Question

1. describe the end behavior of ( f(x) = 0.25x^3 - x^2 - 1 ).

Explanation:

Step1: Analyze the leading term

The leading term of the polynomial \(f(x)=0.25x^{3}-x - 1\) is \(0.25x^{3}\). For a polynomial \(y = ax^{n}+bx^{n - 1}+\cdots+c\), the end - behavior is determined by the leading term \(ax^{n}\). Here \(a = 0.25>0\) and \(n = 3\) (odd).

Step2: Determine the end - behavior

When \(x\to+\infty\), for \(y = 0.25x^{3}\), as \(x\) gets larger and larger, \(y=0.25x^{3}\to+\infty\). Since \(f(x)\) has the same end - behavior as its leading term for large \(|x|\), \(f(x)\to+\infty\) as \(x\to+\infty\).
When \(x\to-\infty\), for \(y = 0.25x^{3}\), since \(x^{3}=x\times x\times x\) and \(x<0\), \(x^{3}<0\). So \(y = 0.25x^{3}\to-\infty\) as \(x\to-\infty\). And \(f(x)\to-\infty\) as \(x\to-\infty\) (because the leading term dominates the behavior of the polynomial for large \(|x|\)).

Answer:

As \(x\to+\infty\), \(f(x)\to+\infty\); as \(x\to-\infty\), \(f(x)\to-\infty\).