Question

Minta bantuannya no 9 dan 10. terima kasih

Answer 1

Anggap y = f(x)

f(x) = ax³ + bx² + c
f'(x) = 3ax² + 2bx

f(-1) = 0
a(-1)³ + b(-1)² + c = 0
-a + b + c = 0

f(0) = 5
a(0)³ + b(0)² + c = 5
c = 5

-a + b + c = 0
-a + b + 5 = 0
b = a - 5

karena sejajar sumbu x, m = 0

f'(x) = 3ax² + 2bx
f'(1) = 0
3a(1)² + 2b(1) = 0
3a + 2b = 0
3a + 2(a - 5) = 0
3a + 2a - 10 = 0
5a = 10
a = 10/5
a = 2

b = a - 5
b = 2 - 5
b = -3

a + b + c = 2 - 3 + 5
= 4
= C.

==========

y = ∛(5 + x)
y = ∛(5 + 3)
y = ∛8
y = 2

kurva melalui (3, 2)

y = ∛(5 + x)
y = (5 + x)^(1/3)
y' = 1/3(5 + x)^(-2/3)
y' = 1/(3(5 + x)^(2/3))

1/(3(5 + 3)^(2/3) = 1/(3(8)^(2/3))
= 1/(3(4))
= 1/12

m = 1/12

y - b = m(x - a)
y - 2 = 1/12(x - 3)
12y - 24 = (x - 3)
12y = x - 3 + 24
12y = x + 21
x - 12y + 21 = 0
= A.

Answer 2

Nomor 10: Terdapat pada lampiran.

Nomor 9.
Kurva y = ax³ + bx² + c
Melalui (-1,0) dan (0,5)

- Melalui (0,5)
5 = a(0)³ + b(0)² + c
c = 5

- Melalui (-1,0)
0 = a(-1)³ + b(-1)² + c
0 = -a + b + 5
Maka, a - b = 5

Garis singgung dengan absis x = 1 sejajar sumbu x (gradien = 0)
Dengan dy/dx = 3ax² + 2bx
Dengan gradien untuk x = 1 pada dy/dx = 0
0 = 3a(1)² + 2b(1)
Maka, 3a + 2b = 0

Eliminasikan a - b = 5, dan 3a + 2b = 0
Yang akan didapat a = 2, dan b = -3, serta c = 5.

Maka:
a + b + c = 2 + (-3) + 5 = 4