# calculus

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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?

• calculus -

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 a-b-c=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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