Summarize all pertinent information obtained by applying the graphing strategy and sketch the graph of $$ \mathrm{y}=\mathrm{f}(\mathrm{x}) $$. $$ \mathrm{f}(\mathrm{x})=\frac{x^{2}+10 \mathrm{x}+24}{x^{2}+8 \mathrm{x}+16} $$ Find any oblique asymptotes of $$ \mathrm{f}(\mathrm{x}) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one oblique asymptote, (Type an equation.) B. The function has two different oblique asymptotes. The oblique asymptote as x approaches - $$ \infty $$ is (Type equations.) C. There are no oblique asymptotes. Find the oblique asymptote as $$ x $$ approaches $$ \infty $$ is A. The function is increasing on $$ \square $$. It is never decreasing. (Type an exact answer. Type your answer in interval notation. Use a comma to separate answers as needed.) B. The function is increasing on $$ \square $$. It is decreasing on $$ \square $$. (Type exact answers. Type your answers in interval notation. Use commas to separate answers as needed.) C. The function is decreasing on $$ \square $$. It is never increasing. Tvne an exart answer Tvne vour answer in interval notation Isse a romma to senarate answers as needed)

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UpStudy AI · Accepted Answer

To analyze the function $$ f(x) = \frac{x^2 + 10x + 24}{x^2 + 8x + 16} $$, we will follow these steps:

1. **Identify the oblique asymptotes**.
2. **Determine the intervals of increase and decrease**.
3. **Sketch the graph based on the findings**.

### Step 1: Finding Oblique Asymptotes

Oblique asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. In this case, both the numerator and denominator are polynomials of degree 2, so we will not have oblique asymptotes. Instead, we will check for horizontal asymptotes.

To find horizontal asymptotes, we can analyze the leading coefficients of the numerator and denominator:

$$
\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2 + 10x + 24}{x^2 + 8x + 16} = \frac{1}{1} = 1
$$

Thus, there is a horizontal asymptote at $$ y = 1 $$.

### Conclusion for Oblique Asymptotes
C. There are no oblique asymptotes.

### Step 2: Finding Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, we need to find the derivative $$ f'(x) $$ and analyze its sign.

Using the quotient rule:

$$
f'(x) = \frac{(g(x)h'(x) - h(x)g'(x))}{(h(x))^2}
$$

where $$ g(x) = x^2 + 10x + 24 $$ and $$ h(x) = x^2 + 8x + 16 $$.

Calculating the derivatives:

- $$ g'(x) = 2x + 10 $$
- $$ h'(x) = 2x + 8 $$

Now applying the quotient rule:

$$
f'(x) = \frac{(x^2 + 8x + 16)(2x + 10) - (x^2 + 10x + 24)(2x + 8)}{(x^2 + 8x + 16)^2}
$$

We will simplify this expression and find the critical points by setting $$ f'(x) = 0 $$.

### Step 3: Analyze the Sign of the Derivative

Let's calculate $$ f'(x) $$ and find the critical points.

I will now compute $$ f'(x) $$ and find the intervals of increase and decrease.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$\left(x^{2}+8x+16\right)\left(2x+10\right)-\left(x^{2}+10x+24\right)\left(2x+8\right)=0$$
- step1: Multiply the terms:
  $$\left(x^{2}+8x+16\right)\left(2x+10\right)+\left(-x^{2}-10x-24\right)\left(2x+8\right)=0$$
- step2: Expand the expression:
  $$-2x^{2}-16x-32=0$$
- step3: Factor the expression:
  $$-2\left(x+4\right)^{2}=0$$
- step4: Divide the terms:
  $$\left(x+4\right)^{2}=0$$
- step5: Simplify the expression:
  $$x+4=0$$
- step6: Move the constant to the right side:
  $$x=0-4$$
- step7: Remove 0:
  $$x=-4$$
The critical point found is $$ x = -4 $$. Now, we will analyze the sign of $$ f'(x) $$ around this critical point to determine the intervals of increase and decrease.

### Step 4: Test Intervals

We will test the intervals $$ (-\infty, -4) $$ and $$ (-4, \infty) $$:

1. **For $$ x < -4 $$** (e.g., $$ x = -5 $$):
   - Substitute $$ x = -5 $$ into $$ f'(x) $$ to check the sign.

2. **For $$ x > -4 $$** (e.g., $$ x = 0 $$):
   - Substitute $$ x = 0 $$ into $$ f'(x) $$ to check the sign.

Let's calculate $$ f'(-5) $$ and $$ f'(0) $$ to determine the behavior of the function in these intervals.

### Step 5: Calculate $$ f'(-5) $$ and $$ f'(0) $$

We will substitute these values into the derivative expression we derived earlier.

First, we will calculate $$ f'(-5) $$ and $$ f'(0) $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$\left(x^{2}+8x+16\right)\left(2x+10\right)-\left(x^{2}+10x+24\right)\left(2x+8\right)$$
- step1: Substitute:
  $$\left(0^{2}+8\times 0+16\right)\left(2\times 0+10\right)-\left(0^{2}+10\times 0+24\right)\left(2\times 0+8\right)$$
- step2: Calculate:
  $$\left(0+8\times 0+16\right)\left(2\times 0+10\right)-\left(0^{2}+10\times 0+24\right)\left(2\times 0+8\right)$$
- step3: Multiply:
  $$\left(0+0+16\right)\left(2\times 0+10\right)-\left(0^{2}+10\times 0+24\right)\left(2\times 0+8\right)$$
- step4: Remove 0:
  $$16\left(2\times 0+10\right)-\left(0^{2}+10\times 0+24\right)\left(2\times 0+8\right)$$
- step5: Multiply:
  $$16\left(0+10\right)-\left(0^{2}+10\times 0+24\right)\left(2\times 0+8\right)$$
- step6: Remove 0:
  $$16\times 10-\left(0^{2}+10\times 0+24\right)\left(2\times 0+8\right)$$
- step7: Calculate:
  $$16\times 10-\left(0+10\times 0+24\right)\left(2\times 0+8\right)$$
- step8: Multiply:
  $$16\times 10-\left(0+0+24\right)\left(2\times 0+8\right)$$
- step9: Remove 0:
  $$16\times 10-24\left(2\times 0+8\right)$$
- step10: Multiply:
  $$16\times 10-24\left(0+8\right)$$
- step11: Remove 0:
  $$16\times 10-24\times 8$$
- step12: Multiply the numbers:
  $$160-24\times 8$$
- step13: Multiply the numbers:
  $$160-192$$
- step14: Subtract the numbers:
  $$-32$$
Substitute $$ x=-5 $$ into the expression $$ (x^2 + 8x + 16)(2x + 10) - (x^2 + 10x + 24)(2x + 8) $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$\left(x^{2}+8x+16\right)\left(2x+10\right)-\left(x^{2}+10x+24\right)\left(2x+8\right)$$
- step1: Substitute:
  $$\left(\left(-5\right)^{2}+8\left(-5\right)+16\right)\left(2\left(-5\right)+10\right)-\left(\left(-5\right)^{2}+10\left(-5\right)+24\right)\left(2\left(-5\right)+8\right)$$
- step2: Multiply the numbers:
  $$\left(\left(-5\right)^{2}-40+16\right)\left(2\left(-5\right)+10\right)-\left(\left(-5\right)^{2}+10\left(-5\right)+24\right)\left(2\left(-5\right)+8\right)$$
- step3: Calculate:
  $$1\times \left(2\left(-5\right)+10\right)-\left(\left(-5\right)^{2}+10\left(-5\right)+24\right)\left(2\left(-5\right)+8\right)$$
- step4: Multiply the numbers:
  $$1\times \left(-10+10\right)-\left(\left(-5\right)^{2}+10\left(-5\right)+24\right)\left(2\left(-5\right)+8\right)$$
- step5: Add the numbers:
  $$1\times 0-\left(\left(-5\right)^{2}+10\left(-5\right)+24\right)\left(2\left(-5\right)+8\right)$$
- step6: Multiply the numbers:
  $$1\times 0-\left(\left(-5\right)^{2}-50+24\right)\left(2\left(-5\right)+8\right)$$
- step7: Calculate:
  $$1\times 0-\left(-1\right)\left(2\left(-5\right)+8\right)$$
- step8: Remove the parentheses:
  $$1\times 0-\left(-\left(2\left(-5\right)+8\right)\right)$$
- step9: Multiply the numbers:
  $$1\times 0-\left(-\left(-10+8\right)\right)$$
- step10: Add the numbers:
  $$1\times 0-\left(-\left(-2\right)\right)$$
- step11: Multiply:
  $$0-\left(-\left(-2\right)\right)$$
- step12: Remove the parentheses:
  $$0-2$$
- step13: Remove 0:
  $$-2$$
The results of the derivative evaluations are as follows:

- $$ f'(-5) = -2 $$ (negative)
- $$ f'(0) = -32 $$ (negative)

### Step 6: Determine Intervals of Increase and Decrease

Since $$ f'(-5) < 0 $$ and $$ f'(0) < 0 $$, the function is decreasing in both intervals:

- **Decreasing on**: $$ (-\infty, -4) $$ and $$ (-4, \infty) $$

### Summary of Findings

1. **Oblique Asymptotes**: 
   - C. There are no oblique asymptotes.

2. **Intervals of Increase and Decrease**:
   - B. The function is increasing on $$ \emptyset $$ (no intervals) and decreasing on $$ (-\infty, -4) $$ and $$ (-4, \infty) $$.

### Sketching the Graph

- The function has a horizontal asymptote at $$ y = 1 $$.
- The function is decreasing everywhere except at the critical point $$ x = -4 $$, where it may change behavior.
- The graph approaches the horizontal asymptote as $$ x $$ approaches $$ \infty $$ and $$ -\infty $$.

### Final Answer

- **Oblique Asymptotes**: C. There are no oblique asymptotes.
- **Intervals of Increase and Decrease**: The function is increasing on $$ \emptyset $$ and decreasing on $$ (-\infty, -4) $$ and $$ (-4, \infty) $$.
There are no oblique asymptotes. The function is decreasing on $$ (-\infty, -4) $$ and $$ (-4, \infty) $$.

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