12. Find a rational function $f(x)$ that meets the following conditions: a) It has a vertical asymptote at $x=0$. b) It has a horizontal asymptote at $y=3$ c) $f(1)=2$ $f(x)=\frac{3 x^{2}-1}{x}$ $f(x)=\frac{6 x+1}{2 x}$ $f(x)=\frac{3 x-1}{x}$ $f(x)=\frac{-8 x^{2}+14}{3 x}$

Step 1 ：First, we need to find a function that meets the given conditions. The function must have a vertical asymptote at x=0, a horizontal asymptote at y=3, and f(1)=2.

Step 2 ：From the horizontal asymptote condition, we know that the degree of the numerator and the denominator must be the same, and the ratio of the leading coefficients must be 3. This is similar to the solution of QuestionA, where the horizontal asymptote was the ratio of the leading coefficients.

Step 3 ：Therefore, the function can be in the form of \(f(x) = \frac{3a x^n + ...}{b x^n + ...}\), where a and b are constants, and n is the degree of the polynomial.

Step 4 ：Next, we consider the vertical asymptote condition. A vertical asymptote at x=0 means that the denominator of the function must be 0 when x=0. Therefore, the denominator must contain the term x.

Step 5 ：So, the function can be further refined to the form \(f(x) = \frac{3a x^n + ...}{x (c x^{n-1} + ...)}\), where c is another constant.

Step 6 ：Finally, we consider the condition f(1)=2. This means that when we substitute x=1 into the function, the result should be 2.

Step 7 ：By substitifying x=1 into the function, we get \(2 = \frac{3a + ...}{c + ...}\).

Step 8 ：Solving this equation will give us the values of a and c, and thus the specific form of the function.

Step 9 ：Among the given options, only \(f(x) = \frac{3x-1}{x}\) meets all these conditions. So, the answer is \(\boxed{f(x) = \frac{3x-1}{x}}\).