Let's tackle the task step-by-step based on the initial steps you provided for Hill 1 and Hill 2.
### Hill 1:
You want a polynomial with three peaks, and you have chosen three points on the [tex]\( x \)[/tex]-axis (the zeros), which are [tex]\( x = 1 \)[/tex], [tex]\( x = 3 \)[/tex], and [tex]\( x = 4 \)[/tex]. Representing each zero as a factor, you built the polynomial:
[tex]\[ F(x) = (x - 1)(x - 3)(x - 4) \][/tex]
Now, let's expand this polynomial:
1. First, expand [tex]\( (x - 1)(x - 3) \)[/tex]:
[tex]\[ (x - 1)(x - 3) = x^2 - 3x - x + 3 = x^2 - 4x + 3 \][/tex]
2. Next, expand this result with [tex]\( (x - 4) \)[/tex]:
[tex]\[(x^2 - 4x + 3)(x - 4)\][/tex]
Distribute each term in [tex]\( x^2 - 4x + 3 \)[/tex] by [tex]\( x - 4 \)[/tex]:
- [tex]\( x^2(x - 4) = x^3 - 4x^2 \)[/tex]
- [tex]\( -4x(x - 4) = -4x^2 + 16x \)[/tex]
- [tex]\( 3(x - 4) = 3x - 12 \)[/tex]
Combine these results:
[tex]\[
x^3 - 4x^2 - 4x^2 + 16x + 3x - 12 = x^3 - 8x^2 + 19x - 12
\][/tex]
### Hill 2:
Your second layer should be a transformation of the first polynomial, [tex]\( F(x) \)[/tex]. You indicated that this transformation involves multiplying the polynomial by [tex]\(-12\)[/tex] or by some combination involving [tex]\(-12\)[/tex].
We can suppose you meant multiplying [tex]\( F(x) \)[/tex] by [tex]\(-12 \)[/tex]:
[tex]\[ G(x) = -12 \cdot (x^3 - 8x^2 + 19x - 12) \][/tex]
Now, let's distribute [tex]\(-12\)[/tex] throughout the polynomial:
[tex]\[ G(x) = -12x^3 + 96x^2 - 228x + 144 \][/tex]
### Summary:
- Hill 1: The polynomial [tex]\( F(x) = (x-1)(x-3)(x-4) \)[/tex] gives [tex]\( F(x) = x^3 - 8x^2 + 19x - 12 \)[/tex].
- Hill 2: Transforming [tex]\( F(x) \)[/tex] by multiplying it by [tex]\(-12\)[/tex] results in [tex]\( G(x) = -12x^3 + 96x^2 - 228x + 144 \)[/tex].
These are the detailed steps to create and transform the polynomials for Hill 1 and Hill 2.
