(2 points) Let \[ f(x)=\frac{x-7}{\sqrt{6 x^{2}+2 x+4}} . \] Find the horizontal and vertical asymptotes of $f(x)$. If there are no asymptotes of a given type, enter None. If there are more than one of a given type, list them separated by commas. Horizontal asymptote(s): $y=\overline{0}$ Vertical asymptote(s): $x=$ NONE

Step 1 ：Find the horizontal asymptotes by taking the limit of the function as x approaches positive and negative infinity. As x approaches positive infinity, we have: \(\lim_{{x \to \infty}} \frac{x-7}{\sqrt{6 x^{2}+2 x+4}} = \lim_{{x \to \infty}} \frac{x}{\sqrt{6 x^{2}}} = \lim_{{x \to \infty}} \frac{1}{\sqrt{6}} = 0\)

Step 2 ：As x approaches negative infinity, we have: \(\lim_{{x \to -\infty}} \frac{x-7}{\sqrt{6 x^{2}+2 x+4}} = \lim_{{x \to -\infty}} \frac{x}{\sqrt{6 x^{2}}} = \lim_{{x \to -\infty}} \frac{1}{\sqrt{6}} = 0\)

Step 3 ：So, the horizontal asymptote is y = 0.

Step 4 ：Find the vertical asymptotes by setting the denominator equal to zero and solving for x. Setting the denominator equal to zero, we have: \(\sqrt{6 x^{2}+2 x+4} = 0\)

Step 5 ：Squaring both sides, we get: \(6 x^{2}+2 x+4 = 0\)

Step 6 ：This is a quadratic equation, but it has no real roots (the discriminant is negative), so there are no vertical asymptotes.

Step 7 ：So, the answer is: Horizontal asymptote(s): \(\boxed{y = 0}\), Vertical asymptote(s): None