Rational Functions and Holes

1.10 Rational Functions and Holes

🏌🏽 Hole in One!

If the multiplicity of the zero 'a' in the numerator is greater than or equal to its multiplicity in the denominator, then the function has a hole at x = a. Remember, multiplicity is the number of times a zero appears in the factorization. For example, if 'a' appears 3 times in the numerator and 2 times in the denominator, there's a hole at x = a.

To find the y-value of the hole, cancel the common factor and then evaluate the simplified function at x = a. Let's look at an example:

r(x) = (x² - 9) / (x - 3)

The numerator has zeros at x = -3 and x = 3, while the denominator has a zero at x = 3. The zero at x=3 has a multiplicity of 2 in the numerator and 1 in the denominator, so there is a hole at x=3. First, factor the numerator:

r(x) = (x + 3)(x - 3) / (x - 3)

Then, cancel the common factor (x - 3):

r(x) = x + 3

Finally, plug in x = 3 to find the y-value of the hole:

r(3) = 3 + 3 = 6

So, there's a hole at (3, 6). 🤓

Graph displaying a hole at the intersection (1,1)

Source: Jed Q

🌉 Connecting to Limits

When a rational function, r(x), has a hole at x = c, the function approaches a finite limit as x approaches c. This limit, L, is the y-coordinate of the hole. Mathematically, this is expressed as: lim r(x) = L as x approaches c. The limit from the left, right, and as x approaches c are all equal to L.

To find the x-coordinate, simply identify the value of 'c' where the hole occurs. The y-coordinate is the value of the limit as x approaches 'c'.


A picture of the function x-4 divided by x-2 of the limit 10

Source: K12 Libretexts

{
  "mcq": [
    {
      "question": "Which of the following functions has a hole at x = 2?",
      "options": [
        "f(x) = (x-2)/(x+2)",
        "g(x) = (x^2 - 4)/(x-2)",
        "h(x) = (x+2)/(x-2)",
        "k(x) = (x^2 + 4)/(x-2)"
      ],
      "answer": "g(x) = (x^2 - 4)/(x-2)"
    },
    {
      "question": "What is the y-coordinate of the hole for the function r(x) = (x^2 - 5x + 6)/(x-2)?",
      "options": [
        "-1",
        "1",
        "2",
        "-2"
      ],
      "answer": "1"
    },
    {
      "question": "The function f(x) = (x^2 - 9)/(x-3) has a hole at x=3. What is the limit of f(x) as x approaches 3?",
      "options": [
        "0",
        "3",
        "6",
        "undefined"
      ],
      "answer": "6"
    }
  ],
  "frq": {
    "question": "Consider the rational function f(x) = (x^3 - 4x^2 + 4x) / (x^2 - 4). \n a) Identify all x-values where the function is undefined. \n b) Determine if there are any holes in the graph of f(x). If so, state the coordinates of the hole(s). \n c) Find the limit of f(x) as x approaches each x-value where a hole exists.",
    "scoring_breakdown": {
      "a": "1 point for correctly identifying x=2 and x=-2 as undefined",
      "b": "1 point for factoring numerator and denominator, 1 point for identifying x=2 as a hole, 1 point for calculating the y-coordinate of the hole",
      "c": "1 point for calculating the limit as x approaches 2"
    },
    "solution":{
       "a":"The function is undefined where the denominator is zero, i.e., x^2 - 4 = 0. This occurs at x = 2 and x = -2.",
       "b":"Factoring both numerator and denominator gives f(x) = x(x-2)(x-2) / (x-2)(x+2). There is a common factor of (x-2), so a hole exists at x=2. After canceling the common factor, the simplified function is f(x) = x(x-2)/(x+2). Plugging in x=2, we get y = 2(2-2)/(2+2) = 0. So, the hole is at (2,0).",
       "c":"The limit of f(x) as x approaches 2 is the y-coordinate of the hole, which is 0. Therefore, lim f(x) as x->2 = 0."
    }
  }
}

Final Exam Focus

Highest Priority Topics:

• Rational Functions and Holes: Understanding how to identify and calculate the coordinates of holes is critical. Expect questions that require factoring, canceling common terms, and evaluating limits.
• Limits: Understanding the concept of limits, especially in the context of holes and asymptotes, is essential. Be prepared to calculate limits both analytically and graphically.

Good luck! You've got this! 💪