Math Problem Statement

Graph f(x)=(-x+3)(x+4)/(x^2-9). Show asymptotes and holes

Solution

The function f(x)=(x+3)(x+4)x29f(x) = \frac{(-x+3)(x+4)}{x^2-9} can be analyzed as follows:

Step 1: Factorize the Denominator

The denominator x29x^2 - 9 is a difference of squares: x29=(x3)(x+3).x^2 - 9 = (x - 3)(x + 3).

Thus, the function becomes: f(x)=(x+3)(x+4)(x3)(x+3).f(x) = \frac{(-x+3)(x+4)}{(x-3)(x+3)}.

Step 2: Simplify the Function

The numerator has a factor of (x+3)(-x + 3), which simplifies to (x3)-(x - 3). Substituting this: f(x)=(x3)(x+4)(x3)(x+3).f(x) = \frac{-(x-3)(x+4)}{(x-3)(x+3)}.

The factor (x3)(x-3) cancels in the numerator and denominator, except at x=3x = 3, where the original function is undefined. This introduces a hole at x=3x = 3.

The simplified function is: f(x)=(x+4)x+3,x3.f(x) = \frac{-(x+4)}{x+3}, \quad x \neq 3.

Step 3: Identify Asymptotes and Holes

  1. Vertical Asymptote: The denominator x+3=0x+3 = 0 implies a vertical asymptote at: x=3.x = -3.

  2. Hole: As noted earlier, there is a removable discontinuity (hole) at: x=3.x = 3. To find the yy-coordinate of the hole, substitute x=3x = 3 into the simplified function: f(x)=(x+4)x+3.f(x) = \frac{-(x+4)}{x+3}. At x=3x = 3: f(3)=(3+4)3+3=76.f(3) = \frac{-(3+4)}{3+3} = \frac{-7}{6}. So, the hole is at (3,76)(3, -\frac{7}{6}).

  3. Horizontal Asymptote: For large values of xx, the highest-degree terms in the numerator and denominator dominate. The degree of the numerator and denominator are the same (x1x^1 in both), so the horizontal asymptote is the ratio of the leading coefficients: y=11=1.y = \frac{-1}{1} = -1.

Step 4: Graph the Function

  • Plot the vertical asymptote at x=3x = -3.
  • Plot the horizontal asymptote at y=1y = -1.
  • Mark the hole at (3,76)(3, -\frac{7}{6}).
  • Sketch the behavior of the simplified function f(x)=(x+4)x+3f(x) = \frac{-(x+4)}{x+3}, taking into account the asymptotes and hole.

Final Notes

Would you like me to generate a visual graph of this function? If yes, I can plot it for you. Let me know!


Related Questions

  1. How do you determine whether a function has a horizontal or oblique asymptote?
  2. What are the steps to identify and classify discontinuities in a function?
  3. Can a function have both removable discontinuities and vertical asymptotes? Provide an example.
  4. How does factoring help simplify rational functions?
  5. What is the significance of asymptotes in real-world applications of functions?

Tip

Always check the original function for undefined points (such as xx values that make the denominator zero) before simplifying to avoid missing holes or discontinuities.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Rational Functions
Asymptotes (Vertical and Horizontal)
Removable Discontinuities (Holes)
Simplification of Rational Expressions

Formulas

Difference of squares: a^2 - b^2 = (a - b)(a + b)
Horizontal asymptote for rational functions: y = coefficient of highest-degree term in numerator / coefficient of highest-degree term in denominator

Theorems

Simplification of Rational Functions
Properties of Asymptotes and Holes

Suitable Grade Level

Grades 10-12