SOLUTION: The function {{{y=ax^3+bx^2+cx+d}}} has a relative maximum at the point (-2,27) and a relative minimum at the point (1,0). Find the values of a, b, c and d. I end up with 4 equa

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Question 1151382: The function has a relative maximum at the point (-2,27) and a relative minimum at the point (1,0). Find the values of a, b, c and d.
I end up with 4 equations the I need to solve simultaneously. Is there a better method?

Answer by ikleyn(52921)   (Show Source): You can put this solution on YOUR website!
.

First derivative is

    y' = 3ax^2 + 2bx + c.


According to the condition, it has zeroes at x= -2 and x= 1.


Hence,   = -  = 1  and   = -2   

(according to Vieta's theorem).



From these equalities,

    b =   and  c = -6a.                (*)



Substitute these values to the equation for y.  You will get

    y =  +  -  + d.         (1)



Substitute x = -2  into (1)  to get y = 27.  It gives you

    27 = -8a + 6a + 12a + d,      or

    27 = 10a + d.                            (2)



Substitute x = 1  into (1)  to get y = 0.  It gives you

    0 = a +  - 6a  + d,      or

    0 = 2a + 3a - 12a + 2d,       or

    0 = -7a + 2d.                            (3)


Thus you have the system of 2 equations in 2 unknowns


    10a + d = 27,    (2')

    -7a + 2d = 0.    (3')


From (2'), express d = 27-10a  and substitute it into (3').  You will get

    -7a + 2*(27-10a) = 0,

    -7a + 54 - 20a   = 0

     54              = 7a + 20a

     54              = 27a

      a              = 54/27 = 2.


Thus  a = 2,  d = 27 - 10a = 27 - 20 = 7.


Finally,  from (*)  

     b =  =  = 3,   and 

     c = -6a = -6*2 = -12.


ANSWER.  a = 2;  b = 3;  c = -12;  d = 7.


Solved.



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