Question

Find a polynomial of the form f(x) = ax^3 + bx^2 + cx + d such that f(0) = −6, f(1) = 4, f(3) = 10, and f(4) = 0.

Answer 1

Answer:

$f(x) = -\frac{x^{3}}{2} - \frac{x^{2}}{3} + \frac{65x}{6} - 6$

Step-by-step explanation:

We use a system of equations to solve this question.

f(0) = −6

This means that when $x = 0, f(x) = -6$. So

$f(x) = ax^{3} + bx^{2} + cx + d$

$-6 = a*0^{3} + b*0^{2} + c*0 + d$

$d = -6$

So

$f(x) = ax^{3} + bx^{2} + cx - 6$

f(1) = 4

This means that when $x = 1, f(x) = 4$

So

$f(x) = ax^{3} + bx^{2} + cx - 6$

$4 = a*1^{3} + b*1^{2} + c*1 - 6$

$a + b + c = 10$

f(3) = 10

This means that when $x = 3, f(x) = 10$

So

$f(x) = ax^{3} + bx^{2} + cx - 6$

$10 = a*3^{3} + b*3^{2} + c*3 - 6$

$27a + 9b + 3c = 16$

f(4) = 0

This means that when $x = 4, f(x) = 0$

So

$f(x) = ax^{3} + bx^{2} + cx - 6$

$0 = a*4^{3} + b*4^{2} + c*4 - 6$

$64a + 16b + 4c = 6$

So we have to solve the following system of equations:

a + b + c = 10

27a + 9b + 3c = 16

64a + 16b + 4c = 6

From the first equation:

a = 10 - b - c

Replacing on the second, and on the third:

$27(10 - b - c) + 9b + 3c = 16$

$270 - 27b - 27c + 9b + 3c = 16$

$18b + 24c = 254$

And

$64(10 - b - c) + 16b + 4c = 6$

$640 - 64b - 64c + 16b + 4c = 6$

$48b + 60c = 634$

So

18b + 24c = 254

48b + 60c = 634

Multiplying the first one by 5, and the second one by -2, to eliminate c

90b + 120c = 1270

-96b - 120c = 1268

$-6b = 2$

$6b = -2$

$b = -\frac{1}{3}$

Then

$18b + 24c = 254$

$24c = 254 - 18b$

$24c = 254 - 18\frac{-1}{3}$

$24c = 260$

$c = \frac{65}{6}$

And

$a = 10 - b - c = 10 -(-\frac{1}{3}) - \frac{65}{6} = -\frac{1}{2}$

So

$f(x) = -\frac{x^{3}}{2} - \frac{x^{2}}{3} + \frac{65x}{6} - 6$