Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

Sketch the graph of \(f\). $$ f(x)=\frac{2 x^{2}-2 x}{x^{2}+x} $$

Step-by-Step Solution

Answer

Sketch with vertical asymptote at \(x = -1\), horizontal asymptote at \(y = 2\), x-intercept at \(x = 1\), and a hole at \(x = 0\).

Step 1: Identify Domain

To find the domain of the function \(f(x)\), we need to determine where the denominator is not zero. Solve \(x^2 + x = 0\) which gives \(x(x + 1) = 0\). Hence, \(x = 0\) and \(x = -1\) are the values that make the denominator zero. The domain is all real numbers except \(x = 0\) and \(x = -1\).

Step 2: Simplify the Function

We simplify the function \(f(x) = \frac{2x^2 - 2x}{x^2 + x}\) by factoring:\(f(x) = \frac{2x(x - 1)}{x(x + 1)}\). By canceling the common factor \(x\) in the numerator and the denominator, we get \(f(x) = \frac{2(x - 1)}{x + 1}, x eq 0\). This simplified form helps in further analysis.

Step 3: Analyze Asymptotes

Horizontal asymptote: Since the degrees of the numerator and denominator are the same, calculate the horizontal asymptote as \(y = \frac{2}{1} = 2\).Vertical asymptote: Set the denominator of the simplified form \(x + 1 = 0\), giving \(x = -1\), which is a vertical asymptote. There's also a hole at \(x = 0\), since the canceled factor was responsible for a zero in the original denominator.

Step 4: Identify x- and y-intercepts

Find the y-intercept by evaluating \(f(0)\). However, since \(x = 0\) is not in the domain, there is no y-intercept. For x-intercepts, set the numerator \(2(x - 1) = 0\) so \(x = 1\) is the x-intercept with \(f(1) = 0\).

Step 5: Analyze Behavior Near Asymptotes

Near \(x = -1\), analyze \(f(x) = \frac{2(x - 1)}{x + 1}\). As \(x\to -1^{+}\), \(f(x)\to \, -\infty\). As \(x\to -1^{-}\), \(f(x)\to \, +\infty\).

Step 6: Sketch the Graph

Plot the asymptotes, \(y = 2\) as a horizontal line and \(x = -1\) as a vertical line.Mark the x-intercept \((1,0)\). There is a hole at \((0, \underset{\rightarrow}{\cdot})\) since there is no value for \(f(x)\) at \(x = 0\).Draw the curve approaching the horizontal asymptote \(y = 2\) as \(|x| \rightarrow \infty\), and behavior diverging to infinity near \(x = -1\).

Key Concepts

Domain of a Function

Understanding the domain of a rational function is crucial. It tells us what values we can plug into the function and expect a real output. In the case of the function \( f(x) = \frac{2x^2 - 2x}{x^2 + x} \), the domain is determined by the denominator \( x^2 + x \). A rational function is undefined wherever the denominator is zero because division by zero is undefined.

To find these values, solve \( x^2 + x = 0 \). Factor this to get \( x(x + 1) = 0 \), giving solutions \( x = 0 \) and \( x = -1 \). These critical points are where the function is undefined. Thus, the domain of our function is all real numbers except \( x = 0 \) and \( x = -1 \).

• Always check where the denominator is zero to find the domain of rational functions.
• Plot the domain to understand where the function is uninterrupted.

Asymptotes

Asymptotes are lines that the graph of a function approaches but never touches. They are vital in understanding the behavior of rational functions as \(|x|\) becomes very large or as the function approaches undefined points.

For our function, we have both horizontal and vertical asymptotes. A horizontal asymptote is found by comparing the degrees of the numerator and denominator. Here, both have the degree of 2, so the horizontal asymptote is \( y = \frac{2}{1} = 2 \). This tells us as \( |x| \) goes to infinity, the function approaches the line \( y = 2 \).

Next, we identify the vertical asymptote by setting the denominator of the simplified function \( x + 1 = 0 \), giving \( x = -1 \). At \( x = 0 \), instead of another asymptote, there is a hole because it was a canceled factor when simplifying the function, indicating a point of discontinuity rather than an asymptote.

• Vertical asymptotes occur where the denominator equals zero and is not canceled out.
• Horizontal asymptotes depend on the degrees of the polynomial terms in the rational function.

x- and y-intercepts

The intercepts of a function are where the graph crosses the axes, vital points for sketching. The y-intercept is particularly easy to find; it's the value of \( f(x) \) when \( x = 0 \). However, in this function, \( x=0 \) is outside our domain due to division by zero. Thus, there isn't a y-intercept.

For x-intercepts, set the numerator equal to zero. Solving \( 2(x - 1) = 0 \) leads to \( x = 1 \). Therefore, the function has an x-intercept at the point \( (1, 0) \).

• Use intercepts to anchor your graph on the axes, providing reference points.
• If a variable value is outside the domain, no intercept can be found at that point.

Graphing Functions

Graphing functions, especially rational ones, involves several key steps to visualize the behavior accurately. Start by identifying significant features: domain, asymptotes, and intercepts.

For \( f(x) \), plot the vertical asymptote at \( x = -1 \) and horizontal asymptote at \( y = 2 \). Remember, the function approaches but doesn't cross these lines. Mark the x-intercept at \( (1, 0) \). Identify any holes; here, there is one at \( x = 0 \), indicating a lack of function value there, visually represented by a small, open circle on the graph.

When sketching, ensure the curve draws near asymptotes as \( x \) gets very large or small, and show behavior near the vertical asymptote, diverging toward positive or negative infinity.

• Use graph paper or graphing tools to assure accuracy.
• Consider symmetry and key points to simplify the graphing.
• Always indicate any holes which mark discontinuities in the function.