SOLUTION: Hi can you plz help me with this qs? I didn't know under which section to post it so im posting it here: the curve y = ax^3 + bx^2 + cx + d has a gradient of -3 at (0, 15/2) and a

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi can you plz help me with this qs? I didn't know under which section to post it so im posting it here: the curve y = ax^3 + bx^2 + cx + d has a gradient of -3 at (0, 15/2) and a      Log On


   

Question 1128653: Hi can you plz help me with this qs? I didn't know under which section to post it so im posting it here:
the curve y = ax^3 + bx^2 + cx + d has a gradient of -3 at (0, 15/2) and a turning point at (3,6). Find a,b,c,d

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


y = ax^3+bx^2+cx+d
y' = 3ax^2+2bx+c
y'' = 6ax+2b

The given information tells us...

y(0) = 15/2 --> c = 15/2 (1)
y(3) = 6 --> 27a+9b+3c+d = 6 (2)
y'(0) = -3 --> c = -3 (3)
y''(3) = 0 --> 18a+2b = 0 (4)

Substitute (1) and (3) in (2):

27a+9b-9+15/2 = 6
27a+9b = 15/2
18a+6b = 5 (5)

Subtract (4) from (5):

4b = 5
b = 5/4 (6)

Substitute (6) in (4):

18a+5/2 = 0
a = -5/36

The polynomial is

%28-5%2F36%29x%5E3%2B%285%2F4%29x%5E2-3x%2B%2815%2F2%29

A graph...