Solve for the numerical coefficients:

y = ax3 + bx2 + cx + d

if it passes through (0,3) and has a point of inflection point and horizontal tangent line at (1,2).

STEP 1: Get y’ and y”

y' = 3ax2 + 2bx + c

y” = 6ax + 2b

STEP 2: Substitute the given point s in the given equation.

Substituting (0,3)…
a(0)3 + b(0)2 + c(0) + d = 3
d = 3 ← Eq (1)

Substituting (1,2)
a(1)3 + b(1)2 + c(1) + d = 3
a + b + c + d = 2 ← Eq (2)

STEP 3: Since we have a horizontal tangent line at pt (1,2) we can use that point at y’ = 0.

Substituting (1,2) in y’ = 0
3a(1)2 + 2b(1) + c = 0
3a + 2b + c = 0 ← Eq (3)

STEP 4: Since (1,2) is a point of inflection, then y” = 0

Substituting (1,2) in y” = 0.
6a(1) + 2b = 0
6a + 2b = 0 ← Eq(4)

Solving simultaneously:

Using Eq(4), transpose 6a to the other side and divide by 2.
2b = -6a
b = -3a

Substitute -3a to the b in Eq(2) and Eq(3) and substitute d =3 from Eq(1) to Eq(2). Solve values of a and c.
a - 3a + c + 3 = 2
-2a + c = -1

3a + 2(-3a) + c = 0
3a – 6a + c= 0
-3a + c = 0

Subtracting:
-2a + c = -1
-3a + c = 0
a = -1

Solving for b and c:
b = -3a
b = -3(-1)
b = 3

-3a + c = 0
c = 3a
c = 3(-1)
c = -3

Therefore we the equation:

y = -x3 + 3x2 - 3x + 3


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