Question

question
the function $f(x)$ is defined below. what is the end behavior of $f(x)$?
$f(x) = 4x^3 + 166x^2 - 2x^4 - 3528 - 168x$
answer
\\(\circ\\) as $x \to -\infty, y \to \infty$ and as $x \to \infty, y \to \infty$
\\(\circ\\) as $x \to -\infty, y \to -\infty$ and as $x \to \infty, y \to \infty$
\\(\circ\\) as $x \to -\infty, y \to \infty$ and as $x \to \infty, y \to -\infty$
\\(\circ\\) as $x \to -\infty, y \to -\infty$ and as $x \to \infty, y \to -\infty$
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Explanation:

Step1: Identify the leading term

First, we need to rewrite the function in standard polynomial form (descending powers of \( x \)):
\( f(x)= - 2x^{4}+4x^{3}+166x^{2}-168x - 3528 \)
The leading term is the term with the highest power of \( x \), which is \( -2x^{4} \). The degree of the polynomial is \( 4 \) (even) and the leading coefficient is \( - 2 \) (negative).

Step2: Analyze end - behavior based on leading term

For a polynomial function \( f(x)=a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{1}x + a_{0} \), the end - behavior is determined by the leading term \( a_{n}x^{n} \).

• When the degree \( n \) is even:
• If the leading coefficient \( a_{n}>0 \), as \( x\to-\infty \) and \( x\to\infty \), \( y = f(x)\to\infty \).
• If the leading coefficient \( a_{n}<0 \), as \( x\to-\infty \) and \( x\to\infty \), \( y = f(x)\to-\infty \).

Since our polynomial has a leading term \( - 2x^{4} \) (degree \( 4 \), even, leading coefficient \( - 2<0 \)), we have:
As \( x\to-\infty \), \( y=f(x)\to-\infty \) (because for large negative \( x \), \( x^{4}\) is positive, and multiplying by \( - 2 \) gives a large negative number) and as \( x\to\infty \), \( x^{4}\) is positive, multiplying by \( - 2 \) gives a large negative number, so \( y = f(x)\to-\infty \).

Answer:

as \( x \to -\infty \), \( y \to -\infty \) and as \( x \to \infty \), \( y \to -\infty \)