Problem 67 - Algebra and Trigonometry

Question

Draw the graph of \(f\) and use it to determine whether the function is one-to- one. $$ f(x)=\frac{x+12}{x-6} $$

Step-by-Step Solution

Answer

The function is not one-to-one.

Step 1: Identify Vertical Asymptote

The function \( f(x) = \frac{x+12}{x-6} \) has a vertical asymptote where the denominator is zero. Set \( x - 6 = 0 \) to find the vertical asymptote, which occurs at \( x = 6 \). This indicates that the function is undefined at \( x = 6 \).

Step 2: Identify Horizontal Asymptote

To find the horizontal asymptote, consider the behavior of the function as \( x \to \infty \) or \( x \to -\infty \). Since the degrees of the numerator and denominator are the same, the horizontal asymptote is at \( y = \frac{1}{1} = 1 \). Hence, the line \( y = 1 \) is the horizontal asymptote.

Step 3: Check for Intercepts

To find the x-intercept, set \( f(x) = 0 \), which gives \( x + 12 = 0 \). Thus, \( x = -12 \) is the x-intercept. For the y-intercept, set \( x = 0 \), which gives \( f(0) = \frac{0 + 12}{0 - 6} = -2 \). Thus, the y-intercept is \( (0, -2) \).

Step 4: Sketch the Graph

Plot the vertical asymptote at \( x = 6 \) and the horizontal asymptote at \( y = 1 \). Plot the intercepts at \( (-12, 0) \) and \( (0, -2) \). Sketch the curve approaching the asymptotes based on these findings. Note that the function will approach the vertical asymptote as \( x \to 6^{+} \) or \( x \to 6^{-} \), and the horizontal asymptote as \( x \to \pm \infty \).

Step 5: Determine if the Function is One-to-One

A function is one-to-one if each horizontal line intersects the graph at most once. Examine the shape of the graph. Since the function approaches the horizontal asymptote \( y = 1 \) from both above and below, each horizontal line intersects the graph more than once, confirming the function is not one-to-one.

Key Concepts

Vertical Asymptotes

Vertical asymptotes occur in a function where the denominator equals zero, leading to undefined points in the graph. In the function \( f(x) = \frac{x+12}{x-6} \), the denominator is \( x-6 \). Setting this expression to zero, we find that the vertical asymptote is located at \( x = 6 \). This does not just create a hole in the graph but forms a line that the graph approaches but never touches or crosses.

This characteristic tells us that at \( x = 6 \), the values of \( f(x) \) will shoot to positive or negative infinity, depending on the direction of approach.

• The graph approaches \( +\infty \) as \( x \) approaches \( 6 \) from the left.
• It approaches \( -\infty \) as \( x \) approaches \( 6 \) from the right.

Understanding this helps you sketch accurate graphs and recognize behavior patterns near undefined points.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a function as \( x \) approaches positive infinity or negative infinity. Unlike vertical asymptotes, which indicate undefined points, horizontal asymptotes suggest a value that the function approaches but rarely touches.

For the function \( f(x) = \frac{x+12}{x-6} \), both the numerator and the denominator are first-degree polynomials (their degrees are equal). This results in a horizontal asymptote. To find it, divide the leading coefficients of the numerator and denominator. This gives us a horizontal asymptote at \( y = 1 \).

• As \( x \to \infty \), \( f(x) \to 1 \).
• As \( x \to -\infty \), \( f(x) \to 1 \).

This asymptote indicates that as the values of \( x \) become very large or very small, the graph of the function will get closer to the line \( y = 1 \). However, it may never actually reach that line, giving it a unique boundary character in the graph's far ends.

Intercepts in Graphing

Finding intercepts in a function's graph helps us understand where it crosses the axes. These points are crucial for accurately plotting the graph and recognizing the function's roots and behavior.

For the function \( f(x) = \frac{x+12}{x-6} \), let's find the intercepts:

• **X-intercept:** Set the function equal to zero: \( x + 12 = 0 \). Solving this gives \( x = -12 \). Thus, the graph crosses the x-axis at \((-12, 0)\).
• **Y-intercept:** Set \( x = 0 \) in the function: \( f(0) = \frac{12}{-6} = -2 \). Hence, the graph crosses the y-axis at \((0, -2)\).

These intercepts are points where the function intersects the x-axis and y-axis. Recognizing the intercepts provides a framework for sketching the basic shape of the graph more accurately, as it shows where the function changes from positive to negative or vice-versa.