Question

directions: for each of the following rational functions, identify and label any values of x where the function has a hole or vertical asymptote.

1. $f(x)=\frac{3x(x - 6)(x + 4)}{(x - 1)(x + 4)}$
2. $g(x)=\frac{(x - 3)(x + 2)}{(x - 3)^2(x - 8)}$
3. $h(x)=\frac{(x - 7)^2(x + 1)}{2x(x - 7)}$
4. $k(x)=\frac{x^2-4x - 12}{x^2-12x + 36}$
5. $r(x)=\frac{x^2-9}{x^3+x^2-6x}$
6. $y=\frac{(x + 3)^5(x - 5)^3}{(x + 3)^3(x - 5)^3}$

Explanation:

Step1: Recall the definition of holes and vertical asymptotes

Holes occur when a factor in the numerator and denominator cancel out. Vertical asymptotes occur at values of \(x\) that make the denominator zero and the factor is not cancelled out.

Step2: Analyze \(f(x)=\frac{3x(x - 6)(x + 4)}{(x - 1)(x+4)}\)

• Hole: Set the common - factor \(x + 4=0\), so \(x=-4\) is a hole.
• Vertical asymptote: Set \(x - 1=0\), so \(x = 1\) is a vertical asymptote.

Step3: Analyze \(g(x)=\frac{(x - 3)(x + 2)}{(x - 3)^2(x - 8)}\)

• Hole: Set the common - factor \(x - 3=0\), so \(x = 3\) is a hole.
• Vertical asymptote: Set \(x - 8=0\), so \(x = 8\) is a vertical asymptote.

Step4: Analyze \(h(x)=\frac{(x - 7)^2(x + 1)}{2x(x - 7)}\)

• Hole: Set the common - factor \(x - 7=0\), so \(x = 7\) is a hole.
• Vertical asymptote: Set \(2x=0\), so \(x = 0\) is a vertical asymptote.

Step5: Analyze \(k(x)=\frac{x^2-4x - 12}{x^2-12x + 36}=\frac{(x - 6)(x+2)}{(x - 6)^2}\)

• Hole: Set the common - factor \(x - 6=0\), so \(x = 6\) is a hole.
• There are no non - cancelled factors in the denominator that make it zero, so no vertical asymptotes.

Step6: Analyze \(r(x)=\frac{x^2-9}{x^3+x^2-6x}=\frac{(x + 3)(x - 3)}{x(x^2+x - 6)}=\frac{(x + 3)(x - 3)}{x(x + 3)(x - 2)}\)

• Hole: Set the common - factor \(x + 3=0\), so \(x=-3\) is a hole.
• Vertical asymptote: Set \(x=0\) and \(x - 2=0\), so \(x = 0\) and \(x = 2\) are vertical asymptotes.

Step7: Analyze \(y=\frac{(x + 3)^5(x - 5)^3}{(x + 3)^3(x - 5)^3}\)

• Hole: Set the common - factors \(x + 3=0\) and \(x - 5=0\), so \(x=-3\) and \(x = 5\) are holes. There are no non - cancelled factors in the denominator that make it zero, so no vertical asymptotes.

Answer:

1. Hole at \(x=-4\), vertical asymptote at \(x = 1\)
2. Hole at \(x = 3\), vertical asymptote at \(x = 8\)
3. Hole at \(x = 7\), vertical asymptote at \(x = 0\)
4. Hole at \(x = 6\), no vertical asymptotes
5. Hole at \(x=-3\), vertical asymptotes at \(x = 0\) and \(x = 2\)
6. Holes at \(x=-3\) and \(x = 5\), no vertical asymptotes