9 2 1 10 2 1 11 1 12 1 13 00 xf(x) s(x) 14 42 15 42 16 3 17 3 18 21 19 21 20 1 21 42 22 42 23 3 24 3 25 21 26 21 27 1 x f(x) s(x) 28 00 29 00 30 42 31 42 32 5 33 5 34 42 35 3 36 3 37 21 38 21 39 1 40 1 41 00

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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=42 0 1 0 -1=41 0 2 4 -2=40 1 3 4 -3=39 1 4 5 -4=38 2 5 5 -5=37 2 6 4 -6=36 3 7 3 -7=35 3 8 3 -8=34 4 9 2 -9=33 5 10 2 -10=32 5 11 1 -11=31 4 12 1 -12=30 4 13 0 -13=29 0 x f(x) -xm(x) 14 4 -14=28 0 15 4 -15=27 1 16 3 -16=26 2 17 3 -17=25 2 18 2 -18=24 3 19 2 -19=23 3 20 1 -20=22 4 21 4 -21=21 4 22 4 -22=20 1 23 3 -23=19 2 24 3 -24=18 2 25 2 -25=17 3 26 2 -26=16 3 27 1 -27=15 4 x f(x) -xm(x) 28 0 -28=14 4 29 0 -29=13 0 30 4 -30=12 1 31 4 -31=11 1 32 5 -32=10 2 33 5 -33=9 2 34 4 -34=8 3 35 3 -35=7 3 36 3 -36=6 4 37 2 -37=5 5 38 2 -38=4 5 39 1 -39=3 4 40 1 -40=2 4 41 0 -41=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) 0 00 1 00 2 4-4 3 4 4 5-5 5 5 6 4-4 7 3-3 8 3 9 2-2 10 2-2 11 1 12 1 13 00 xf(x) p(x) 14 4-4 15 4-4 16 3-3 17 3-3 18 2-2 19 2-2 20 1 21 4-4 22 4-4 23 3-3 24 3-3 25 2-2 26 2-2 27 1 x f(x) p(x) 28 00 29 00 30 4-4 31 4-4 32 5-5 33 5-5 34 4-4 35 3-3 36 3-3 37 2-2 38 2-2 39 1 40 1 41 00

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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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