Math Problem Statement

Find all holes of the following function. If a hole exists, write your answer as a coordinate point in simplest form. f(x) = (x + 3) / (x^2 - 6x - 27)

Solution

To find the holes of the function:

f(x)=x+3x26x27f(x) = \frac{x+3}{x^2 - 6x - 27}

Step 1: Factor the Denominator

Factor x26x27x^2 - 6x - 27:

x26x27=(x9)(x+3)x^2 - 6x - 27 = (x - 9)(x + 3)

Thus, the function simplifies to:

f(x)=x+3(x9)(x+3)f(x) = \frac{x+3}{(x-9)(x+3)}

Step 2: Identify the Hole

A hole occurs where a common factor cancels in both the numerator and denominator. Since x+3x+3 is a common factor, it cancels out, leaving:

f(x)=1x9,x3f(x) = \frac{1}{x - 9}, \quad x \neq -3

So, there is a hole at x=3x = -3.

Step 3: Find the y-Coordinate of the Hole

To find the corresponding yy-value, substitute x=3x = -3 into the simplified function:

f(3)=139=112=112f(-3) = \frac{1}{-3 - 9} = \frac{1}{-12} = -\frac{1}{12}

Final Answer

The hole is at:

(3,112)\left(-3, -\frac{1}{12}\right)


Would you like a further breakdown of any step?

Here are five related questions you might find useful:

  1. How do you determine vertical asymptotes from a rational function?
  2. What is the difference between a hole and a vertical asymptote?
  3. How do you find the domain of a rational function?
  4. What happens if there are multiple removable discontinuities (holes)?
  5. Can a rational function have both a hole and a horizontal asymptote?

Tip: Holes occur when a factor cancels, while vertical asymptotes remain when factors do not cancel.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Factoring
Removable Discontinuities

Formulas

Factoring Quadratic Expressions
Finding Holes in Rational Functions

Theorems

Removable Discontinuity Theorem

Suitable Grade Level

Grades 10-12