Question

question 9 (1 point)
which of the following functions has exactly one vertical asymptote when graphed?
a) ( f(x)=\frac{x}{x^{2}-4} )
b) ( f(x)=\frac{6 x}{x^{2}-3 x} )
c) ( f(x)=\frac{6 x^{2}-18 x}{x} )
d) b and c

Explanation:

Step1: Analyze function \(a\)

For \(f(x)=\frac{x}{x^{2}-4}=\frac{x}{(x + 2)(x-2)}\), the denominator is zero when \(x=-2\) or \(x = 2\). So there are two vertical asymptotes \(x=-2\) and \(x = 2\).

Step2: Analyze function \(b\)

First, simplify \(f(x)=\frac{6x}{x^{2}-3x}=\frac{6x}{x(x - 3)}\). Cancel out the common factor \(x\) (note \(x
eq0\)). The function is equivalent to \(f(x)=\frac{6}{x - 3}\) for \(x
eq0\). The denominator is zero when \(x = 3\). Since \(x = 0\) is a removable discontinuity (a hole), there is one vertical asymptote \(x=3\).

Step3: Analyze function \(c\)

Simplify \(f(x)=\frac{6x^{2}-18x}{x}\). Cancel out the common factor \(x\) (\(x
eq0\)). We get \(f(x)=6x-18\) for \(x
eq0\). This is a linear function with a removable discontinuity at \(x = 0\), no vertical asymptote.

Answer:

B. \(f(x)=\frac{6x}{x^{2}-3x}\)