Question

consider the answers to a) and b). what is happening to f(x) as x → ±∞?

consider the answers to c) and d). what is happening to f(x) as the limits are approaching ±∞?

consider the answer to f). what is happening to f(x) as x → 2?

we can conclude that a line y = b is a horizontal asymptope of a function f(x) if either ____________ or __________. similarly, a line x = a is a vertical asymptope of a function f(x) if either __________ or __________. this is also where the denominator equals ____ when f(x) is in the most simplified form.

1. using the concepts from above, confirm that the function f(x) = \\(\frac{3x^2 - 2x - 8}{x^2 - 4}\\) has a horizontal asymptote at y = 3 and a vertical asymptote at x = −2 using calculus methods.

Explanation:

Part about Asymptote Definitions

For horizontal asymptote \( y = b \), by definition, it occurs when \( \lim_{x \to \infty} f(x)=b \) or \( \lim_{x \to -\infty} f(x)=b \). For vertical asymptote \( x = a \), it occurs when \( \lim_{x \to a^+} f(x)=\pm\infty \) or \( \lim_{x \to a^-} f(x)=\pm\infty \). The vertical asymptote occurs where the denominator is \( 0 \) in the simplified form of \( f(x) \) (since if numerator and denominator have a common factor, we cancel it, and the remaining denominator zero gives vertical asymptote).

Step 1: Confirm Horizontal Asymptote (\( y = 3 \))

To find horizontal asymptote, we compute \( \lim_{x \to \pm\infty} f(x) \) for \( f(x)=\frac{3x^2 - 2x - 8}{x^2 - 4} \). For limits at \( \pm\infty \) of rational functions, we divide numerator and denominator by the highest power of \( x \) in the denominator (here \( x^2 \)):
\[
\lim_{x \to \pm\infty} \frac{3x^2 - 2x - 8}{x^2 - 4}=\lim_{x \to \pm\infty} \frac{\frac{3x^2}{x^2}-\frac{2x}{x^2}-\frac{8}{x^2}}{\frac{x^2}{x^2}-\frac{4}{x^2}}=\lim_{x \to \pm\infty} \frac{3-\frac{2}{x}-\frac{8}{x^2}}{1 - \frac{4}{x^2}}
\]
As \( x \to \pm\infty \), \( \frac{1}{x}\to 0 \) and \( \frac{1}{x^2}\to 0 \). So:
\[
\frac{3 - 0 - 0}{1 - 0}=3
\]
Thus, \( \lim_{x \to \infty} f(x)=3 \) and \( \lim_{x \to -\infty} f(x)=3 \), so \( y = 3 \) is a horizontal asymptote.

Step 2: Confirm Vertical Asymptote (\( x = -2 \))

First, simplify \( f(x) \). Factor numerator and denominator:

• Numerator: \( 3x^2 - 2x - 8 \). Using quadratic formula or factoring, \( 3x^2 - 2x - 8=(3x + 4)(x - 2) \)
• Denominator: \( x^2 - 4=(x - 2)(x + 2) \)

So, \( f(x)=\frac{(3x + 4)(x - 2)}{(x - 2)(x + 2)} \), \( x
eq 2 \). After canceling \( (x - 2) \) (since \( x
eq 2 \)), \( f(x)=\frac{3x + 4}{x + 2} \), \( x
eq 2 \). Now, find \( \lim_{x \to -2} f(x) \). Compute left - hand limit (\( x\to - 2^- \)) and right - hand limit (\( x\to - 2^+ \)):

For \( \lim_{x \to -2^-} \frac{3x + 4}{x + 2} \): As \( x\to - 2^- \), \( x+2\to 0^- \) (negative and approaching 0), and \( 3x + 4=3(-2)+4=-6 + 4=-2 \). So \( \frac{-2}{0^-}\to+\infty \) (since negative divided by negative approaching 0 is positive infinity).

For \( \lim_{x \to -2^+} \frac{3x + 4}{x + 2} \): As \( x\to - 2^+ \), \( x + 2\to 0^+ \) (positive and approaching 0), and \( 3x+4=-2 \). So \( \frac{-2}{0^+}\to-\infty \).

Since \( \lim_{x \to -2^-} f(x)=\infty \) and \( \lim_{x \to -2^+} f(x)=-\infty \), \( x=-2 \) is a vertical asymptote.

Answer:

We can conclude that a line \( y = b \) is a horizontal asymptope of a function \( f(x) \) if either \( \boldsymbol{\lim_{x \to \infty} f(x) = b} \) or \( \boldsymbol{\lim_{x \to -\infty} f(x) = b} \). Similarly, a line \( x = a \) is a vertical asymptope of a function \( f(x) \) if either \( \boldsymbol{\lim_{x \to a^+} f(x) = \pm\infty} \) or \( \boldsymbol{\lim_{x \to a^-} f(x) = \pm\infty} \). This is also where the denominator equals \( \boldsymbol{0} \) when \( f(x) \) is in the most simplified form.

Problem 5: Confirm Horizontal and Vertical Asymptotes