Question

15.
deg:
coeff:
justify:

Explanation:

Step1: Determine the degree (deg)

The end - behavior of a polynomial function is determined by the leading term. For a polynomial function \(y = a_nx^n+\cdots+a_1x + a_0\), the degree \(n\) and the leading coefficient \(a_n\) determine the end - behavior. The graph of the polynomial has ends that go in the same direction (both downwards in this case). For a polynomial, if the leading coefficient is negative and the degree is even, the ends of the graph will both point downwards. Also, the number of turning points (local maxima and minima) of a polynomial of degree \(n\) is at most \(n - 1\). Looking at the graph, we can see that there are 3 turning points (a local maximum, a local minimum, and another local maximum). So, if the number of turning points is \(n-1\), and \(n - 1=3\), then \(n = 4\)? Wait, no. Wait, let's re - examine. Wait, the end - behavior: as \(x ightarrow+\infty\) and \(x ightarrow-\infty\), the graph goes down. For a polynomial, when the degree is even, the ends have the same direction. When the degree is odd, the ends have opposite directions. Since the ends here are the same (both down), the degree must be even. Also, the number of turning points: the graph has 3 turning points (let's count: starting from the left, it goes up, then down, then up, then down). The maximum number of turning points of a polynomial of degree \(n\) is \(n - 1\). So if there are 3 turning points, \(n-1\geq3\), so \(n\geq4\). But also, the end - behavior: when \(x ightarrow\pm\infty\), \(y ightarrow-\infty\). For a polynomial \(y = a_nx^n+\cdots\), if \(n\) is even and \(a_n<0\), then as \(x ightarrow\pm\infty\), \(y ightarrow-\infty\) (because \(x^n\) is positive for large \(|x|\) when \(n\) is even, and multiplying by a negative \(a_n\) gives negative values). Now, let's think about the degree. Let's assume the degree is 4? Wait, no, if we have 3 turning points, the degree could be 4 (since \(n - 1 = 3\) implies \(n = 4\))? Wait, no, actually, the number of turning points is at most \(n - 1\). So if there are 3 turning points, the degree is at least 4. But looking at the graph, the end - behavior is both ends down, so degree is even. Let's consider the leading coefficient (coeff). Since the ends are down, and degree is even, the leading coefficient is negative.

Wait, maybe a better way: For a polynomial function, the end - behavior is determined by \(\lim_{x ightarrow\pm\infty}f(x)\). If the degree \(n\) is even:
- If \(a_n>0\), \(\lim_{x ightarrow\pm\infty}f(x)=+\infty\)
- If \(a_n<0\), \(\lim_{x ightarrow\pm\infty}f(x)=-\infty\)

If the degree \(n\) is odd:
- If \(a_n>0\), \(\lim_{x ightarrow+\infty}f(x)=+\infty\), \(\lim_{x ightarrow-\infty}f(x)=-\infty\)
- If \(a_n<0\), \(\lim_{x ightarrow+\infty}f(x)=-\infty\), \(\lim_{x ightarrow-\infty}f(x)=+\infty\)

In our graph, as \(x ightarrow+\infty\) and \(x ightarrow-\infty\), the function goes to \(-\infty\), so the degree is even and the leading coefficient is negative.

Now, the number of turning points: the graph has 3 turning points. The maximum number of turning points of a polynomial of degree \(n\) is \(n - 1\). So if there are 3 turning points, \(n-1\geq3\), so \(n\geq4\). But let's assume the simplest case where the number of turning points is \(n - 1\). So if \(n-1 = 3\), then \(n = 4\)? Wait, no, 3 turning points would mean \(n=4\) (since \(n - 1=3\)). Wait, no, for example, a quartic (degree 4) polynomial can have up to 3 turning points. So let's say the degree is 4 (even), and the leading coefficient is negative.

Step2: Determine the leading coefficient (coeff)

As we saw from the end - behavior, since as \(x ightarrow\pm\infty\), \(y ightarrow-\infty\) and the degree is even, the leading coefficient \(a_n\) must be negative.

Step3: Justify

- **Degree**: The graph has the same end - behavior (both ends down), which implies the degree is even. Also, the number of turning points (3) implies that the degree \(n\) satisfies \(n-1\geq3\), so \(n\geq4\). The simplest even degree with 3 turning points is 4 (since a degree 4 polynomial can have up to 3 turning points).
- **Coefficient**: Since the ends of the graph go down (as \(x ightarrow\pm\infty\), \(y ightarrow-\infty\)) and the degree is even, the leading coefficient must be negative (because for an even - degree polynomial \(y = a_nx^n+\cdots\), if \(a_n<0\), \(x^n\) is positive for large \(|x|\), so \(y\) is negative for large \(|x|\)).

Answer:

• deg: 4 (even, with 3 turning points, so \(n = 4\))
• coeff: negative (since end - behavior is \(y

ightarrow-\infty\) as \(x
ightarrow\pm\infty\) and degree is even)

• justify: Degree is even (same end - behavior) and at least 4 (3 turning points, \(n-1\geq3\)). Leading coefficient is negative (ends down, even degree).