Question

Tolong bantu isiin nomor 9,10 butuh cepat..

Answer 1

Verified answer

Jawab

y = x^3  -  3x +4
m = y' = 3x^2  - 3
m sejajar  garis y = 9x - 8 --> m = 9
3x^2 - 3 = 9
3x^2 = 12
x^2 = 4
x = 2  atau x = - 2

titik singgung 
x = 2 --> y = x^3 - 3x + 4 --> y = 8-6+4 = 6 --> A(2,6)
x = - 2 --> y = (-2)^3 - 3(-2) + 4 --> y= -8 +6+4= 2  --> B(-2,2)

pers garis singgung 
(i) m = 9  di A(2,6)
y - 6 = 9(x  - 2) --> y = 9x - 18 +6 --> y = 9x  -12

(ii) m = 9 , di B(-2,2)
y - 2 = 9(x + 2)--> y = 9x +18 +2 --> y = 9x + 20


10)
y = x^3 - 5x^2  + 5x + 3
m = y' = 3x^2 -10x  + 5 
garis y = -1/2 x  --> m1 = -1/2
 m tegaklurus --> m2 = 2

3x^2 -10x + 5 = 2
3x^2 -10x + 3 = 0
(3x - 1)(x - 3) = 0
x= 1/3  atau x = 3

y= x^3 - 5x^2 + 5x + 3
x= 1/3 --> y = (1/3)^3 -5(1/3)^2 + 5(1/3) + 3 = 112/27
pers garis singgung  y - 112/27 = 2(x - 1/3) 
y  = 2x - 2/3 + 112/27
27y = 54 x - 18 + 112
27y= 54 x + 94

x= 3 --> y = 3^3 - 5. 3^2 + 5. 3 + 3  = 27 - 45 +15 + 3  =0
y = 9(x  - 3) --> y= 9x  + 27