Question

end behavior
x→-∞, y→-∞; x→∞, y→∞
x→-∞, y→-∞; x→∞, y→-∞
x→-∞, y→∞; x→∞, y→∞
x→-∞, y→∞; x→∞, y→-∞

Explanation:

Step1: Analyze the leading coefficient and degree of the polynomial

For a polynomial \(y = a_nx^n+a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\), the end - behavior is determined by the leading term \(a_nx^n\).
If \(n\) is odd:

• When \(a_n>0\), as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to\infty\), \(y\to\infty\)
• When \(a_n < 0\), as \(x\to-\infty\), \(y\to\infty\) and as \(x\to\infty\), \(y\to-\infty\)

If \(n\) is even:

• When \(a_n>0\), as \(x\to\pm\infty\), \(y\to\infty\)
• When \(a_n < 0\), as \(x\to\pm\infty\), \(y\to-\infty\)

Looking at the graph (a non - even - degree polynomial since the left and right ends go in different directions), and from the general shape (the left end \(x\to-\infty\), \(y\to-\infty\) and the right end \(x\to\infty\), \(y\to\infty\)), we can assume the leading coefficient \(a_n>0\) and the degree \(n\) is odd.

Step2: Match with the given options

The option \(x\to-\infty,y\to-\infty;x\to\infty,y\to\infty\) (the first option) is consistent with the analysis of a polynomial with a positive leading coefficient and an odd degree.

Answer:

The first option \(x\to-\infty,y\to-\infty;x\to\infty,y\to\infty\)