Question

1. find a formula for a cubic function $f$ if $f(1) = 6$ and $f(-1) = f(0) = f(2) = 0$.

Explanation:

Step1: Use roots to set form

Since $f(-1)=f(0)=f(2)=0$, the cubic function has roots $x=-1, 0, 2$. So we write:
$$f(x)=ax(x+1)(x-2)$$
where $a$ is a non-zero constant.

Step2: Solve for $a$ using $f(1)=6$

Substitute $x=1$ and $f(1)=6$ into the function:
$$6=a(1)(1+1)(1-2)$$
Simplify the right-hand side:
$$6=a(1)(2)(-1)$$
$$6=-2a$$
Solve for $a$:
$$a=\frac{6}{-2}=-3$$

Step3: Expand to get final formula

Substitute $a=-3$ back into the factored form and expand:
$$f(x)=-3x(x+1)(x-2)$$
First multiply $(x+1)(x-2)=x^2-2x+x-2=x^2-x-2$, then:
$$f(x)=-3x(x^2-x-2)=-3x^3+3x^2+6x$$

Answer:

$f(x)=-3x^3+3x^2+6x$ (or factored form $f(x)=-3x(x+1)(x-2)$)