Question

The aryanptote to the curve \(y^{2}(4-x)=x^{3}\) is, it.

Short answer

The curve has no true vertical or oblique asymptotes.

Step-by-step solution

Simplify and Rearrange the Equation

The given equation is \( y^2 (4 - x) = x^3 \). First, isolate \( y^2 \) by dividing both sides by \( 4-x \), resulting in \( y^2 = \frac{x^3}{4-x} \).

Identify Horizontal Asymptotes

For horizontal asymptotes, consider the behavior as \( x \to \infty \) or \( x \to -\infty \). As \( x \to \infty \), \( \frac{x^3}{4-x} \approx x^2 \) which implies \( y^2 \approx x^2 \), and thus \( y \approx x \) or \( y \approx -x \). Therefore, potential horizontal asymptotes don't exist since \( y \to \infty \).

Identify Vertical Asymptotes

Check the values of \( x \) where the denominator becomes zero, in \( y^2 = \frac{x^3}{4-x} \). This happens where \( 4-x = 0 \), i.e., \( x = 4 \). However, if \( y^2 \) approaches infinity for finite \( x \), there exists no vertical asymptote since \( x^3 \to \infty \) cancels it out.

Evaluate Oblique Asymptotes

For oblique (slant) asymptotes, divide \( x^3 \) by \( 4-x \) using polynomial long division. After performing the division, you find the quotient is \( x^2 + 4x + 16 + \frac{64}{4-x} \), meaning as \( x \to \infty \) or \( x \to -\infty \), the asymptote is influenced by the term \( x^2 + 4x + 16 + \frac{64}{4-x} \).

Conclusion on the Asymptotes

By observing the polynomial from the division, find that the dominant behavior is quadratic, and due to the division remainder, there is no true oblique asymptote.

Key concepts

horizontal asymptotes

Horizontal asymptotes show how a function behaves as the input values reach very large positive or negative numbers. To find them, observe what happens to a function like \( y = \frac{x^3}{4-x} \) as \( x \) approaches infinity. If a horizontal asymptote exists, the function's values will approach a constant line. In this case, as \( x \to \infty \), the expression \( \frac{x^3}{4-x} \) simplifies roughly to \( x^2 \), resulting in \( y^2 \approx x^2 \). Solving for \( y \), we get \( y \approx x \) and \( y \approx -x \), indicating no horizontal lines getting approached by \( y \). Here, as \( x \to \infty \), \( y \) does not settle to a constant value, and hence, we conclude no horizontal asymptotes.

vertical asymptotes

Vertical asymptotes occur at values of \( x \) where the function's value tends to become infinite. These are the values where the denominator in the function, like \( y^2 = \frac{x^3}{4-x} \), equals zero since this makes the fraction undefined or infinite. Thus, set the denominator \( 4-x = 0 \) leading to \( x = 4 \). If \( y^2 \) were to approach infinity at this point, it would hint at a vertical asymptote. However, for \( x = 4 \), the accompanying term \( x^3 \to \infty \) also means \( y^2 \) doesn't move to infinity in a straightforward manner, preventing a typical vertical asymptote. Hence, \( x = 4 \) doesn't produce an infinite \( y^2 \), dismissing the presence of a vertical asymptote.

oblique asymptotes

Oblique asymptotes, or slant asymptotes, typically appear in rational functions where the degree of the numerator is one more than the denominator. Here, analyzing \( y^2 = \frac{x^3}{4-x} \), we perform polynomial long division of the expression \( \frac{x^3}{4-x} \). This operation might indicate a slanted line, but instead reveals a result of \( x^2 + 4x + 16 \) plus a remainder term, \( \frac{64}{4-x} \). As \( x \to \infty \), %% \( x \) dominates and suggests a quadratic nature in the function. Although a type of slope occurs, the behavior suggests complex interaction, not aligning with a simple linear slant. As such, the situation doesn't yield a true oblique asymptote but rather, a dominant quadratic behavior consistent with a parabola-like curve.