For.the given rational function $$ f(x)=\frac{12-4 x^{2}}{x^{2}-4} $$ (A) Find any intercepts for the graph. (B) Find any vertical and horizontal asymptotes for the graph. (C) Graph the function. A. There is one vertical asymptote. Its equation is $$ \square $$ B. There are two vertical asymptotes. In order from left to right on the graph of the function, the equations of the vertical asymptotes are (B) Identify the vertical asymptotes, if there are any. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. C. There are three vertical asymptotes, In order from left to right on the graph of the function, the equations of the vertical asymptotes are D. There are no vertical asymptotes. A. and A. A. $$ \square $$ B $$ \square $$ B. There is no horizontal asymptote.

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**Step 1. Simplify the Function**

We start with
$$
f(x)=\frac{12-4x^2}{x^2-4}.
$$
Factor the numerator and denominator:
- Numerator: $$12-4x^2=4(3-x^2)$$. Notice that this can also be written as
  $$
  4(3-x^2)=-4(x^2-3).
  $$
- Denominator: $$x^2-4=(x-2)(x+2)$$.

Thus, the function becomes
$$
f(x)=\frac{-4(x^2-3)}{(x-2)(x+2)}.
$$

---

**Step 2. Find the Intercepts**

*X-intercepts:* Set the numerator equal to zero (provided the denominator is not zero).
$$
-4(x^2-3)=0\quad\Longrightarrow\quad x^2-3=0\quad\Longrightarrow\quad x^2=3\quad\Longrightarrow\quad x=\pm \sqrt{3}.
$$
So, the x-intercepts are $$\left(-\sqrt{3},\,0\right)$$ and $$\left(\sqrt{3},\,0\right)$$.

*Y-intercept:* Set $$x=0$$.
$$
f(0)=\frac{12-4(0)^2}{0^2-4}=\frac{12}{-4}=-3.
$$
Thus, the y-intercept is $$(0,-3)$$.

---

**Step 3. Find the Vertical and Horizontal Asymptotes**

*Vertical Asymptotes:* Find the values for which the denominator is zero, and verify these are not removable discontinuities.

Set
$$
x^2-4=0\quad\Longrightarrow\quad (x-2)(x+2)=0\quad\Longrightarrow\quad x=2\quad \text{or}\quad x=-2.
$$
Check that these do not cancel with the numerator:
- For $$x=2$$: $$12-4(2)^2=12-16=-4\neq 0$$.
- For $$x=-2$$: $$12-4(-2)^2=12-16=-4\neq 0$$.

Thus, there are vertical asymptotes at
$$
x=-2 \quad \text{and} \quad x=2.
$$

*Horizontal Asymptote:* Compare the degrees of the numerator and denominator. Both are degree 2.
- The leading term in the numerator is $$-4x^2$$.
- The leading term in the denominator is $$x^2$$.

Thus, the horizontal asymptote is given by
$$
y=\frac{-4x^2}{x^2}=-4.
$$

---

**Step 4. Graphing the Function**

Key features for the graph:
- **X-intercepts:** $$\left(-\sqrt{3},\,0\right)$$ and $$\left(\sqrt{3},\,0\right)$$.
- **Y-intercept:** $$(0,-3)$$.
- **Vertical Asymptotes:** $$x=-2$$ and $$x=2$$.
- **Horizontal Asymptote:** $$y=-4$$.

When graphing:
- Plot the intercepts.
- Draw dashed vertical lines at $$x=-2$$ and $$x=2$$ to indicate the vertical asymptotes.
- Draw a dashed horizontal line at $$y=-4$$ to indicate the horizontal asymptote.
- The graph will approach but not cross these asymptotes. Sketch the behavior in the intervals separated by the vertical asymptotes.

---

**Answer to the Given Multiple Choice Question**

- **Vertical Asymptotes:** There are two vertical asymptotes. In order, from left to right, they are
  $$
x=-2\quad \text{and}\quad x=2.
  $$
- **Horizontal Asymptote:** The horizontal asymptote is
  $$
y=-4.
  $$

Thus, the correct selections are:

- For vertical asymptotes: **B. There are two vertical asymptotes. In order from left to right on the graph of the function, the equations of the vertical asymptotes are $$x=-2$$ and $$x=2$$.**
- For the horizontal asymptote: **$$y=-4$$.**
There are two vertical asymptotes at $$x = -2$$ and $$x = 2$$, and a horizontal asymptote at $$y = -4$$.

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