Summarize all pertinent information obtained by applying the graphing strategy and sketch the graph of $$ y=f(x) $$. $$ f(x)=\frac{x^{2}+10 x+24}{x^{2}+8 x+16} $$ Find any horizontal asymptotes of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one horizontal asymptote, $$ \square $$ (Type an equation.) B. The function has two horizontal asymptotes. The top asymptote is $$ \square $$ and the bottom asymptote is $$ \square $$ . (Type equations.) C. There are no horizontal asymptotes. Find any vertical asymptotes of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one vertical asymptote, $$ \square $$ (Type an equation.) B. The function has two vertical asymptotes. The leftmost asymptote is $$ \square $$ and the rightmost asymptote is $$ \square $$ . (Type equations.) C. There are no vertical asymptotes.

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UpStudy AI · Accepted Answer

**Step 1. Factor the numerator and denominator.**

We have
$$
f(x)=\frac{x^2+10x+24}{x^2+8x+16}.
$$
Factor the quadratic expressions:
$$
x^2+10x+24 = (x+4)(x+6),
$$
$$
x^2+8x+16 = (x+4)^2.
$$
Thus,
$$
f(x)=\frac{(x+4)(x+6)}{(x+4)^2},
$$
with the note that $$x\neq -4$$ since the original denominator must be nonzero.

---

**Step 2. Simplify the expression and note the discontinuity.**

Cancel a common factor $$(x+4)$$ (keeping in mind that $$x=-4$$ is not in the domain):
$$
f(x)=\frac{x+6}{x+4}\quad \text{for } x\neq -4.
$$
Even after cancellation, the factor in the denominator remains, so there is still a discontinuity at $$x=-4$$.

---

**Step 3. Determine the horizontal asymptote.**

For rational functions where the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients.

The simplified form $$\frac{x+6}{x+4}$$ has leading coefficients 1 (in the numerator and denominator).

Thus,
$$
\lim_{x\to \pm\infty} f(x)=\frac{1}{1}=1.
$$
There is one horizontal asymptote:
$$
y=1.
$$

*Answer choice for horizontal asymptote:*  
**A. The function has one horizontal asymptote, $$y=1$$.**

---

**Step 4. Determine the vertical asymptote.**

Set the denominator of the simplified function equal to zero:
$$
x+4=0\quad \Rightarrow \quad x = -4.
$$
Since the numerator, evaluated at $$x=-4$$, is $$ (-4)+6=2 \neq 0$$, the discontinuity at $$x=-4$$ is a vertical asymptote.

*Answer choice for vertical asymptote:*  
**A. The function has one vertical asymptote, $$x=-4$$.**

---

**Step 5. Additional characteristics from the graphing strategy.**

- **Domain:** $$x\in\mathbb{R}$$ except $$x=-4$$.
- **$$x$$-intercepts:** Set $$f(x)=0$$. In the factored form $$\frac{(x+4)(x+6)}{(x+4)^2}$$, the numerator is zero when $$x=-4$$ or $$x=-6$$. However, $$x=-4$$ is not in the domain. Therefore, there is an $$x$$-intercept at $$x=-6$$, i.e., at $$(-6, 0)$$.
- **$$y$$-intercept:** Set $$x=0$$:
  $$
  f(0)=\frac{0^2+10(0)+24}{0^2+8(0)+16}=\frac{24}{16}=\frac{3}{2}.
  $$
- **Sketch Information:**  
  - The graph approaches the line $$y=1$$ as $$x\to\pm\infty$$.  
  - There is a vertical asymptote at $$x=-4$$.  
  - The function has an $$x$$-intercept at $$(-6,0)$$ and a $$y$$-intercept at $$\left(0,\frac{3}{2}\right)$$.

---

**Final Answer:**

- **Horizontal Asymptote:**  
  **A.** The function has one horizontal asymptote, $$y=1$$.

- **Vertical Asymptote:**  
  **A.** The function has one vertical asymptote, $$x=-4$$.
**Horizontal Asymptote:** A. The function has one horizontal asymptote, $$y=1$$. **Vertical Asymptote:** A. The function has one vertical asymptote, $$x=-4$$.

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