Graph the function. Plot all necessary asymptotes. Plot at least 4 points. For vertical asymptotes, make sure there are at least two points on each side. $$ v(x)=\frac{2 x^{2}-3 x-2}{x^{2}+1} $$

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**Step 1. Identify the Domain**

The function is  
$$
v(x)=\frac{2x^2-3x-2}{x^2+1}.
$$
Since the denominator is  
$$
x^2+1,
$$
which is always positive (because $$x^2 \ge 0$$ and thus $$x^2+1 \ge 1$$), there are no restrictions on $$x$$. Hence, the domain is all real numbers, $$\mathbb{R}$$.

---

**Step 2. Determine the Asymptotes**

*Vertical Asymptotes*  
Vertical asymptotes typically occur when the denominator equals zero. Since  
$$
x^2+1=0
$$
has no real solutions, there are **no vertical asymptotes**.

*Horizontal Asymptote*  
For a rational function where the degrees of the numerator and denominator are the same, the horizontal asymptote is determined by the ratio of the leading coefficients. Here, the numerator has a leading coefficient $$2$$ and the denominator has a leading coefficient $$1$$. Therefore, the horizontal asymptote is  
$$
y= \frac{2}{1}=2.
$$

---

**Step 3. Find the Intercepts**

*x-intercepts (where $$v(x)=0$$)*  
Set the numerator equal to zero:
$$
2x^2-3x-2=0.
$$
This quadratic can be solved using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2-4ac}}{2a},
$$
with $$a=2$$, $$b=-3$$, and $$c=-2$$. Thus,
$$
x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9+16}}{4} = \frac{3 \pm 5}{4}.
$$

So, the solutions are:
$$
x = \frac{3+5}{4} = 2 \quad \text{and} \quad x = \frac{3-5}{4} = -\frac{1}{2}.
$$
Therefore, the x-intercepts are at  
$$\left(2,0\right)$$ and $$\left(-\frac{1}{2},0\right)$$.

*y-intercept (where $$x=0$$)*  
Evaluate:
$$
v(0)=\frac{2(0)^2-3(0)-2}{(0)^2+1}=\frac{-2}{1}=-2.
$$
So, the y-intercept is at $$(0,-2)$$.

---

**Step 4. Evaluate Additional Points**

To understand the behavior of the function, compute a few more points.

1. For $$x=1$$:
   $$
   v(1)=\frac{2(1)^2-3(1)-2}{1^2+1}=\frac{2-3-2}{2}=\frac{-3}{2}=-1.5.
   $$
   So, point: $$(1,-1.5)$$.

2. For $$x=-1$$:
   $$
   v(-1)=\frac{2(-1)^2-3(-1)-2}{(-1)^2+1}=\frac{2+3-2}{1+1}=\frac{3}{2}=1.5.
   $$
   So, point: $$(-1,1.5)$$.

3. For $$x=3$$:
   $$
   v(3)=\frac{2(3)^2-3(3)-2}{(3)^2+1}=\frac{18-9-2}{9+1}=\frac{7}{10}=0.7.
   $$
   So, point: $$(3, 0.7)$$.

4. For $$x=-2$$:
   $$
   v(-2)=\frac{2(-2)^2-3(-2)-2}{(-2)^2+1}=\frac{8+6-2}{4+1}=\frac{12}{5}=2.4.
   $$
   So, point: $$(-2, 2.4)$$.

---

**Step 5. Graph Behavior and Asymptotic Approach**

- **Horizontal Asymptote $$y=2$$:**  
  As $$x \to \pm\infty$$, the function approaches $$y=2$$.

- **No Vertical Asymptotes:**  
  The function is continuous for all $$x$$.

---

**Step 6. Sketch the Graph**

1. **Plot asymptote:**  
   Draw a dashed horizontal line at $$y=2$$.

2. **Plot intercepts:**  
   - x-intercepts: $$\left(2,0\right)$$ and $$\left(-\frac{1}{2},0\right)$$.  
   - y-intercept: $$(0,-2)$$.

3. **Plot Additional Points:**  
   - $$(1, -1.5)$$  
   - $$(-1, 1.5)$$  
   - $$(3, 0.7)$$  
   - $$(-2, 2.4)$$

4. **Graph Behavior:**  
   - For very negative $$x$$, the function approaches $$y=2$$ from below or above as indicated by evaluated points.
   - For very positive $$x$$, the function also approaches $$y=2$$.

5. **Smooth Curve:**  
   Sketch a smooth curve that passes through the plotted points and approaches the horizontal asymptote $$y=2$$ as $$x$$ tends to $$\pm\infty$$.

Since there are no vertical asymptotes, there is no requirement to plot any points on either side of a vertical line.

---

**Conclusion**

When graphing  
$$
v(x)=\frac{2x^{2}-3x-2}{x^{2}+1},
$$
remember:
- Domain: All real numbers.
- Horizontal asymptote at $$y=2$$.
- No vertical asymptotes.
- X-intercepts at $$x=2$$ and $$x=-\frac{1}{2}$$.
- Y-intercept at $$(0,-2)$$.
- Additional points such as $$(1,-1.5)$$, $$(-1,1.5)$$, $$(3,0.7)$$, and $$(-2,2.4)$$ help shape the graph.

Plot the function with these details to capture its overall behavior.
To graph the function $$ v(x) = \frac{2x^{2} - 3x - 2}{x^{2} + 1} $$, follow these steps: 1. **Domain**: All real numbers since the denominator $$ x^{2} + 1 $$ is always positive. 2. **Asymptotes**: - **Horizontal Asymptote**: $$ y = 2 $$ (since the degrees of numerator and denominator are equal, and the leading coefficients ratio is $$ \frac{2}{1} = 2 $$). - **Vertical Asymptotes**: None, as the denominator never equals zero. 3. **Intercepts**: - **X-intercepts**: $$ x = 2 $$ and $$ x = -\frac{1}{2} $$. - **Y-intercept**: $$ (0, -2) $$. 4. **Additional Points**: - $$ (1, -1.5) $$ - $$ (-1, 1.5) $$ - $$ (3, 0.7) $$ - $$ (-2, 2.4) $$ 5. **Graph Behavior**: - The function approaches the horizontal asymptote $$ y = 2 $$ as $$ x $$ approaches both positive and negative infinity. - Plot the intercepts and additional points to sketch the curve, ensuring it approaches $$ y = 2 $$ without crossing it. 6. **Final Graph**: Draw a horizontal dashed line at $$ y = 2 $$, plot all intercepts and points, and sketch a smooth curve that reflects the function's behavior based on the asymptote and intercepts. This summary provides a clear, simplified overview of graphing the given function without making judgments.

Graph the function. Plot all necessary asymptotes. Plot at least 4 points. For vertical asymptotes, make sure there are at least two points on each side. \[ v(x)=\frac{2 x^{2}-3 x-2}{x^{2}+1} \]

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