Question

consider the following polynomial.
$f(x) = x^2(x + 5)(x - 9)$
step 2 of 2: describe the behavior of the graph of $f(x)$ as $x \
ightarrow \pm \infty$.
answer 4 points
$f(x) \
ightarrow \square$ as $x \
ightarrow -\infty$
$f(x) \
ightarrow \square$ as $x \
ightarrow \infty$

Explanation:

Step1: Determine the leading term

First, expand the polynomial to find the leading term. Multiply the leading terms of each factor: \(x^2 \cdot x \cdot x = x^{2 + 1 + 1}=x^4\). The leading term is \(x^4\), and the leading coefficient is \(1\) (positive).

Step2: Analyze end - behavior for \(x

ightarrow\infty\)
For a polynomial function \(y = a_nx^n+\cdots+a_0\), when \(n\) (the degree of the polynomial) is even and \(a_n>0\) (the leading coefficient is positive), as \(x
ightarrow\infty\), \(y
ightarrow\infty\). Since the degree of \(f(x)\) is \(4\) (even) and the leading coefficient is \(1>0\), as \(x
ightarrow\infty\), \(f(x)
ightarrow\infty\).

Step3: Analyze end - behavior for \(x

ightarrow-\infty\)
When \(n\) is even and \(a_n > 0\), as \(x
ightarrow-\infty\), we have \(y=a_nx^n+\cdots+a_0\). Since \(x^n\) (with \(n\) even) is positive when \(x\) is negative (because a negative number raised to an even power is positive), and the leading coefficient \(a_n = 1>0\), as \(x
ightarrow-\infty\), \(f(x)=x^4+\cdots
ightarrow\infty\) (because \((-\infty)^4=\infty\) and the other terms become negligible as \(|x|\) becomes very large).

Answer:

For \(f(x)
ightarrow\) \(\boldsymbol{\infty}\) as \(x
ightarrow-\infty\)

For \(f(x)
ightarrow\) \(\boldsymbol{\infty}\) as \(x
ightarrow\infty\)