Problem 80 - Algebra and Trigonometry

Question

Graph the rational function, and find all vertical asymptotes, x- and y-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$ y=\frac{x^{4}}{x^{2}-2} $$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = \pm\sqrt{2}\). Intercepts at \((0,0)\). No local extrema. Polynomial end behavior: \(y = x^2\).

Step 1: Identifying Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Set the denominator \(x^2 - 2 = 0\) to find these points: solving \(x^2 = 2\), we find \(x = \pm\sqrt{2}\). Thus, the vertical asymptotes are at \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).

Step 2: Finding x-intercepts

The x-intercepts occur where the numerator is zero, provided that these points are not also vertical asymptotes. Set \(x^4 = 0\). Solving this equation gives \(x = 0\). Thus, there is an x-intercept at the point \((0,0)\).

Step 3: Finding the y-intercept

The y-intercept occurs where the graph crosses the y-axis, which is when \(x = 0\). Use the function \(y = \frac{x^4}{x^2 - 2}\) with \(x = 0\), giving \(y = 0\). Thus, the y-intercept is also the point \((0,0)\).

Step 4: Determining Local Extrema

Find the first derivative \(f'(x) = \frac{d}{dx}(\frac{x^4}{x^2-2})\). Use the quotient rule: \(f'(x) = \frac{(x^2 - 2)(4x^3) - x^4(2x)}{(x^2 - 2)^2}\). Simplify this to get \(f'(x) = \frac{2x^3(x^2+4)}{(x^2 - 2)^2}\). Set \(f'(x) = 0\) to find critical points: \(2x^3(x^2+4) = 0\), so \(x = 0\). After analysis, verify this is neither maximum nor minimum with the second derivative test or by examining the sign change in \(f'(x)\).

Step 5: Long Division for End Behavior

Perform polynomial long division of the numerator by the denominator: divide \(x^4\) by \(x^2 - 2\). The quotient obtained is \(x^2\) with remainder \(2x^2\). Hence, the polynomial \(x^2\) captures the end behavior of the rational function.

Step 6: Graphing the Functions

Graph both \(y = \frac{x^4}{x^2-2}\) and \(y = x^2\) using a sufficiently large viewing rectangle. Check that as \(x\) approaches \(+\infty\) or \(-\infty\), both functions exhibit the same tendency to rise without bound due to their similar end behavior.

Key Concepts

Vertical Asymptotes

When graphing rational functions, vertical asymptotes are critical to understand. These asymptotes occur where the denominator of the rational function equals zero, but the numerator does not. This means the function approaches infinity at these points, creating a vertical line on the graph.In the case of the function given by \( y = \frac{x^4}{x^2 - 2} \), the denominator is \( x^2 - 2 \). Setting this equal to zero gives \( x^2 = 2 \), leading to two solutions: \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). Therefore, the vertical asymptotes are at these values of \( x \).Vertical asymptotes help us understand that the function will spike upwards or downwards near these \( x \)-values, drastically affecting the graph's shape. Keep in mind, vertical asymptotes are never crossed by the graph.

X- and Y-intercepts

The intercepts of a graph are points where it crosses the axes, which are crucial for understanding the graph's positioning. **X-intercepts** For x-intercepts, the value of \( y \) is zero. This happens where the numerator of the function equals zero, as long as these x-values do not result in a zero denominator (which would indicate a vertical asymptote).For \( y = \frac{x^4}{x^2 - 2} \), setting the numerator \( x^4 \) to zero gives \( x = 0 \). As the denominator is non-zero at this point, we get an x-intercept at (0,0).**Y-intercept** Y-intercepts occur when \( x = 0 \). If substituting into the function gives a real value, it will be the y-intercept. For our function, substituting \( x = 0 \) yields \( y = 0 \). Hence, the y-intercept is also at (0,0).These points are essential for sketching the graph, providing a base for where the function intersects the axes.

Local Extrema

Local extrema of a function refer to the points where the function reaches a local minimum or maximum. These can be determined using the derivative of the function, which identifies critical points. For the rational function \( y = \frac{x^4}{x^2 - 2} \), we apply the quotient rule to derive its derivative: \[ f'(x) = \frac{ (x^2 - 2)(4x^3) - x^4(2x) }{(x^2 - 2)^2} \] This simplifies to \[ f'(x) = \frac{2x^3(x^2+4)}{(x^2 - 2)^2} \]. Setting \( f'(x) = 0 \) gives the critical point \( x = 0 \). However, examining this with a second derivative test, or checking the sign changes in \( f'(x) \) around x = 0, shows no local maximum or minimum at this point.Local extrema offer insights into the peaks and valleys of a graph, guiding the visual representation and behavior around levels of higher or lower values.