To determine the equation of a 5th degree polynomial that meets the given conditions, we need to follow a systematic approach. Let's denote the polynomial by $P(x)$.

Conditions given:

1. The polynomial touches the x-axis at $(-3, 0)$. This means $x = -3$ is a root with at least multiplicity 2.
2. The polynomial crosses the y-axis at $(0, -4.5)$. This means $P(0) = -4.5$.
3. The polynomial crosses the x-axis at $(1, 0)$. This means $x = 1$ is a simple root.

Step-by-step solution:

1. Identify the roots and their multiplicities:

   • Since the polynomial touches the x-axis at $x = -3$, $(-3)$ must be a root with multiplicity at least 2.
   • Since the polynomial crosses the x-axis at $x = 1$, $1$ must be a simple root.

   Therefore, the polynomial can be written in the form: $P(x) = a(x + 3)^2(x - 1)Q(x)$ where $Q(x)$ is a polynomial of degree 2 because we need a polynomial of degree 5 in total.

2. Determine $Q(x)$: Since we need a 5th degree polynomial and we already have 3 factors, $Q(x)$ must be a quadratic polynomial. Assume: $Q(x) = (x - r_1)(x - r_2)$ where $r_1$ and $r_2$ are the remaining roots.

3. Use the y-intercept condition: The polynomial crosses the y-axis at $(0, -4.5)$. Thus, $P(0) = a(0 + 3)^2(0 - 1)(0 - r_1)(0 - r_2) = -4.5$ Simplifying, we get: $9a(-1)(-r_1)(-r_2) = -4.5$ $9a r_1 r_2 = -4.5$ $a r_1 r_2 = -\frac{4.5}{9}$ $a r_1 r_2 = -\frac{1}{2}$

4. Determine the unknowns: Since we need additional information to determine $r_1$ and $r_2$, let's proceed assuming specific roots to solve the problem, or express them symbolically.

   Suppose $r_1$ and $r_2$ are such that $r_1 = r$ and $r_2 = s$. Then we have: $P(x) = a(x + 3)^2(x - 1)(x - r)(x - s)$ with $a \cdot r \cdot s = -\frac{1}{2}$

5. Solve for the roots $r$ and $s$: Without additional constraints, assume simplest case:

   • Let $r = 2$, then $s$ would be determined by: $a \cdot 2 \cdot s = -\frac{1}{2}$ $a \cdot s = -\frac{1}{4}$ If we assume $a = 1$: $s = -\frac{1}{4}$

6. Construct the polynomial: $P(x) = (x + 3)^2(x - 1)(x - 2)\left(x + \frac{1}{4}\right)$ If we include a constant $a$ to adjust for scaling: $P(x) = a(x + 3)^2(x - 1)(x - 2)(4x + 1)$ Substitute $P(0) = -4.5$ to find $a$: $-4.5 = a(3)^2(-1)(-2)(1) \Rightarrow -4.5 = a(9)(2)$ $-4.5 = 18a \Rightarrow a = -\frac{4.5}{18} = -\frac{1}{4}$

7. Final polynomial: $P(x) = -\frac{1}{4} (x + 3)^2(x - 1)(x - 2)(4x + 1)$

This is the polynomial of degree 5 that satisfies the given conditions.

Do you want more details or have any questions?

Relative questions:

1. What is a polynomial's degree and why is it important?
2. What is the difference between a root's multiplicity and a simple root?
3. How do you determine a polynomial equation from its roots?
4. How can you use the y-intercept to find a polynomial's leading coefficient?
5. What are the implications of a polynomial touching the x-axis at a point?
6. How can we verify that a given polynomial meets specific conditions?
7. What role does the constant coefficient play in the polynomial equation?
8. How do you graph a polynomial to visualize its roots and behavior?

Tip:

To understand the behavior of a polynomial function, always analyze its roots and their multiplicities, as well as how it intersects the axes.