Math Problem Statement

For the function Upper F left parenthesis x right parenthesis equals StartFraction 1 Over left parenthesis x minus 9 right parenthesis squared EndFraction ​, ​(a) graph the rational function using​ transformations, (b) use the final graph to find the domain and​ range, and​ (c) use the final graph to list any​ vertical, horizontal, or oblique asymptotes. What is the domain of the given​ function?

Solution

Let's analyze the function F(x)=1(x9)2F(x) = \frac{1}{(x - 9)^2} step by step to answer the questions.

(a) Graph the rational function using transformations

We start with the basic function f(x)=1x2f(x) = \frac{1}{x^2}, which has the following characteristics:

  • It's always positive for all x0x \neq 0, with a vertical asymptote at x=0x = 0.
  • As x0x \to 0, the function approaches ++\infty.
  • As x±x \to \pm \infty, the function approaches 00 (horizontal asymptote at y=0y = 0).

Transformations applied to f(x)=1x2f(x) = \frac{1}{x^2}:

  1. Horizontal shift: F(x)=1(x9)2F(x) = \frac{1}{(x - 9)^2} shifts the graph of 1x2\frac{1}{x^2} 9 units to the right. So, the vertical asymptote moves to x=9x = 9.

This gives us the graph of F(x)F(x). It retains the same shape as f(x)=1x2f(x) = \frac{1}{x^2}, but it is centered at x=9x = 9.

(b) Use the final graph to find the domain and range

Domain:

The domain of F(x)=1(x9)2F(x) = \frac{1}{(x - 9)^2} is all real numbers except where the denominator is zero. The denominator is zero when x=9x = 9, so the function is undefined there. Thus, the domain is: (,9)(9,+)\boxed{(-\infty, 9) \cup (9, +\infty)}

Range:

Since (x9)2(x - 9)^2 is always positive for x9x \neq 9, and the function 1(x9)2\frac{1}{(x - 9)^2} outputs positive values, the graph of F(x)F(x) stays above the x-axis, approaching zero but never reaching it. Thus, the range is: (0,+)\boxed{(0, +\infty)}

(c) Asymptotes

Vertical Asymptote:

  • There is a vertical asymptote at x=9x = 9, as the function tends to infinity as x9x \to 9.

Horizontal Asymptote:

  • As x±x \to \pm \infty, F(x)=1(x9)20F(x) = \frac{1}{(x - 9)^2} \to 0, so there is a horizontal asymptote at y=0y = 0.

Oblique Asymptote:

  • There are no oblique asymptotes because the degree of the numerator is less than the degree of the denominator.

Summary:

  1. Domain: (,9)(9,+)(-\infty, 9) \cup (9, +\infty)
  2. Range: (0,+)(0, +\infty)
  3. Vertical Asymptote: x=9x = 9
  4. Horizontal Asymptote: y=0y = 0
  5. No oblique asymptote

Would you like a detailed graph of the function?


Here are 5 questions to explore:

  1. What is the behavior of the function near x=9x = 9?
  2. How does the transformation affect the graph's symmetry?
  3. Can you determine the limits as x9x \to 9^- and x9+x \to 9^+?
  4. What happens if we change the power in the denominator from 2 to 1?
  5. How would a vertical shift affect the graph?

Tip: Asymptotes give critical insights into the long-term behavior of a function! Always check for them when dealing with rational functions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Rational Functions
Transformations
Asymptotes
Domain and Range

Formulas

F(x) = 1 / (x - 9)^2
Basic form: f(x) = 1 / x^2
Transformation: horizontal shift by 9 units

Theorems

Asymptote Theorem for Rational Functions
Transformation Rules for Functions

Suitable Grade Level

Grades 10-12