Question

f is a polynomial of degree 3. f has a root of multiplicity 2 at u=4, f(−4)=0, and f(0)=−28.8. Find an algebraic equation for f. Round all answers to 3 decimal place as/if needed

Answer 1

Answer:

$\displaystyle f(u) = -0.45(u-4)^2 (u+4)$

Step-by-step explanation:

We are given that function f is a polynomial of degree 3. It has a root of multiplicity 2 at u = 4, f(-4) = 0, and f(0) = -28.8. We want to determine an algebraic equation for f.

First, since it has a root of multiplicity 2 at u = 4, one factor is:

$\displaystyle (u - (4)) ^2 = (u-4)^2$

Furthermore, since f(-4) = 0, f has another root at u = -4. Since the degree of f is 3, the multiplicity of this root must be 1. Hence, the other factor is:

$\displaystyle (u-(-4))^1 = (u + 4) ^1 = (u+4)$

Hence, our function is:

$\displaystyle f(u) = a(u-4)^2(u+4)$

Since f(0) = -28.8, f = -28.8 when u = 0. Find a:

$\displaystyle \begin{aligned}(-28.8) & = a((0) - 4)^2 ((0) + 4) \\ \\ -28.8 & = a(-4)^2(4) \\ \\ -28.8 & = 64a \\ \\ a & = -0.45 \end{aligned}$

In conclusion, our function will be:

$\displaystyle f(u) = -0.45(u-4)^2 (u+4)$