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Fungsi, Persamaan Fungsi Linear dan Fungsi Kuadrat

FUNGSI. Fungsi, Persamaan Fungsi Linear dan Fungsi Kuadrat. FUNCTION. Functions, Linear Function Equation and Quadratic Function. RELASI. Ada 3 cara dalam menyatakan suatu relasi : Diagram panah Himpunan pasangan berurutan Diagram Cartesius Contoh:

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Fungsi, Persamaan Fungsi Linear dan Fungsi Kuadrat

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  1. FUNGSI Fungsi, Persamaan Fungsi Linear dan Fungsi Kuadrat

  2. FUNCTION Functions, Linear Function Equation and Quadratic Function

  3. RELASI Ada 3 cara dalam menyatakan suatu relasi : • Diagram panah • Himpunan pasangan berurutan • Diagram Cartesius Contoh: Diketahui himpunan A = {1,2,3,4,5} dan himpunan B = {becak, mobil, sepeda, motor,bemo}. Relasi yang menghubungkan himpunan A ke himpunan B adalah “banyak roda dari”. Tunjukkan relasi tersebut dengan: • Diagram panah • Himpunan pasangan berurutan • Diagram Cartesius FUNGSI

  4. RELATION There are 3 ways to state a relation : • Arrow Diagram • Set of Ordered pairs • Cartesian Diagram Example: Given a set of A = {1,2,3,4,5} and set B = {pedicab, car, bike cycle, motor cycle, bemo}. The relation that relate set of A to B is “the quantity of the wheel”. Show those relations with: • Arrow Diagram • Set of Ordered pairs • Cartesian Diagram FUNGSI

  5. RELASI c. DiagramCartesius Jawab: a. Diagram panah Y “banyak roda dari” 1. becak . becak • 2. • mobil . mobil 3. motor • . motor 4. sepeda . sepeda • 5. . bemo bemo • X O 1 2 3 4 A B b. Himpunan pasangan berurutan = {(2, sepeda), (2, motor), (3, becak) (3, bemo), (4, mobil )} FUNGSI

  6. RELATION c. Cartesian Diagram Answer: a. Arrow Diagram Y “the quantity of the wheel” 1. pedicab . pedicab • 2. • car . car 3. motor • . Motor 4. Bike cycle . Bike cycle • 5. . bemo bemo • X O 1 2 3 4 A B b. Set of ordered pairs = {(2,bike cycle), (2, motor), (3, pedicab) (3, bemo), (4, car )} FUNGSI

  7. Pengertian Fungsi : FUNGSI Suatu fungsi f dari himpunan A ke himpunan B adalah suatu relasi yang memasangkan setiap elemen dari A secara tunggal , dengan elemen pada B . . . . . . . . . . . A B f FUNGSI

  8. The definition of Function: FUNCTION A function f of set A to set B is a relation that match every element of A as a single to element B . . . . . . . . . . . A B f FUNGSI

  9. Beberapa cara penyajian fungsi : Dengan diagram panah f : D  K. Lambang fungsi tidak harus f.Misalnya, un = n2 + 2n atau u(n) = n2 + 2n Dengan diagram Kartesius Himpunan pasangan berurutan Dalam bentuk tabel FUNGSI FUNGSI

  10. There are few ways to state function: With arrow diagram f : D  K. The symbol of function not always f. Example, un = n2 + 2n or u(n) = n2 + 2n With Cartesian diagram The set of ordered pairs In table FUNCTION FUNGSI

  11. Contoh : grafik fungsi 4 disebut bayangan (peta) dari 2 dan juga dari –2. – 2 dan 2 disebut prapeta dari 4, dan dilambangkan f–1(4) = 2 atau – 2. Grafik Kartesius merupakan grafik fungsi y=f(x) hanya apabila setiap garis sejajar sumbu- Y yang memotong grafik hanya memotong di tepat satu titik saja. FUNGSI Gambarlah grafiksebuah fungsi: f: x  f(x) = x2 dengan Df = {–2, –1, 0, 1, 2}, Rf = {0, 1, 4}. Y (–2,4) (2,4) (–1,1) (1,1) X (0,0) O FUNGSI

  12. Example :function graph 4 is also called the shadow (map) of 2 and also from –2. – 2 and 2 is called pre map of 4 and symbolized by f–1(4) = 2 or – 2. Cartesian graph is a function graph of y=f(x) only if every line is parallel with Y-axis that intersecting the graph in one point only. FUNCTION Draw a graph of a function: f: x  f(x) = x2 With Df = {–2, –1, 0, 1, 2}, Rf = {0, 1, 4}. Y (–2,4) (2,4) (–1,1) (1,1) X (0,0) O FUNGSI

  13. Beberapa Fungsi Khusus 1). Fungsi Konstan 2). Fungsi Identitas 3). Fungsi Modulus 4). Fungsi Genap dan Fungsi Ganjil Fungsi genap jika f(x) = f(x), dan Fungsi ganjil jika f(x) = f(x) 5).Fungsi Tangga dan Fungsi Nilai Bulat Terbesar [[ x ] = {b | b  x < b + 1, b bilangan bulat, xR} Misal, jika 2  x < 1 maka [[x] = 2 6).Fungsi Linear 7). Fungsi Kuadrat 8). Fungsi Turunan FUNGSI FUNGSI

  14. Special Functions 1). Constant Function 2). Identity Function 3). Modulus Function 4). Even and Odd Function Even function if f(x) = f(x), and Odd function if f(x) = f(x) 5).Ladder Function and The Biggest Integer Value Function [[ x ] = {b | b  x < b + 1, b integer number, xR} example,if2  x < 1 then [[x] = 2 6).Linear Function 7). Quadratic Function 8). Differential Function FUNCTION FUNGSI

  15. Jenis Fungsi 1. Injektif ( Satu-satu)Fungsi f:AB adalah fungsi injektif apabila setiap dua elemen yang berlainan di A akan dipetakan pada dua elemen yang berbeda di B. Misalnya Fungsi f(x) = 2x adalah fungsi satu-satu dan f(x) = x2bukan suatu fungsi satu-satu sebab f(-2) = f(2). FUNGSI 2. Surjektif (Onto)Fungsi f: AB maka apabila f(A)  B dikenal fungsi into. Jika f(A) = B maka f adalah suatu fungsi surjektif. Fungsi f(x) = x2 bukan fungsi yang onto 3. Bijektif (Korespondensi Satu-satu)Apabila f: A B merupakan fungsi injektif dan surjektif maka “f adalah fungsi yang bijektif” FUNGSI

  16. Kinds of Function 1. Injective ( one by one)Function f:AB is an injective function if every two different elements in A will be mapped into two different element in B. Example: Function f(x) = 2x is one by one function and f(x) = x2 is not one by one function because f(-2) = f(2). FUNGSI 2. Surjective (Onto)Functioni f: AB then iff(A)  B it is known as into function. If f(A) = B then f is a surjective function. Function f(x) = x2 it is not onto function 3. Bijective (one by one correspondence)If f: A B is injective and surjective function then “f is bijective function” FUNGSI

  17. FUNGSI LINEAR 1.Bentuk Umum Fungsi Linear Fungsi ini memetakan setiap x R kesuatu bentuk ax + b dengan a ≠ 0, a dan b konstanta. Grafiknya berbentuk garis lurus yang disebut grafik fungsi linear dengan Persamaan y = mx + c, m disebut gradien dan c konstanta 2. Grafik Fungsi Linear Cara menggambar grafik fungsi linear ada 2 : 1. Dengan tabel 2. Dengan menentukan titik- titik potong dengan sumbu x dan sumbu y FUNGSI

  18. LINEAR FUNCTION 1.General Form of Linear Function This function map every x R into the form of ax + b with a ≠ 0, a and b Constanta. The graph is in straight line which is called linear function graph with the equation of y = mx + c, m is called gradient and c is Constanta 2. Linear Function Graph there are two ways to draw linear function graph: 1. by table 2. by determining the intersection points with x-axis and y-axis FUNGSI

  19. FUNGSI LINEAR Contoh : Suatu fungsi linear ditentukan oleh y = 4x – 2 dengan daerah asal • Buat tabel titik-titik yangmemenuhi persamaan diatas . • Gambarlah titik-titik tersebut dalam diagram Cartesius. • Tentukan titik potong grafik dengan sumbu X dan sumbu Y. {x \-1 x 2, x R}. Jawab a. Ambil sembarang titik pada domain X -1 0 1 2 Y = 4x-2 -6 -2 2 6 Jadi, grafik fungsi melalui titik-titik (-1,-6), (0,-2), (1,2), (2,6) FUNGSI

  20. LINEAR FUNCTION Example : A linear function is determine by y = 4x – 2 with the domain • Make a table of points that fulfill the equation above. • Draw the points in Cartesians diagram • Determine the intersection point of the graph with X-axis and Y-axis. {x \-1 x 2, x R}. Answer a. Take any points in the domain X -1 0 1 2 Y = 4x-2 -6 -2 2 6 Then, the function graph through these points (-1,-6), (0,-2), (1,2), (2,6) FUNGSI

  21. FUNGSI LINEAR Y c. Titik potong dengan sumbu x ( y= 0 ) y = 4x – 2 0 = 4x - 2 2 = 4x x = b. 6 • 2 • Jadi titik potong dengan sumbu X adalah ( ½,0) X 1 2 O Titik potong dengan sumbu Y ( x = 0 ) y = 4x – 2 y = 4(0) – 2 y = -2 Jadi titik potong dengan sumbu Y adalah (0,-2) -2 -1 • -2 • -6 FUNGSI

  22. LINEAR FUNCTION Y c. Intersection points with x-axis ( y= 0 ) y = 4x – 2 0 = 4x - 2 2 = 4x x = b. 6 • 2 • Then, the intersection points with x-axis is ( ½,0) X 1 2 O Intersection points with y-axis ( x = 0 ) y = 4x – 2 y = 4(0) – 2 y = -2 Intersection points with y-axis is (0,-2) -2 -1 • -2 • -6 FUNGSI

  23. FUNGSI LINEAR 3. Gradien Persamaan Garis Lurus Cara menentukan gradien : (i). Persamaan bentuk y = mx+c, gradiennya adalah m. (ii). Persamaan bentuk ax+by+c=0 atau ax+by=-c adalah m= (iii). Persamaan garis lurus melalui dua titik (x1,y1) dan (x2,y2), gradiennya adalah m = • Contoh : • Tentukan gradien persamaan garis berikut • a. y = 3x – 4 • b. 2x – 5y = 7 • 2. Tentukan gradien garis yang melalui pasangan titik (-2,3) dan (1,6) FUNGSI

  24. LINEAR FUNCTION 3. The Gradient of Straight Line Equation How to determine gradient: (i). The equation of y = mx+c, the gradient is m. (ii). The equation of ax+by+c=0 or ax+by=-c is m= (iii). Straight line equation through two points (x1,y1) and (x2,y2), the gradient is m = • Example : • Define the gradient of the line equation below: • a. y = 3x – 4 • b. 2x – 5y = 7 • 2. Define the gradient of the line which through the points pairs of (-2,3) and (1,6) FUNGSI

  25. FUNGSI LINEAR Jawab : 1a. Y = 3x – 4 gradien = m = 3 b. 2x - 5y = 7, a = 2 dan b = - 5 m = = - 2. m = = = = 1 FUNGSI

  26. LINEAR FUNCTION answer : 1a. Y = 3x – 4 gradient = m = 3 b. 2x - 5y = 7, a = 2 and b = - 5 m = = - 2. m = = = = 1 FUNGSI

  27. FUNGSI LINEAR 4. Menentukan Persamaan Garis Lurus • Persamaan garis melalui sebuah titik (x1,y1) dan gradien m adalah y – y1 = m ( x – x1 ) • Persamaan garis melalui dua titik (x1,y1) dan (x2,y2) adalah = Contoh 1 : Tentukan persamaan garis yang melalui titik ( -2, 1 ) dan gradien -2 Jawab : y – y1 = m ( x – x1 ) y – 1 = -2 ( x – (-2)) y - 1 = -2x – 4 y = -2x - 3 FUNGSI

  28. LINEAR FUNCTION 4. Determine the straight line equation • Line equation through a point (x1,y1) and gradient m is y – y1 = m ( x – x1 ) • Line equation through two points (x1,y1) and (x2,y2) is = Example 1 : Define the line equation that through point ( -2, 1 ) and gradient -2 Answer : y – y1 = m ( x – x1 ) y – 1 = -2 ( x – (-2)) y - 1 = -2x – 4 y = -2x - 3 FUNGSI

  29. FUNGSI LINEAR Contoh2 : Tentukan persamaan garis yang melalui titik P(-2, 3) dan Q(1,4) Jawab : = = = 3(y – 3) = 1(x + 2) 3y – 9 = x + 2 3y - x – 11 = 0 FUNGSI

  30. LINEAR FUNCTION Example2 : Determine the line equation that through point P(-2, 3) and Q(1,4) Answer : = = = 3(y – 3) = 1(x + 2) 3y – 9 = x + 2 3y - x – 11 = 0 FUNGSI

  31. FUNGSI LINEAR 5. Kedudukan dua garis lurus • Dua garis saling berpotongan jika m1 ≠ m2 • Dua garis saling sejajar jika m1 = m2 • Dua garis saling tegak lurus jika m1. m2 = -1 atau m1 = - • Contoh : • Tentukan persamaan garis lurus yang melalui titik (2,-3) dan sejajar dengan garis x – 2y + 3 = 0 • Tentukan persamaan garis lurus yang melalui titik (-3,5) dan tegak lurus pada 6x – 3y – 10 = 0 FUNGSI

  32. LINEAR FUNCTION 5. The Position of Two Straight line • Two lines are intersecting if m1 ≠ m2 • Two lines are parallel if m1 = m2 • Two lines are perpendicular if m1. m2 = -1 or m1 = - • Example : • Determine the straight line equation that through point (2,-3) and parallel with line x – 2y + 3 = 0 • Determine the straight line equation that through point (-3,5) and perpendicular to 6x – 3y – 10 = 0 FUNGSI

  33. FUNGSI LINEAR Jawab : 1. Diketahui persamaan garis x – 2y + 3 = 0 maka Persamaan garis melalui titik (2,-3) dan gradien adalah y – y1 = m ( x – x1) y + 3 = ½ ( x – 2 ) y + 3 = ½ x – 1 2y + 6 = x – 2 x – 2y – 8 = 0 Jadi persamaan garis lurus yang sejajar dengan garis x – 2y + 3 = 0 dan melalui titik (2,-3) adalah x – 2y – 8 = 0 FUNGSI

  34. LINEAR FUNCTION Answer : 1. Known the line equation x – 2y + 3 = 0 then The line equation through point (2,-3) and gradient is y – y1 = m ( x – x1) y + 3 = ½ ( x – 2 ) y + 3 = ½ x – 1 2y + 6 = x – 2 x – 2y – 8 = 0 Then the straight line equation that parallel with line x – 2y + 3 = 0 and through point (2,-3) is x – 2y – 8 = 0 FUNGSI

  35. FUNGSI LINEAR 2. Diketahui persamaan garis 6x – 3y – 10 = 0. Persamaan garis lurus yang dicari melalui titik (-3,5) dan bergradien -½, maka persamaannya adalah y – y1 = m(x – x1) y – 5 = -½ (x + 3) y – 5 = -½x - 2y – 10 = -x – 3 x + 2y – 10 + 3 = 0 x + 2y – 7 = 0 Jadi, persamaan garis lurus yang melalui titik (-3,5) dan tegak lurus garis 6x – 3y – 10 = 0 adalah x + 2y – 7 = 0. FUNGSI

  36. LINEAR FUNCTION 2. Known the line equation 6x – 3y – 10 = 0. the straight line equation that is found through point (-3,5) and has gradient -½, then the equation is y – y1 = m(x – x1) y – 5 = -½ (x + 3) y – 5 = -½x - 2y – 10 = -x – 3 x + 2y – 10 + 3 = 0 x + 2y – 7 = 0 so, the straight line equation that through point (-3,5) and perpendicular to line 6x – 3y – 10 = 0 is x + 2y – 7 = 0. FUNGSI

  37. FUNGSI KUADRAT 1.Bentuk umum fungsi kuadrat y = f(x) ax2+bx+c dengan a,b, c  R dan a  0 Grafik fungsi kuadrat berbentuk parabola simetris 2. Sifat-sifat Grafik Fungsi Kuadrat Berdasarkan nilai a (i) Jika a > 0 (positif), maka grafik terbuka ke atas. Fungsi kuadrat memiliki nilai ekstrim minimum, dinotasikan ymin atau titik balik minimum. (ii) Jika a < 0 (negatif), maka grafik terbuka ke bawah. Fungsi kuadrat memiliki nilai ekstrim maksimum, dinotasikan ymaks atau titik balik maksimum. FUNGSI

  38. QUADRATIC FUNCTION 1. The General Form of Quadratic Function y = f(x) ax2+bx+c with a,b, c  R and a  0The Graph of Quadratic Function is in the form of symmetrical parabola 2. The properties of quadratic function Graph Based on value a (i) If a > 0 (positive), then the graph will be up side. The quadratic function has extreme minimum value. It is denoted by ymin or minimum turning point (ii) if a < 0 (negative), then the graph will up side down. The quadratic function has extreme maximum value. It is denoted by ymaks or maximum turning point. FUNGSI

  39. FUNGSI KUADRAT Berdasarkan NilaiDiskriminan (D) Nilai diskriminan suatu persamaan kuadrat adalah D = b2 – 4ac Hubungan antara D dengan titik potong grafik dengan sumbu X • Jika D > 0 maka grafik memotong sumbu X di dua titik yang berbeda. • Jika D = 0 maka grafik menyinggung sumbu X di sebuah titik. • Jika D < 0 maka grafik tidak memotong dan tidak menyinggung sumbu X. FUNGSI

  40. QUADRATIC FUNCTION Based on discriminant value (D) Discriminantvalue of a quadratic equation is D = b2 – 4ac The relation between D and intersection point of a graph with X-axis • If D > 0 then the graph will intersects x-axis in two different points. • If D = 0 then the graph will on the x-axis in a point. • If D < 0 then the graph will not intersect and not on the x-axis. FUNGSI

  41. Kedudukan Grafik Fungsi Kuadrat Terhadap Sumbu X (ii) (iii) X X X X (v) (vi) (iv) X (i) FUNGSI KUADRAT a > 0 D = 0 a > 0 D < 0 a > 0 D > 0 X a < 0 D = 0 a < 0 D > 0 a < 0 D < 0 FUNGSI

  42. The Position of Quadratic Function Graph Towards x-axis (ii) (iii) X X X X (v) (vi) (iv) X (i) QUADRATIC FUNCTION a > 0 D = 0 a > 0 D < 0 a > 0 D > 0 X a < 0 D = 0 a < 0 D > 0 a < 0 D < 0 FUNGSI

  43. 3. Menggambar Grafik Fungsi Kuadrat Langkah-langkah menggambar grafik fungsi kuadrat : (i) Menentukan titik potong dengan sumbu X (y = 0) (ii) Menentukan titik potong dengan sumbu Y (x = 0) (iii) Menentukan sumbu simentri dan koordinat titik balik Persamaan sumbu simetri adalah x = Koordinat titik puncak / titik balik adalah (iv) Menentukan beberapa titik bantu lainnya (jika di perlukan) FUNGSI KUADRAT FUNGSI

  44. 3. Drawing Quadratic Function Graph The steps to draw quadratic function graph : (i) Define the intersection point with x-axis (y = 0) (ii) Define the intersection point with y-axis Y (x = 0) (iii) Define symmetrical axis and turning coordinate The symmetrical axis equation is x = Vertex /turning coordinate is (iv) Define other points if necessary QUADRATIC FUNCTION FUNGSI

  45. FUNGSI KUADRAT Contoh : Gambarlah grafik fungsi kuadrat y = x2 – 4x – 5. Jawab : (i) Titik potong dengan sumbu X (y = 0) x2 – 4x – 5 = 0 (x + 1)(x – 5) = 0 x = -1 atau x = 5 Jadi, titik potong grafik dengan sumbu X adalah titik (-1, 0) dan (5, 0). • Titik potong dengan sumbu Y (x = 0) • y = 02 – 4(0) – 5 • y = -5 • Jadi titik potong dengan sumbu Y adalah titik ( 0, -5 ) FUNGSI

  46. QUADRATIC FUNCTION Example : Draw a graph of quadratic function y = x2 – 4x – 5. Jawab : (i) The intersection point with X-axis (y = 0) x2 – 4x – 5 = 0 (x + 1)(x – 5) = 0 x = -1 or x = 5 So, the intersection point with x-axis is (-1, 0) and (5, 0). • The intersection points with axis Y (x = 0) • y = 02 – 4(0) – 5 • y = -5 • So, the intersection points with Y-axis is ( 0, -5 ) FUNGSI

  47. FUNGSI KUADRAT (iii) Sumbu simetri dan koordinat titik balik Jadi, sumbu simetrinya x = 2 dan koordinat titik baliknya (2, -9). (iv) Menentukan beberapa titik bantu. Misal untuk x = 1, maka y = -8. Jadi, titik bantunya (1, -8). FUNGSI

  48. QUADRATIC FUNCTION (iii) Symmetrical axis and turning coordinate So, the symmetrical axis is x = 2 and the turning coordinate is (2, -9). (iv) Determine the helping points. For example, for x = 1, then y = -8. Then, the helping point is (1, -8). FUNGSI

  49. FUNGSI KUADRAT Grafiknya : Y • • X -1 0 1 2 3 4 5 -1 -2 -3 -4 -5 -6 -7 -8 -9 • • • • • FUNGSI

  50. QUADRATIC FUNCTION THE GRAPH : Y • • X -1 0 1 2 3 4 5 -1 -2 -3 -4 -5 -6 -7 -8 -9 • • • • • FUNGSI

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