Problem 22 - Precalculus: Functions and Graphs

Question

Sketch the graph of \(f\) $$f(x)=\frac{-3 x^{2}-3 x+6}{x^{2}-9}$$

Step-by-Step Solution

Answer

The graph of \(f(x)\) has vertical asymptotes at \(x = 3\) and \(x = -3\), a horizontal asymptote at \(y = -3\), y-intercept at \((0, -\frac{2}{3})\), and x-intercepts at \((-2,0)\) and \((1,0)\).

Step 1: Identify the Domain

To find the domain of the function, we need to find where the denominator is not equal to zero. Set \(x^2 - 9 = 0\) to find these points. Solving \(x^2 - 9 = 0\) gives \((x - 3)(x + 3) = 0\). Therefore, \(x = 3\) and \(x = -3\) are where the function is undefined. Thus, the domain is all real numbers except \(x = 3\) and \(x = -3\).

Step 2: Determine Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero and the numerator does not. We already established \(x = 3\) and \(x = -3\) from the domain. Since the numerator \(-3x^2 - 3x + 6\) does not equal zero at these points, both \(x = 3\) and \(x = -3\) are vertical asymptotes.

Step 3: Calculate Horizontal Asymptote

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Both are degree 2. The horizontal asymptote is the ratio of the leading coefficients: \(\frac{-3}{1} = -3\). Thus, the horizontal asymptote is \(y = -3\).

Step 4: Determine Intersections with the Axis

For the y-intercept, set \(x = 0\) in the function: \(f(0) = \frac{-3(0)^2 - 3(0) + 6}{0^2 - 9} = \frac{6}{-9} = -\frac{2}{3}\). So, the y-intercept is \((0, -\frac{2}{3})\). To find the x-intercept, set the numerator equal to zero: \(-3x^2 - 3x + 6 = 0\), and solve \(3x^2 + 3x - 6 = 0\). Simplifying gives \(x = -2\) and \(x = 1\), providing x-intercepts \((-2,0)\) and \((1,0)\).

Step 5: Analyze and Sketch the Graph

Plot the vertical asymptotes at \(x = 3\) and \(x = -3\), and the horizontal asymptote at \(y = -3\). Mark the intercepts at \((0, -\frac{2}{3})\), \((-2,0)\), and \((1,0)\). Sketch the curve, respecting asymptotes, intercepts, and general shape behavior (e.g., knowing signs in intervals). The graph should approach vertical asymptotes and the horizontal asymptote \(y = -3\) as \(x\) goes to infinity.

Key Concepts

Asymptotes

In the world of rational functions, asymptotes are like invisible boundaries that the function approaches but never quite touches. Let's first break down the two types of asymptotes: **vertical** and **horizontal**.
Vertical asymptotes occur where the denominator of the rational function equals zero, but only if the numerator does not also equal zero at these points. In our example, the denominator is zero at both \(x = 3\) and \(x = -3\), as we saw from the factorization process \((x - 3)(x + 3) = 0\). Since the numerator does not equal zero at these points, these are the locations of our vertical asymptotes. The function's graph will rise or fall steeply near these lines, approaching them but never crossing.
**Horizontal asymptotes** provide information on the behavior of the function as \(x\) becomes extremely large or small. When the degrees of the numerator and denominator are the same, like in our function where both are degree 2, the horizontal asymptote is determined by the ratio of their leading coefficients. Therefore, the horizontal asymptote is \(y = -3\), resulting from the ratio \(\frac{-3}{1}\).

• Vertical asymptotes: \(x = 3\) and \(x = -3\)
• Horizontal asymptote: \(y = -3\)

Through knowing these bounds, you can predict how the graph behaves near these lines.

Graphing Functions

Graphing a rational function involves piecing together intercepts, asymptotes, and the general behavior in between. It's like solving a puzzle where each piece shows a bit more of the story.
Start with the **vertical asymptotes** at \(x = 3\) and \(x = -3\). These will guide the apparent boundaries of your graph. The graph will approach these lines, from above or below, as it gets closer to them.
Next, draw the **horizontal asymptote** at \(y = -3\). This helps represent the behavior as \(x\) becomes incredibly large or small. The function will get closer and closer to this line but will never quite reach it.
The **intercepts** are crucial as well for anchoring your graph. In our example, you have the y-intercept at \( (0, -\frac{2}{3}) \), and x-intercepts at \((-2, 0)\) and \((1, 0)\). These points will provide the exact spots where the curve crosses the axes.After plotting these key features, sketch the curve. It might help to test more points for accuracy between intercepts and asymptotes. Keep in mind the function's behavior: it may move above or below the asymptotes, reflecting each point slightly differently.

• Begin with asymptotes
• Mark intercepts on the graph
• Sketch the curve, respecting limits

Intercepts

Intercepts are points where the function crosses the axes and are paramount in sketching graphs of rational functions. Let's walk through the method of finding these points.
- **Y-intercept**: This is where the graph meets the y-axis, which happens when \(x = 0\). For our function, substituting \(x = 0\) into \(f(x)\) gives \(f(0) = -\frac{2}{3}\). Thus, the y-intercept is at the point \((0, -\frac{2}{3})\).
- **X-intercepts**: These occur where the graph crosses the x-axis, which means the function must equal zero at these points. For this, we set the numerator equal to zero since the function equals zero when the numerator is zero, assuming the denominator is not zero. Solving \(-3x^2 - 3x + 6 = 0\), we find points at \(x = -2\) and \(x = 1\), thus the x-intercepts are \((-2, 0)\) and \((1, 0)\).Intercepts help define the core structure of the graph, giving you tangible spots where the curve has definitive heights or depths along the axes.

• Y-intercept: \((0, -\frac{2}{3})\)
• X-intercepts: \((-2, 0)\), \((1, 0)\)