Answer
To graph the rational function \( f(x) = \frac{2x^{2} - 8x}{x^{2} - 6x + 8} \), follow these steps:
1. **Simplify the Function:**
\[
f(x) = \frac{2x(x - 4)}{(x - 2)(x - 4)} = \frac{2x}{x - 2}, \quad x \neq 4
\]
There's a hole at \(x = 4\).
2. **Determine the Domain:**
\[
x \neq 2, \quad x \neq 4
\]
3. **Find Intercepts:**
- **Y-Intercept:** \((0, 0)\)
- **X-Intercept:** \((0, 0)\)
4. **Identify Asymptotes:**
- **Vertical Asymptote:** \(x = 2\)
- **Horizontal Asymptote:** \(y = 2\)
5. **Plot Additional Points:**
- \((1, -2)\)
- \((3, 6)\)
- \((5, \approx 3.33)\)
- \((-1, \approx 0.67)\)
6. **Graph Behavior:**
- Approaches \(y = 2\) as \(x\) approaches \(\pm\infty\).
- Approaches \(-\infty\) as \(x\) approaches \(2\) from the left.
- Approaches \(+\infty\) as \(x\) approaches \(2\) from the right.
- Has a hole at \((4, 4)\).
**Summary:**
- **Domain:** All real numbers except \(x = 2\) and \(x = 4\).
- **Intercepts:** \((0, 0)\).
- **Asymptotes:** Vertical at \(x = 2\) and horizontal at \(y = 2\).
- **Hole:** At \((4, 4)\).
- **Additional Points:** \((1, -2)\), \((3, 6)\), \((5, \approx 3.33)\), \((-1, \approx 0.67)\).
Plot these points and sketch the asymptotes to accurately graph the function.
Solution
To graph the rational function
\[
f(x) = \frac{2x^{2} - 8x}{x^{2} - 6x + 8},
\]
follow these steps to understand its key features, including intercepts, asymptotes, and behavior around critical points.
### 1. **Simplify the Function**
First, factor both the numerator and the denominator to see if any terms cancel out:
- **Numerator:**
\[
2x^{2} - 8x = 2x(x - 4)
\]
- **Denominator:**
\[
x^{2} - 6x + 8 = (x - 2)(x - 4)
\]
After factoring, the function becomes:
\[
f(x) = \frac{2x(x - 4)}{(x - 2)(x - 4)} = \frac{2x}{x - 2}, \quad \text{for } x \neq 4
\]
**Note:** The factor \((x - 4)\) cancels out, indicating a **removable discontinuity** (a hole) at \(x = 4\).
### 2. **Determine the Domain**
The function is undefined where the denominator is zero:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
**Domain:** All real numbers except \(x = 2\) and \(x = 4\).
\[
\text{Domain: } \mathbb{R} \setminus \{2, 4\}
\]
### 3. **Find Intercepts**
- **Y-Intercept:**
Set \(x = 0\):
\[
f(0) = \frac{2(0)}{0 - 2} = 0
\]
**Y-Intercept:** \((0, 0)\)
- **X-Intercept:**
Set the numerator equal to zero:
\[
2x = 0 \quad \Rightarrow \quad x = 0
\]
**X-Intercept:** \((0, 0)\)
### 4. **Identify Asymptotes**
- **Vertical Asymptote:**
The function has a vertical asymptote where the denominator is zero **and** the numerator is non-zero.
\[
x = 2
\]
**Vertical Asymptote:** \(x = 2\)
- **Horizontal Asymptote:**
Compare the degrees of the numerator and the denominator.
- Degree of numerator after simplification: 1
- Degree of denominator after simplification: 1
Since they're equal, the horizontal asymptote is the ratio of the leading coefficients:
\[
y = \frac{2}{1} = 2
\]
**Horizontal Asymptote:** \(y = 2\)
### 5. **Locate the Hole**
There is a hole at \(x = 4\). To find its coordinates, substitute \(x = 4\) into the simplified function:
\[
f(4) = \frac{2(4)}{4 - 2} = \frac{8}{2} = 4
\]
**Hole:** \((4, 4)\)
### 6. **Plot Additional Points**
For a more accurate graph, calculate \(f(x)\) at several values of \(x\):
| \(x\) | \(f(x) = \frac{2x}{x - 2}\) |
|-------|-----------------------------|
| 1 | \(\frac{2(1)}{1 - 2} = -2\) |
| 3 | \(\frac{2(3)}{3 - 2} = 6\) |
| 5 | \(\frac{2(5)}{5 - 2} \approx 3.33\) |
| -1 | \(\frac{2(-1)}{-1 - 2} = \frac{-2}{-3} \approx 0.67\) |
**Additional Points:**
- \((1, -2)\)
- \((3, 6)\)
- \((5, \approx 3.33)\)
- \((-1, \approx 0.67)\)
### 7. **Analyze Behavior Near Asymptotes**
- **As \(x\) approaches \(2\) from the left (\(x \to 2^-\)):**
\[
f(x) = \frac{2x}{x - 2} \to -\infty
\]
- **As \(x\) approaches \(2\) from the right (\(x \to 2^+\)):**
\[
f(x) = \frac{2x}{x - 2} \to +\infty
\]
- **As \(x\) approaches \(\pm\infty\):**
\[
f(x) \to 2
\]
### 8. **Graph Overview**
Based on the above analysis, here's a description of the graph:
- **Intercepts:** The graph passes through the origin \((0, 0)\).
- **Hole:** There's a hole at \((4, 4)\).
- **Asymptotes:** The graph has a vertical asymptote at \(x = 2\) and a horizontal asymptote at \(y = 2\).
- **Behavior:**
- **For \(x < 2\):** The graph approaches the horizontal asymptote \(y = 2\) from below as \(x\) moves to \(-\infty\) and dives down to \(-\infty\) as it nears \(x = 2\).
- **For \(2 < x < 4\):** The graph rises from \(+\infty\) at \(x = 2\) and passes through \((3, 6)\), approaching the hole at \((4, 4)\).
- **For \(x > 4\):** After the hole, the graph continues from \((4, 4)\), passes through \((5, \approx 3.33)\), and approaches the horizontal asymptote \(y = 2\) from above as \(x\) increases to \(+\infty\).
### 9. **Visual Representation**
While I can't provide a visual graph directly, here's an ASCII approximation to help you visualize the function's behavior:
```
y
|
| * (3,6)
| /|
| / |
| / |
| / |
|-----*----*----*--- y=2 (Horizontal Asymptote)
| (0,0) | \
| | \
| | \
| | * (5, ~3.33)
| |
| |
| |
| x=2 (Vertical Asymptote)
|
```
**Note:** The asterisks (*) represent key points like intercepts and plotted points. The dashed line indicates the horizontal asymptote \(y = 2\), and the vertical asymptote at \(x = 2\) is shown as a solid line. The hole at \((4, 4)\) would be indicated by an open circle on a detailed graph.
### 10. **Summary**
- **Domain:** \(\mathbb{R} \setminus \{2, 4\}\)
- **X-Intercept:** \((0, 0)\)
- **Y-Intercept:** \((0, 0)\)
- **Vertical Asymptote:** \(x = 2\)
- **Horizontal Asymptote:** \(y = 2\)
- **Hole:** \((4, 4)\)
- **Additional Points:** \((1, -2)\), \((3, 6)\), \((5, \approx 3.33)\), \((-1, \approx 0.67)\)
By plotting these points and considering the asymptotic behavior, you can sketch the graph of the function accurately.
Answered by UpStudy AI and reviewed by a Professional Tutor
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