1. a) f (x) = ax + b

f (1) = 1 , f (3) = 7

1 = a(1) + b

1 = a + b ...... persamaan (1)

7 = a(3) + b

7 = 3a + b ......persamaan (2)

eliminasi (1) dan (2)

1 = a + b

7 = 3a + b

-------------- -

-6= -2a

masukkan ke persamaan (1)

1 = 3 + b

b = 1 - 3 = -2

1. b) persamaan ==>> f (x) = 3x -2

2. a) f (x) = x + 2

domain {0, 1, 2, 3, 4, 5}

x | 0 | 1 | 2 | 3 | 4 | 5 |

---|-----|-----|-----|-----|-----|------|

y | 2 | 3 | 4 | 5 | 6 | 7 |

2.b ) grafik fungsi => gambar di foto ya

titik potong sumbu x => y = 0

0 = x + 2

x = -2 => koordinat = (-2 , 0)

titik potong sumbu y => x = 0

y = 0 + 2 = 2 ==> koordinat = (0 , 2)