In Problem 79, we are tasked with finding a polynomial function whose graph is shown, with the smallest degree possible.

Observations from the graph:

1. X-intercepts (roots): The graph crosses the x-axis at $x = -3$ and $x = 2$, indicating these are roots of the polynomial.
2. Behavior at the intercepts:
   • At $x = -3$, the graph touches the x-axis and turns around. This indicates a double root at $x = -3$, implying a multiplicity of 2.
   • At $x = 2$, the graph crosses the x-axis, suggesting a root with multiplicity 1.
3. Y-intercept: The graph passes through $(0, -72)$, which can be used to find the constant factor.

General form of the polynomial:

Since we have a double root at $x = -3$ and a single root at $x = 2$, the polynomial can be written as: $f(x) = a(x + 3)^2(x - 2)$ where $a$ is a constant to be determined.

Finding $a$ using the y-intercept:

We know that $f(0) = -72$. Substituting $x = 0$ into the polynomial: $f(0) = a(0 + 3)^2(0 - 2) = a(3)^2(-2) = a(9)(-2) = -18a$ Setting this equal to $-72$, we can solve for $a$: $-18a = -72 \quad \Rightarrow \quad a = \frac{-72}{-18} = 4$

Final polynomial:

Thus, the polynomial function is: $f(x) = 4(x + 3)^2(x - 2)$

Would you like further details on how the polynomial was derived, or do you have any questions?

Here are some related questions you might consider:

1. How do we determine the degree of a polynomial from its graph?
2. What is the significance of the multiplicity of a root in the shape of a graph?
3. How does changing the constant factor $a$ affect the graph of the polynomial?
4. What would the graph look like if both roots had multiplicity 1?
5. How can we use additional points on the graph to further verify the equation?

Tip: When finding polynomials from graphs, always check the behavior at the intercepts to determine the root multiplicities.