we know that y passes through (0,4) and (1,20)

y = ax^3 + bx^2 + cx + d

=> 4 = a(0)^3 + b(0)^2 + c(0) + d

=> d = 4

=> 20 = a(1)^3 + b(1)^2 + c(1) + d

=> 20 = a + b + c + 4

=> 16 = a + b + c

that is, a + b + c = 16

call this equation (1)

so a + b + c = 16 .....................(1)

we also know that the derivatve gives the slope at any value of x right? so let's start there

y = ax^3 + bx^2 + cx + d

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

we have extrema occuring at the points where y' = 0

set y' = 0

=> 3ax^2 + 2bx + c = 0

we know that x = -1 and x = -3, so we have the simultaneous equations:

3a(-1)^2 + 2b(-1) + c = 0 ....................(2)

3a(-3)^2 + 2b(-3) + c = 0 ....................(3)

3a - 2b + c = 0 ....................(2)

27a - 6b + c = 0 ..................(3)

so we know d, to find a,b, and c we must solve the system:

a + b + c = 16 .....................(1)

3a - 2b + c = 0 ....................(2)

27a - 6b + c = 0 ..................(3)

=> 2a - 3b = -16 ...................(4) = (2) - (1)

=> 24a - 4b = 0 .....................(5) = (3) - (2)

=> 24a - 36b = -192 ...............(4)*12

=> 24a - 4b = 0 .....................(5)

=> 32b = 192 .........................(5) - (4)

=> b = 6

but 2a - 3b = -16

=> 2a - 18 = -16

=> 2a = 2

=> a = 1

but a + b + c = 16

=> 1 + 6 + c = 16

=> c = 16 - 7 = 9

so we have a = 1, b = 6, c = 9, d = 4

so the curve is y = x^3 + 6x^2 + 9x + 4