Question

lessons 2.4 to 2.9 - review
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1. describe the behavior of $g(x) = -3x^4 + 5^3 - 8x^2 + 1$ as $x$ approaches $\infty$ and as $x$ approaches $-\infty$.

explain.
lesson 2.4

Explanation:

Step1: Identify Leading Term

The leading term of \( g(x)= -3x^4 + 5^3 - 8x^2 + 1 \) is \( -3x^4 \) (the term with the highest degree, which is 4 here).

Step2: Analyze Degree and Leading Coefficient

The degree of the polynomial (4) is even, and the leading coefficient (-3) is negative. For polynomials, when the degree is even:

• If the leading coefficient is positive, as \( x\to\pm\infty \), \( f(x)\to\infty \).
• If the leading coefficient is negative, as \( x\to\pm\infty \), \( f(x)\to-\infty \).

Since our leading term is \( -3x^4 \) (even degree, negative leading coefficient), we use this rule.

Step3: Determine Behavior as \( x\to\infty \)

As \( x \) approaches \( \infty \), we look at the leading term \( -3x^4 \). As \( x \) gets very large positive, \( x^4 \) is positive, and multiplying by -3 makes it negative and very large in magnitude. So \( g(x)\to-\infty \) as \( x\to\infty \).

Step4: Determine Behavior as \( x\to-\infty \)

As \( x \) approaches \( -\infty \), \( x^4 = (-x)^4 \) (since any real number to an even power is positive), so \( x^4 \) is positive when \( x \) is negative (because the exponent is even). Then \( -3x^4 \) is still negative (negative times positive) and very large in magnitude. So \( g(x)\to-\infty \) as \( x\to-\infty \).

Answer:

As \( x \to \infty \), \( g(x) \to -\infty \); as \( x \to -\infty \), \( g(x) \to -\infty \). This is because the leading term is \( -3x^4 \) (even degree, negative leading coefficient), so for both large positive and large negative \( x \), the function values approach negative infinity.