Part (a): Determine the Polynomial Function

To find the polynomial function of least degree that passes through the given points, we can assume a polynomial of the form:

$p(x) = ax^4 + bx^3 + cx^2 + dx + e$

Since there are 5 points given, a polynomial of degree 4 is needed to uniquely determine the function. We will use the points to set up a system of equations to solve for the coefficients $a$, $b$, $c$, $d$, and $e$.

1. For $(-2, 28)$:

$p(-2) = 16a - 8b + 4c - 2d + e = 28$

2. For $(-1, 0)$:

$p(-1) = a - b + c - d + e = 0$

3. For $(0, -6)$:

$p(0) = e = -6$

4. For $(1, -8)$:

$p(1) = a + b + c + d + e = -8$

5. For $(2, 0)$:

$p(2) = 16a + 8b + 4c + 2d + e = 0$

Substituting $e = -6$ into the other equations:

1. $16a - 8b + 4c - 2d - 6 = 28$ → $16a - 8b + 4c - 2d = 34$
2. $a - b + c - d - 6 = 0$ → $a - b + c - d = 6$
3. $a + b + c + d - 6 = -8$ → $a + b + c + d = -2$
4. $16a + 8b + 4c + 2d - 6 = 0$ → $16a + 8b + 4c + 2d = 6$

This gives us a system of four equations:

16a - 8b + 4c - 2d & = 34 \\ a - b + c - d & = 6 \\ a + b + c + d & = -2 \\ 16a + 8b + 4c + 2d & = 6 \\ \end{align*}$$ Now, we can solve this system to find the values of $$a$$, $$b$$, $$c$$, and $$d$$. ### Solving the System Let's solve these equations step by step. 1. From equation $$e = -6$$, we've already found $$e$$. 2. We can use elimination or substitution methods to find $$a$$, $$b$$, $$c$$, and $$d$$. I'll solve these equations.The solution to the system of equations gives the coefficients: $$a = 1, \quad b = -1, \quad c = 1, \quad d = -3, \quad \text{and} \quad e = -6$$ So, the polynomial function $$p(x)$$ of least degree that