Problem 53 - Algebra and Trigonometry Real Mathematics, Real People

Question

Sketch the graph of the rational function by hand. As sketching aids, check for intercepts, vertical asymptotes, and slant asymptotes. $$g(x)=\frac{x^{3}}{2 x^{2}-8}$$

Step-by-Step Solution

Answer

The graph of the function \(g(x)=\frac{x^{3}}{2 x^{2}-8}\) has x-intercepts and a y-intercept at \(x=0\), vertical asymptotes at \(x=-2\) and \(x=2\), and a slant asymptote at \(y = \frac{1}{2}x\).

Step 1: Identify the Intercepts

The x-intercepts are found by setting the numerator equal to zero and solving for x. \(\ x^{3} = 0 \rightarrow x = 0\). The y-intercept is found by placing \(x = 0\), which gives \(y = g(0) = 0\). So the y-intercept is also at 0.

Step 2: Identify the Vertical Asymptotes

Vertical asymptotes occur when the denominator of a rational function equals zero. Thus, we find the vertical asymptotes by setting \(2x^{2} - 8 = 0\). Solving this equation gives \(x = -2, x = 2\) as the locations of the vertical asymptotes.

Step 3: Identify the Slant Asymptotes

To find any slant asymptotes, recognize that a rational function \(f(x) = \frac{P(x)}{Q(x)}\) has a slant asymptote if the degree of the polynomial \(P(x)\) is one more than the degree of \(Q(x)\). Here, because the degree of the numerator is only one more than the degree of the denominator, the function has a slant asymptote. Perform the division of \(x^{3}\) by \(2x^{2} - 8\) using synthetic division to find the equation of the slant asymptote which is \(y = \frac{1}{2}x\).

Key Concepts

X-Intercepts of Rational Functions

Finding the x-intercepts of a rational function is an essential step in graphing, as they indicate the points where the function crosses the x-axis. For the function given, \(g(x)=\frac{x^{3}}{2 x^{2}-8}\), the x-intercepts are determined by setting the numerator equal to zero and solving for \(x\). In our case, solving \(x^{3} = 0\) reveals that there is a single x-intercept at \(x = 0\).

An important detail here, which enhances understanding, is that if the numerator can be factored further, each factor that equals zero may provide additional x-intercepts. In our example, the cubic term cannot be factored further to yield additional roots, thus confirming \(x = 0\) as the sole x-intercept.

Vertical Asymptotes

Vertical asymptotes are like the 'boundaries' of a rational function that the function approaches but never touches or crosses. They occur at the values of \(x\) for which the denominator is zero (assuming the numerator isn't also zero at those points, as that could indicate a hole instead of an asymptote). For our function, \(g(x)\), setting the denominator \(2x^{2}-8\) equal to zero, \(2x^{2} - 8 = 0\), helps us find the vertical asymptotes. Factoring the denominator as \(2(x^{2} - 4) = 0\) and applying the difference of squares, \(x^{2} - 4 = (x + 2)(x - 2)\), indicates that the function has vertical asymptotes at \(x = -2\) and \(x = 2\).

Understanding vertical asymptotes aids in sketching the graph correctly, as it reveals the function's behavior around these 'forbidden' x-values, where the function will rise or fall indefinitely.

Slant Asymptotes

Slant asymptotes, also known as oblique asymptotes, come into play when the numerator of a rational function has a degree exactly one higher than the denominator. Unlike horizontal asymptotes, which are y-values that the function approaches as \(x\) goes to infinity or negative infinity, slant asymptotes follow a diagonal line that the function approaches but never reaches. For our function \(g(x)=\frac{x^{3}}{2 x^{2}-8}\), we compare the degrees of the numerator and denominator and confirm that a slant asymptote exists since the numerator's degree is one higher.

We find the equation of the slant asymptote by performing long division or synthetic division of the numerator by the denominator. For \(g(x)\), dividing \(x^{3}\) by \(2x^{2} - 8\) yields the line \(y = \frac{1}{2}x\) as the slant asymptote. Knowing the slant asymptote helps to predict and sketch the end behavior of the function far from the origin.