Question

asymptotes, if any, of each rational function.

1. $h(x) = \frac{x^3 - 8}{x^2 - 5x + 6}$ \t48. $g(x) = \frac{x^3 + 1}{x^2 - 5x - 14}$
2. $q(x) = \frac{2x^2 - 5x - 12}{3x^2 - 11x - 4}$ \t52. $f(x) = \frac{x^2 + 6x + 5}{2x^2 + 7x + 5}$
3. $g(x) = \frac{x^4 - 1}{x^3 - x}$ \t56. $f(x) = \frac{x^4 - 16}{x^2 - 2x}$

Explanation:

Let's solve for the asymptotes of \( H(x)=\frac{x^{3}-8}{x^{2}-5x + 6} \) (Problem 47) as an example.

Step 1: Factor numerator and denominator

First, factor the numerator \( x^{3}-8 \) using the difference of cubes formula \( a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2}) \). Here, \( a=x \), \( b = 2 \), so \( x^{3}-8=(x - 2)(x^{2}+2x + 4) \).

Next, factor the denominator \( x^{2}-5x + 6 \). We need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3. So, \( x^{2}-5x + 6=(x - 2)(x - 3) \).

So, \( H(x)=\frac{(x - 2)(x^{2}+2x + 4)}{(x - 2)(x - 3)} \), \( x
eq2,3 \).

Step 2: Simplify the function

Cancel out the common factor \( (x - 2) \) (since \( x
eq2 \), we can do this). So, \( H(x)=\frac{x^{2}+2x + 4}{x - 3} \), \( x
eq2,3 \).

Step 3: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points) in the simplified function. Set the denominator \( x - 3=0 \), so \( x = 3 \). We check the numerator at \( x = 3 \): \( 3^{2}+2(3)+4=9 + 6+4=19
eq0 \). So, \( x = 3 \) is a vertical asymptote.

Step 4: Find horizontal or oblique asymptotes

To find horizontal or oblique asymptotes, we compare the degrees of the numerator and denominator. The degree of the numerator \( x^{2}+2x + 4 \) is 2, and the degree of the denominator \( x - 3 \) is 1. Since the degree of the numerator is greater than the degree of the denominator by 1, we have an oblique (slant) asymptote. We find it by performing polynomial long division of the numerator by the denominator.

Divide \( x^{2}+2x + 4 \) by \( x - 3 \):

\( x^{2}+2x + 4=(x - 3)(x + 5)+19 \)

So, \( \frac{x^{2}+2x + 4}{x - 3}=x + 5+\frac{19}{x - 3} \). As \( x
ightarrow\pm\infty \), \( \frac{19}{x - 3}
ightarrow0 \), so the oblique asymptote is \( y=x + 5 \).

Step 5: Check for holes

A hole occurs where a common factor cancels out. Here, we canceled \( (x - 2) \), so there is a hole at \( x = 2 \). To find the y - coordinate of the hole, substitute \( x = 2 \) into the simplified function \( \frac{x^{2}+2x + 4}{x - 3} \): \( \frac{2^{2}+2(2)+4}{2 - 3}=\frac{4 + 4+4}{-1}=\frac{12}{-1}=-12 \). So, there is a hole at \( (2,-12) \).

Answer:

For \( H(x)=\frac{x^{3}-8}{x^{2}-5x + 6} \):

• Vertical Asymptote: \( x = 3 \)
• Oblique Asymptote: \( y=x + 5 \)
• Hole: \( (2,-12) \)

(If you want to solve other problems, follow a similar process: factor numerator and denominator, simplify, find vertical asymptotes, horizontal/oblique asymptotes, and holes)