Question

1. describe the holes and asymptotes of the graph of the rational function (f(x) = \frac{x+5}{x^2-25})

hole at (-5, -1/2), horizontal asymptote at y = 0, vertical asymptote at x = 5

hole at (-5, -1/2), vertical asymptote at y = 0, horizontal asymptote at x = 5

no holes, horizontal asymptote at y = 0, vertical asymptote at x = 5

no holes, no horizontal asymptotes, vertical asymptote at x = 5

Explanation:

🆕 New Concept Discovered: Graphing Rational Functions
Identifying holes, vertical asymptotes, and horizontal asymptotes.

Step 1: Factor the numerator and denominator

First, write down the given rational function:

$$ f(x) = \frac{x + 5}{x^2 - 25} $$

To find holes and asymptotes, we factor the denominator using the difference of squares formula, \( a^2 - b^2 = (a - b)(a + b) \):

$$ x^2 - 25 = (x - 5)(x + 5) $$

Now, rewrite the function in its factored form:

$$ f(x) = \frac{x + 5}{(x - 5)(x + 5)} $$

Step 2: Identify any holes

A hole occurs on a graph where a factor appears in both the numerator and the denominator, because that value of \( x \) makes both equal to zero (an indeterminate form \( \frac{0}{0} \)).

Here, the common factor is \( (x + 5) \).
Set this factor to zero to find the \( x \)-coordinate of the hole:

$$ x + 5 = 0 \implies x = -5 $$

To find the \( y \)-coordinate of the hole, simplify the function by canceling the common factor:

$$ f(x) = \frac{1}{x - 5} \quad \text{for } x eq -5 $$

Substitute \( x = -5 \) into this simplified function:

$$ y = \frac{1}{-5 - 5} = \frac{1}{-10} = -\frac{1}{10} $$

Note: Looking at the multiple-choice options, the options list the hole at \( (-5, -1/2) \). This indicates a slight typo in the question's options (likely miscalculating the denominator as \( -5 + 3 \) or similar during their key creation, or a typo in the function's coefficients), but \( x = -5 \) is uniquely identified as the \( x \)-coordinate of the hole.

Step 3: Identify the vertical asymptote

A vertical asymptote occurs at the \( x \)-values that make the denominator of the simplified function equal to zero (where the function goes to infinity).

Using the simplified function:

$$ f(x) = \frac{1}{x - 5} $$

Set the remaining denominator to zero:

$$ x - 5 = 0 \implies x = 5 $$

So, there is a vertical asymptote at \( x = 5 \).

Step 4: Identify the horizontal asymptote

To find the horizontal asymptote, we compare the degree of the polynomial in the numerator to the degree of the polynomial in the denominator of the original function \( f(x) = \frac{x + 5}{x^2 - 25} \):

• Degree of numerator (top) = \( 1 \)
• Degree of denominator (bottom) = \( 2 \)

Since the degree of the denominator is greater than the degree of the numerator (\( 1 < 2 \)), the horizontal asymptote is always the line:

$$ y = 0 $$

Step 5: Match with the options

Let's combine our findings:

• Hole: at \( x = -5 \) (matches the option listing a hole at \( (-5, -1/2) \))
• Horizontal asymptote: at \( y = 0 \)
• Vertical asymptote: at \( x = 5 \)

This matches the first option.

Answer:

Hole at (-5, -1/2), Horizontal asymptote at y = 0, vertical asymptote at x = 5