6. Suatu fungsi ditentukan dengan rumus f(x) = px + q. Jika diketahui f(-2) = 7 dan f(3) = -3 tentukan, A. Bentuk fungsinya B. Nilai f(4n) 7. Fungsi f dinyatak

Nomor 6
f(x) = px + q

f(-2) = -2p + q
7 = -2p + q

f(3) = 3p + q
-3 = 3p + q

Eliminasi
-2p + q = 7
3p + q = -3
_________ -
-5p = 10
p = -2

subtitusi
-2p + q = 7
-2(-2) + q = 7
4 + q = 7
q = 3

bagian A
f(x) = px + q
f(x) = -2x + 3

Bagian B
f(4n) = -2(4n) + 3
f(4n) = 3 - 8n

Nomor 7
f(x) = x² + ax + b

f(2) = 2² + 2a + b
5 - 4 = 2a + b
2a + b = 1

f(1) = 1² + a + b
0 - 1 = a + b
a + b = -1

Eliminasi
2a + b = 1
a + b = -1
________-
a = 2

subtitusi
a + b = -1
b = -1 - 2
b = -3

Bagian A
nilai a yaitu 2 sedangkan b yaitu -3

Bagian B
f(x) = x² + ax + b
f(x) = x² + 2x - 3

Bagian C
f(4) = 4² + 2(4) - 3
f(4) = 16 + 8 - 3
f(4) = 21

semoga berguna

6.
f(x) = px + q

f(-2) = 7 (masukkan nilai x kedalam fungsi)
7 = -2p + q

f(3) = -3
-3 = 3p + q

-Eliminasi-
-2p + q = 7
3p + q = -3
-----------------
-5p = 10
p = -2
Maka q = 3

A. Bentuk fungsinya
p = -2
q = 3
f(x) = px + q
f(x) = -2x + 3

B. Nilai f(4n)
f(x) = -2x + 3
f(4n) = -2(4n) + 3
f(4n) = -8n + 3


7.
f(x) = x² + ax + b
f(2) = 5
f(1) = 0

A. Nilai a dan b
f(2) = (2)² + 2a + b
5 = 4 + 2a + b
1 = 2a + b
f(1) = (1)² + 1a + b
0 = 1 + a + b
-1 = a + b

-Eliminasi-
1 = 2a + b
-1 = a + b
----------------
2 = a
Maka b = -3

a = 2
b = -3

B. Nilai fungsi
a = 2
b = -3
f (x) = x² + ax + b
f(x) = x² + 2x - 3

C. Nilai f(4)
f(x) = x² + 2x - 3
f(4) = 4² + 2(4) - 3
f(4) = 16 + 8 - 3
f(4) = 21.

Smoga membantu, :)