Find the cubic equation: ch5 and 6 math problem for which f(-1)=3, f(1)=1, f(2)=6, and f(3)=7.

Question

Find the cubic equation:

 

for which f(-1)=3, f(1)=1, f(2)=6, and f(3)=7.

Answer 1

Since f(1) = 1, a + b + c + d = 1

Since f(-1) = 3, -a + b - c + d = 3

Since f(2) = 6, 8a + 4b + 2c + d = 6

Since f(3) = 7, 27a + 9b + 3c + d = 7

Subtract the second equation from the first, the third equation from the first, and the fourth equation from the first:

2a + 2c = -2

-7a - 3b -c = -5

-26a - 8b - 2c = -6

From the first equation, a + c = -1. So, c = -1 - a

Plugging into the other 2 equations, we get:

-7a - 3b - (-1-a) = -5

-26a - 8b - 2(-1 - a) = -6

Simplify:

-6a - 3b = -6

-24a - 8b = -8

Multiply the first equation by -4:

24a + 12b = 24

-24a - 8b = -8

Add the equations to get 4b = 16. So, b = 4

Since -6a - 3b = -6 and b = 4, -6a - 12 = -6

-6a = 6. So, a = -1.

c = -1 - a. So, c = 0

a + b + c + d = 1, so -1 + 4 + 0 + c + d = 1

d = -2

f(x) = -x³ + 4x² - 2