Question

determine the equations of any vertical asymptotes and the values of x for any holes in the graph in the following rational function
f(x) = \frac{x - 2}{x^2 - x - 2}
\bigcirc asymptote: x = 2; no holes
\bigcirc asymptote: x = 2, x = -1; no holes
\bigcirc asymptote: x = -1; hole: x = 2
\bigcirc asymptote: x = 2; hole: x = -1

Explanation:

Step1: Factor the denominator

First, factor the quadratic in the denominator. The denominator is \(x^2 - x - 2\). We need two numbers that multiply to \(-2\) and add to \(-1\). Those numbers are \(-2\) and \(1\). So, \(x^2 - x - 2=(x - 2)(x + 1)\). So the function becomes \(f(x)=\frac{x - 2}{(x - 2)(x + 1)}\).

Step2: Identify holes and vertical asymptotes

A hole occurs when there is a common factor in the numerator and the denominator. Here, \((x - 2)\) is a common factor. We can cancel out \((x - 2)\) (for \(x eq2\)), so there is a hole at \(x = 2\) (we set the canceled factor equal to zero to find the hole). After canceling, the function simplifies to \(f(x)=\frac{1}{x + 1}\) (for \(x eq2\)). Now, to find the vertical asymptote, we set the denominator of the simplified function equal to zero. So, \(x + 1 = 0\) gives \(x=-1\). Wait, no, wait. Wait, original factoring: denominator is \((x - 2)(x + 1)\), numerator is \(x - 2\). So when we cancel \(x - 2\), the domain excludes \(x = 2\) (hole) and \(x=-1\) (vertical asymptote)? Wait, no, let's re - do.

Wait, the steps:

1. Factor numerator and denominator:
- Numerator: \(x - 2\) (can't be factored further).
- Denominator: \(x^2 - x - 2=(x - 2)(x + 1)\) (as above).

2. Find values that make denominator zero: \(x = 2\) and \(x=-1\).

3. Find values that make both numerator and denominator zero (holes): For \(x = 2\), numerator \(x - 2=0\) and denominator \((x - 2)(x + 1)=0\). So \(x = 2\) is a hole (since it makes both numerator and denominator zero). For \(x=-1\), numerator \(x - 2=-1 - 2=-3 eq0\), so \(x=-1\) is a vertical asymptote (since it makes denominator zero but not numerator). Wait, no, that's the mistake earlier. Let's correct:

Hole: A hole is at \(x = a\) if \(f(a)\) is undefined (denominator zero) but the limit as \(x\to a\) exists (because numerator and denominator both have a factor of \((x - a)\)). So here, at \(x = 2\), numerator is \(0\), denominator is \(0\), so we can cancel \((x - 2)\), so the limit as \(x\to2\) of \(f(x)\) is \(\lim_{x\to2}\frac{x - 2}{(x - 2)(x + 1)}=\lim_{x\to2}\frac{1}{x + 1}=\frac{1}{3}\), so there is a hole at \(x = 2\).

Vertical asymptote: At \(x=-1\), denominator is zero (since \((-1)^2-(-1)-2=1 + 1-2 = 0\)) and numerator is \(-1 - 2=-3 eq0\), so as \(x\to - 1\), the function \(f(x)=\frac{x - 2}{(x - 2)(x + 1)}\) (before canceling) will have a vertical asymptote. Wait, no, after canceling, the function is \(\frac{1}{x + 1}\) (for \(x eq2\)), so as \(x\to - 1\), \(\frac{1}{x + 1}\to\pm\infty\), so vertical asymptote at \(x=-1\), and hole at \(x = 2\). Wait, but let's check the options. The options are:

- asymptote: \(x = 2\), no holes

- asymptote: \(x = 2,x=-1\), no holes

- asymptote: \(x=-1\); hole: \(x = 2\)

- asymptote: \(x = 2\); hole: \(x=-1\)

Wait, I think I made a mistake in the earlier step. Let's re - analyze:

Original function: \(f(x)=\frac{x - 2}{x^2 - x - 2}=\frac{x - 2}{(x - 2)(x + 1)}\).

- For hole: The common factor is \(x - 2\), so we set \(x - 2 = 0\), so \(x = 2\) is the hole (because when \(x = 2\), both numerator and denominator are zero, so we can remove the factor \(x - 2\) from numerator and denominator, meaning there's a hole at \(x = 2\)).

- For vertical asymptote: After removing the hole (canceling \(x - 2\)), the function becomes \(f(x)=\frac{1}{x + 1}\) (for \(x eq2\)). To find the vertical asymptote, we set the denominator of the simplified function equal to zero. So \(x + 1=0\) gives \(x=-1\). Wait, no, that's not right. Wait, the vertical asymptote is found by setting the denominator of the original function (after removing holes) equal to zero. Wait, the original denominator is \((x - 2)(x + 1)\), but we removed the factor \(x - 2\) (because of the hole), so the remaining factor is \(x + 1\), so the vertical asymptote is at \(x=-1\), and the hole is at \(x = 2\). So the correct option is "asymptote: \(x=-1\); hole: \(x = 2\)".

Answer:

asymptote: \(x=-1\); hole: \(x = 2\) (corresponding to the option "asymptote: \(x=-1\); hole: \(x = 2\)")