(2 points)

Let
$$f(x)=\frac{x-7}{\sqrt{6x^2+2x+4}}.$$

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): $y=\bar{0}$

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: $\underset{x\to \mathrm{\infty }}{lim}\frac{x-7}{\sqrt{6x^2+2x+4}}=\underset{x\to \mathrm{\infty }}{lim}\frac{x}{\sqrt{6x^2}}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{6}}=0$

Step 2 ：As x approaches negative infinity, we have: $\underset{x\to -\mathrm{\infty }}{lim}\frac{x-7}{\sqrt{6x^2+2x+4}}=\underset{x\to -\mathrm{\infty }}{lim}\frac{x}{\sqrt{6x^2}}=\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{\sqrt{6}}=0$

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: $\sqrt{6x^2+2x+4}=0$

Step 5 ：Squaring both sides, we get: $6x^2+2x+4=0$

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): $\boxed{{\displaystyle y=0}}$, Vertical asymptote(s): None