Question

consider the following polynomial.
$s(x) = 4x^2(x + 5)(x - 8)$
step 2 of 2: describe the behavior of the graph of $s(x)$ as $x \
ightarrow \pm\infty$.
answer 4 points
$s(x) \
ightarrow \square$ as $x \
ightarrow -\infty$
$s(x) \
ightarrow \square$ as $x \
ightarrow \infty$

Explanation:

Step1: Determine the leading term

To find the end - behavior of a polynomial, we first need to find the leading term. The leading term of a polynomial is the term with the highest degree. For the polynomial \(s(x)=4x^{2}(x + 5)(x - 8)\), we expand the polynomial (or at least find the degree and the leading coefficient).

First, multiply out the factors. The degree of \(x^{2}\) is 2, the degree of \((x + 5)\) is 1, and the degree of \((x - 8)\) is 1. When we multiply polynomials, the degree of the product is the sum of the degrees of the factors. So the degree of \(s(x)\) is \(2+1 + 1=4\).

Now, find the leading coefficient. The leading coefficient of \(x^{2}\) is 4, the leading coefficient of \((x + 5)\) is 1, and the leading coefficient of \((x - 8)\) is 1. When we multiply the leading terms: \(4x^{2}\times x\times x=4x^{4}\). So the leading term of \(s(x)\) is \(4x^{4}\).

Step2: Analyze the end - behavior based on the leading term

For a polynomial \(y = a_nx^n+\cdots+a_1x + a_0\), the end - behavior is determined by the leading term \(a_nx^n\).

• If the degree \(n\) is even:
• If the leading coefficient \(a_n>0\), then as \(x

ightarrow\infty\), \(y
ightarrow\infty\) and as \(x
ightarrow-\infty\), \(y
ightarrow\infty\) (because raising a negative number to an even power gives a positive number, and multiplying by a positive leading coefficient keeps it positive).

In our case, the leading term is \(4x^{4}\), where \(n = 4\) (even) and \(a_n=4>0\).

So, as \(x
ightarrow-\infty\), \(s(x)=4x^{4}+\cdots
ightarrow\infty\) (since \((-\infty)^4=\infty\) and \(4\times\infty=\infty\)).

As \(x
ightarrow\infty\), \(s(x)=4x^{4}+\cdots
ightarrow\infty\) (since \(\infty^4=\infty\) and \(4\times\infty=\infty\)).

Answer:

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

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