Questions: Find the horizontal and vertical asymptotes of the following curve. Be sure to use a limit statement to explain why there is an asymptote, or why is not an asymptote. y=(x^2-3 x^3)/(2 x^3-8 x)

Solution Steps

To find the horizontal and vertical asymptotes of the given curve, we need to analyze the behavior of the function as \( x \) approaches infinity and the points where the denominator is zero. For horizontal asymptotes, evaluate the limits as \( x \to \infty \) and \( x \to -\infty \). For vertical asymptotes, find the values of \( x \) that make the denominator zero and check the limits around those points.

We are given the function:

\[ y = \frac{x^2 - 3x^3}{2x^3 - 8x} \]

First, factor both the numerator and the denominator:

• Numerator: \(x^2 - 3x^3 = x^2(1 - 3x)\)
• Denominator: \(2x^3 - 8x = 2x(x^2 - 4) = 2x(x - 2)(x + 2)\)

The function simplifies to:

\[ y = \frac{x^2(1 - 3x)}{2x(x - 2)(x + 2)} \]

Vertical asymptotes occur where the denominator is zero and the numerator is not zero.

Set the denominator equal to zero:

\[ 2x(x - 2)(x + 2) = 0 \]

This gives the solutions \(x = 0\), \(x = 2\), and \(x = -2\).

Check the numerator at these points:

• At \(x = 0\): \(x^2(1 - 3x) = 0\)
• At \(x = 2\): \(x^2(1 - 3x) = 4(1 - 6) = -20\)
• At \(x = -2\): \(x^2(1 - 3x) = 4(1 + 6) = 28\)

The vertical asymptotes are at \(x = 2\) and \(x = -2\).

Horizontal asymptotes are found by evaluating the limits as \(x\) approaches infinity or negative infinity.

\[ \lim_{x \to \infty} \frac{x^2 - 3x^3}{2x^3 - 8x} = \lim_{x \to \infty} \frac{-3x^3 + x^2}{2x^3 - 8x} \]

Divide every term by \(x^3\):

\[ = \lim_{x \to \infty} \frac{-3 + \frac{1}{x}}{2 - \frac{8}{x^2}} \]

As \(x \to \infty\), \(\frac{1}{x} \to 0\) and \(\frac{8}{x^2} \to 0\):

\[ = \frac{-3}{2} \]

Thus, the horizontal asymptote is \(y = -\frac{3}{2}\).

Final Answer