a) x-   0 | 1 | 2 | 3
   y-  24 |12 | 6 | 3

That means it contains the 4 points 

(0,24), (1,12), (2,6) and (3,3)

The polynomial of least degree which has 4 
coefficients is a third degree (cubic) polynomial:

y = ax³ + bx² + cx + d

Subnstitute each of those points in:

24 = a(0)³ + b(0)² + c(0) + d
24 = d

12 = a(1)³ + b(1)² + c(1) + d
12 = a + b + c + d

6 = a(2)³ + b(2)² + c(2) + d
6 = 8a + 4b + 2c + d

3 = a(3)³ + b(3)² + c(3) + d
3 = 27a + 9b + 3c + d

So you have the system of 4 equations 
in 4 unknowns:

24 = d
12 = a + b + c + d
6 = 8a + 4b + 2c + d 
3 = 27a + 9b + 3c + d

Can you solve that? If not post again.

a = -1/2, b = 9/2, c = -16, d = 24.

So the formula

y = ax³ + bx² + cx + d

becomes

y = x³ + x² - 16x + 24

or letting y = f(x)

f(x) = x³ + x² - 16x + 24

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This is done the exact same way:

b) x-   0 | 3 | 6 | 9
   y-   5 |10 |20 |40

That means it contains the 4 points 

(0,5), (3,10), (6,20) and (9,40)

The polynomial of least degree which has 4 
coefficients is a third degree (cubic) polynomial:

y = ax³ + bx² + cx + d

Substitute each of those points in:

5 = a(0)³ + b(0)² + c(0) + d
5 = d

10 = a(3)³ + b(3)² + c(3) + d
10 = 27a + 9b + 3c + d

20 = a(6)³ + b(6)² + c(6) + d
20 = 216a + 36b + 6c + d

40 = a(9)³ + b(9)² + c(9) + d
40 = 729a + 81b + 9c + d

So you have the system of 4 equations 
in 4 unknowns:

 5 = d
10 = 27a + 9b + 3c + d
20 = 216a + 36b + 6c + d
40 = 729a + 81b + 9c + d

Can you solve that? If not post again.

a = 5/162, b = 0, c = 25/18, d = 5.

So the formula

y = ax³ + bx² + cx + d

becomes

y = x³ + 0x² - x + 5

or letting y = f(x)

f(x) = x³ - x + 5

Edwin