Problem 29 - Precalculus with Calculus Previews

Question

Find the vertical and slant asymptotes for the graph of the given rational function. Find \(x\) - and \(y\) -intercepts of the graph. Sketch the graph \(f\). $$ f(x)=\frac{x^{3}-8}{x^{2}-x} $$

Step-by-Step Solution

Answer

Vertical asymptotes: \( x = 0, x = 1 \); slant asymptote: \( y = x + 1 \); x-intercept: \( (2,0) \); no y-intercept.

Step 1: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero at the same point. For the function \( f(x) = \frac{x^3 - 8}{x^2 - x} \), we need to set the denominator equal to zero: \[ x^2 - x = 0 \]This factors as \( x(x - 1) = 0 \), giving solutions \( x = 0 \) and \( x = 1 \). We will check if these points are indeed vertical asymptotes by ensuring the numerator is non-zero. For \( x = 0 \), \( x^3 - 8 = -8 \) which is not zero, confirming \( x = 0 \) is a vertical asymptote. Similarly, for \( x = 1 \), \( 1^3 - 8 = -7 \) is also not zero, confirming \( x = 1 \) is a vertical asymptote.

Step 2: Finding Slant Asymptote

Slant (or oblique) asymptotes occur when the degree of the numerator is one more than the degree of the denominator. Here \( f(x) = \frac{x^3 - 8}{x^2 - x} \) has a numerator of degree 3 and a denominator of degree 2, meeting the condition for a slant asymptote.To find the slant asymptote, perform polynomial long division dividing \( x^3 - 8 \) by \( x^2 - x \). The division gives a quotient of \( x + 1 \), which means the slant asymptote is \( y = x + 1 \). The remainder of this division is not considered in the asymptote.

Step 3: Find X-Intercepts

X-intercepts occur when \( f(x) = 0 \), which happens when the numerator is zero and the denominator is not zero. Set \( x^3 - 8 = 0 \), which gives \( x^3 = 8 \), thus \( x = 2 \). At \( x = 2 \), the denominator \( x^2 - x = 2^2 - 2 = 2 \) is not zero, confirming an x-intercept at \( (2, 0) \).

Step 4: Find Y-Intercept

The y-intercept is found by evaluating \( f(x) \) at \( x = 0 \). Substitute into the function: \[ f(0) = \frac{0^3 - 8}{0^2 - 0} \]However, the denominator becomes zero, so there is no y-intercept (since there is a vertical asymptote at \( x = 0 \)).

Step 5: Sketch the Graph

To sketch the graph, consider both the asymptotes and intercepts calculated:- Vertical asymptotes at \( x = 0 \) and \( x = 1 \).- A slant asymptote along \( y = x + 1 \).- An x-intercept at \( (2, 0) \).Near the vertical asymptotes, the function approaches \( \pm \infty \), and overall, the function approaches the slant asymptote \( y = x + 1 \) for large \( |x| \).

Key Concepts

Vertical Asymptotes

Vertical asymptotes can be thought of as lines that the graph of a function gets very close to but never actually touches or crosses. For rational functions, a vertical asymptote occurs where the denominator is zero, and the numerator isn't zero at that same point. To find these asymptotes in the function \[f(x) = \frac{x^3 - 8}{x^2 - x},\] we set the denominator equal to zero to find possible vertical asymptotes:

• \(x^2 - x = 0\)
• This factors into \(x(x - 1) = 0\)
• The solutions are \(x = 0\) and \(x = 1\)

Next, ensure these values don't make the numerator zero:

• At \(x = 0\), the numerator is \(-8\), which is not zero.
• At \(x = 1\), the numerator is \(-7\), also not zero.

Therefore, both \(x = 0\) and \(x = 1\) are vertical asymptotes.

Slant Asymptotes

Slant asymptotes, also known as oblique asymptotes, occur when the degree of the polynomial in the numerator is exactly one higher than the denominator in a rational function. Unlike vertical asymptotes, slant asymptotes don't appear as vertical lines but as slanted lines on the graph.
To find these, we perform long division on:

• Divide \(x^3 - 8\) by \(x^2 - x\)
• The quotient is \(x + 1\), which gives the equation of the slant asymptote: \[y = x + 1\]

Remember, slant asymptotes show how the function behaves for very large or very small values of \(x\). Near the ends of the graph, the function will closely follow this line.

Intercepts

Intercepts are essential for understanding where the graph meets the axes.
**X-intercepts** occur where the graph crosses the x-axis, meaning the function equals zero at those points. This happens when the numerator equals zero but not the denominator.

• Set \(x^3 - 8 = 0\)
• Solve for \(x\): \(x^3 = 8\)
• \(x = 2\)

Check if the denominator remains non-zero:

• \(x^2 - x \) at \(x = 2\) is 2, which isn't zero.

The x-intercept is at \((2, 0)\).
**Y-intercepts** occur where the graph crosses the y-axis, meaning \(x = 0\).

• Plug \(x = 0\) into the function.
• The denominator becomes 0, so there's no y-intercept as it indicates a vertical asymptote at this point.

Polynomial Division

Polynomial division is a technique to simplify expressions or find slant asymptotes. It is similar to long division with numbers.
For the function \(\frac{x^3 - 8}{x^2 - x}\), you perform polynomial long division as follows:

• Divide \(x^3\) by \(x^2\) to get \(x\).
• Multiply \(x\) by \(x^2 - x\) and subtract from \(x^3 - 8\).
• Bring down the next term to continue the division.
  This process yields a quotient of \(x + 1\) and shows the remaining function behavior.

Polynomial division helps us determine slant asymptotes, providing a simplified perspective on how the function's graph behaves as \(x\) becomes very large or very negative.