Questions: Find all asymptotes and/or holes in the graph of (f(x)=fracx^2-2 x-8x^2-x-6).

Find all asymptotes and/or holes in the graph of (f(x)=fracx^2-2 x-8x^2-x-6).
Transcript text: Find all asymptotes and/or holes in the graph of $f(x)=\frac{x^{2}-2 x-8}{x^{2}-x-6}$.
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Solution

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Solution Steps

To find the asymptotes and holes in the graph of the function \( f(x) = \frac{x^2 - 2x - 8}{x^2 - x - 6} \), we need to:

  1. Factor both the numerator and the denominator.
  2. Identify any common factors between the numerator and the denominator, which will indicate holes.
  3. Determine the vertical asymptotes by setting the denominator equal to zero and solving for \( x \).
  4. Determine the horizontal asymptote by comparing the degrees of the numerator and the denominator.
Step 1: Factor the Numerator and Denominator

The function \( f(x) = \frac{x^2 - 2x - 8}{x^2 - x - 6} \) can be factored as follows:

  • Numerator: \( x^2 - 2x - 8 = (x - 4)(x + 2) \)
  • Denominator: \( x^2 - x - 6 = (x - 3)(x + 2) \)
Step 2: Identify Common Factors (Holes)

The common factor between the numerator and the denominator is \( x + 2 \). This indicates that there is a hole in the graph at \( x = -2 \).

Step 3: Determine Vertical Asymptotes

To find the vertical asymptotes, we set the denominator equal to zero: \[ x - 3 = 0 \implies x = 3 \] Thus, there is a vertical asymptote at \( x = 3 \).

Step 4: Determine Horizontal Asymptote

Since the degrees of the numerator and denominator are both 2, we find the horizontal asymptote by taking the limit as \( x \) approaches infinity: \[ \text{Horizontal Asymptote} = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{1}{1} = 1 \]

Final Answer

  • Hole: \( x = -2 \)
  • Vertical Asymptote: \( x = 3 \)
  • Horizontal Asymptote: \( y = 1 \)

Thus, the final answers are: \[ \boxed{x = -2} \quad \text{(hole)} \] \[ \boxed{x = 3} \quad \text{(vertical asymptote)} \] \[ \boxed{y = 1} \quad \text{(horizontal asymptote)} \]

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