Problem 15 - Precalculus with Calculus Previews

Question

Find the vertical and horizontal asymptotes for the graph of the given rational function. Find \(x\) - and \(y\) -intercepts of the graph. Sketch the graph of \(f\). $$ f(x)=\frac{x}{x^{2}-1} $$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = 1, -1\); horizontal asymptote at \(y = 0\); intercepts at \((0,0)\).

Step 1: Determine Vertical Asymptotes

To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\) because vertical asymptotes occur where the denominator is zero and the numerator is not zero. Given the function \[ f(x) = \frac{x}{x^2 - 1} \]Set the denominator equal to zero:\[ x^2 - 1 = 0 \]Factor the quadratic equation:\[ (x - 1)(x + 1) = 0 \]This gives the solutions:\(x = 1\) and \(x = -1\). Hence, there are vertical asymptotes at \(x = 1\) and \(x = -1\).

Step 2: Determine Horizontal Asymptotes

For horizontal asymptotes, compare the degrees of the numerator and denominator. Here, the numerator has degree 1 and the denominator has degree 2. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\). Thus, there is one horizontal asymptote at \(y = 0\).

Step 3: Find x-intercepts

To find the \(x\)-intercepts, set the numerator equal to zero and solve for \(x\). This is because the \(x\)-intercepts occur where the graph of the function crosses the \(x\)-axis, and this happens when the output \(f(x) = 0\). Given the function:\[ \frac{x}{x^2 - 1} = 0 \]The numerator \(x\) must be zero, thus:\[ x = 0 \]So, the \(x\)-intercept is at \((0, 0)\).

Step 4: Find y-intercepts

To find the \(y\)-intercept, substitute \(x = 0\) into the function to find \(f(0)\). This is because the \(y\)-intercept is where the graph crosses the \(y\)-axis.\[ f(x) = \frac{x}{x^2 - 1} \]Substitute \(x = 0\):\[ f(0) = \frac{0}{0^2 - 1} = 0 \]Thus, the \(y\)-intercept is also at \((0, 0)\).

Step 5: Sketch the Graph

Now that we know the behavior of the function, we can sketch the graph. Remember, there are vertical asymptotes at \(x = 1\) and \(x = -1\), a horizontal asymptote at \(y = 0\), and both \(x\)- and \(y\)-intercepts at \((0, 0)\).The graph approaches the vertical asymptotes, becoming undefined at \(x = 1\) and \(x = -1\), and approaches the horizontal asymptote as \(x\) moves towards \(\pm\infty\). The graph will pass through the origin \((0, 0)\).

Key Concepts

Asymptotes

Rational functions often feature asymptotes, which are lines that the graph of the function approaches but never actually crosses. There are two kinds of asymptotes important to this function: vertical and horizontal.
Vertical asymptotes occur where the denominator of a rational function equals zero, and they signify locations where the function's value becomes undefined or blows up to infinity. To find them for our function, \(f(x) = \frac{x}{x^2 - 1}\), we set the denominator zero:

• \(x^2 - 1 = 0\)
• This factors into \((x - 1)(x + 1) = 0\)
• Solving for \(x\) gives the vertical asymptotes at \(x = 1\) and \(x = -1\).

Horizontal asymptotes provide information about the end behavior of the function. They are determined by the degrees of the numerator and the denominator. In our case, the degree of the numerator (1) is less than that of the denominator (2), which indicates a horizontal asymptote at \(y = 0\).
This means as \(x\) heads towards positive or negative infinity, the graph approaches but does not touch \(y = 0\). Understanding these asymptotes aids in comprehensively analyzing the function's graph.

Intercepts

Intercepts are essential to sketching the graph as they show where the graph crosses the axes.
To find the **x-intercept**, set the numerator equal to zero and solve for \(x\). This occurs where the function \(f(x)\) equals zero. For our function, \(f(x) = \frac{x}{x^2 - 1}\), solving gives:

• \(x = 0\)

This tells us the x-intercept is at the origin \((0, 0)\).
Finding the **y-intercept** involves substituting \(x = 0\) into the function. This represents where the graph intersects the y-axis. With our function:

• \(f(0) = \frac{0}{0^2 - 1} = 0\)

So the y-intercept is also at \((0, 0)\). These intercept points are crucial as they tether the graph to real points on the coordinate plane and help in validating the graph when sketching.

Graph Sketching

Graph sketching combines all previously gathered information into a visual representation of the function. Start by plotting the asymptotes and intercepts you have calculated. Vertical asymptotes at \(x = 1\) and \(x = -1\) are drawn as dashed lines. They serve as boundaries the graph approaches but never crosses. The horizontal asymptote at \(y = 0\) is a horizontal dashed line, indicating the graph levels off as \(x\) moves toward infinity.
Next, mark the intercepts at \((0, 0)\). This point is particularly significant as it's where both axes intersect, and the graph crosses here too.

• Consider the areas marked off by the vertical asymptotes. These will hint at how the curve behaves within those sections.
• The graph will approach each vertical asymptote but will veer away rapidly, following the directional cues set by the behavior near the asymptotes.
• As \(x\) increases beyond the vertical asymptotes, the graph follows toward the horizontal asymptote at \(y = 0\).

Put all these together on a coordinate plane, and you’ll see how the asymptotes and intercepts inform the overall shape and direction of the graph. With a thorough understanding and plotting, you'll have a clear and complete graph of the rational function.