calculus
posted by Anonymous .
Find the cubic function of the form y=ax^3 + bx^2 + cx + d which has a relative maximum point at (0,2) and a point of inflection at (1,2). How do I even start this one?

First of all, both given points are on the curve, so
for
(0,2) > 2 = 0+0+0+d
so we know d=2
(1,2) > 2 = a(1) + b(1) + c(1) + d
a + b + c + 2 = 2
a  b  c = 4
y' = 3ax^2 + 2bx + c
since (0,2) is a max, we know y' = 0 when x = 0
0 = 3a(0) + 2b(0) + c
c = 0
y'' = 6ax + 2b
we know y'' = 0 when x = 1
0 = 6a + 2b
2b = 6a
b = 3a
so in abc=4
a  3a  0 = 4
2a = 4
a = 2 and b=6 , c=0 , d=2
then the function is
y = 2x^3  6x^2 + 2
check my arithmetic.
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