question 9 (1 point)\nwhich of the following functions has exactly one vertical asymptote when graphed?\n a)…

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

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

Answer

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\neq0)). The function is equivalent to (f(x)=\frac{6}{x - 3}) for (x\neq0). 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\neq0)). We get (f(x)=6x-18) for (x\neq0). This is a linear function with a removable discontinuity at (x = 0), no vertical asymptote.

Answer:

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