Question

f(x)=\frac{-3x^{2}+9x}{x^{3}-x^{2}-6x}
vertical asymptotes:
horizontal asymptotes:
holes:

Explanation:

Step1: Factor numerator and denominator

Numerator: \(-3x^{2}+9x=-3x(x - 3)\)
Denominator: \(x^{3}-x^{2}-6x=x(x^{2}-x - 6)=x(x - 3)(x + 2)\)
So \(f(x)=\frac{-3x(x - 3)}{x(x - 3)(x + 2)}\)

Step2: Simplify the function

Cancel out the common factors \(x\) and \((x - 3)\) (for \(x
eq0\) and \(x
eq3\)). The simplified function is \(f(x)=\frac{-3}{x + 2}\) (with \(x
eq0\) and \(x
eq3\))

Step3: Find vertical asymptotes

Set the denominator of the simplified function (excluding canceled - out factors) equal to zero.
For \(y=\frac{-3}{x + 2}\), set \(x+2 = 0\), then \(x=-2\)

Step4: Find horizontal asymptotes

Since the degree of the numerator (\(n = 0\)) is less than the degree of the denominator (\(m=1\)) in the simplified function \(y=\frac{-3}{x + 2}\), the horizontal asymptote is \(y = 0\)

Step5: Find holes

The holes occur at the values of \(x\) that make the canceled - out factors equal to zero.
Set \(x=0\) and \(x = 3\) (from the canceled factors \(x\) and \(x - 3\))

Answer:

Vertical Asymptotes: \(x=-2\)
Horizontal Asymptotes: \(y = 0\)
Holes: \(x = 0\) and \(x=3\)