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Sagot :
To graph the function [tex]\( f(x) = \frac{x^2 - 2x + 6}{x - 3} \)[/tex], we should follow these steps:
1. Identify the domain:
The denominator cannot be zero, so we set [tex]\( x - 3 = 0 \)[/tex]. This gives [tex]\( x = 3 \)[/tex] as a point of discontinuity. The function is not defined at [tex]\( x = 3 \)[/tex].
2. Find the asymptotes:
- Vertical Asymptote: Since the function is undefined at [tex]\( x = 3 \)[/tex], there is a vertical asymptote at [tex]\( x = 3 \)[/tex].
- Oblique Asymptote: Because the degree of the numerator (2) is one more than the degree of the denominator (1), there is an oblique asymptote. To find it, we perform polynomial long division:
[tex]\( x^2 - 2x + 6 \div (x - 3) \)[/tex]:
[tex]\[ x^2 - 2x + 6 = (x - 3)(x + 1) + 9 \][/tex]
This simplifies to:
[tex]\[ \frac{x^2 - 2x + 6}{x - 3} = x + 1 + \frac{9}{x - 3} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( \frac{9}{x - 3} \)[/tex] approaches zero. Hence, the oblique asymptote is:
[tex]\[ y = x + 1 \][/tex]
3. Plot key points:
Evaluate the function at various points to understand its behavior away from [tex]\( x = 3 \)[/tex]:
- [tex]\( f(0) = \frac{0^2 - 2 \cdot 0 + 6}{0 - 3} = -2 \)[/tex]
- [tex]\( f(1) = \frac{1^2 - 2 \cdot 1 + 6}{1 - 3} = -\frac{2.5}{2} \)[/tex]
- [tex]\( f(2) = \frac{2^2 - 2 \cdot 2 + 6}{2 - 3} = -2 \)[/tex]
- [tex]\( f(4) = \frac{4^2 - 2 \cdot 4 + 6}{4 - 3} = 6 \)[/tex]
- [tex]\( f(5) = \frac{5^2 - 2 \cdot 5 + 6}{5 - 3} = 8 \)[/tex]
4. Sketch the asymptotes and key points:
Draw the vertical asymptote at [tex]\( x = 3 \)[/tex] and the oblique asymptote [tex]\( y = x + 1 \)[/tex].
5. Plot the function:
- As [tex]\( x \)[/tex] approaches 3 from the left ([tex]\( x \to 3^- \)[/tex]), [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches 3 from the right ([tex]\( x \to 3^+ \)[/tex]), [tex]\( f(x) \to \infty \)[/tex].
6. Combine information for the graph:
- Draw the curve approaching [tex]\(-\infty\)[/tex] as [tex]\( x \)[/tex] approaches 3 from the left, and approaching [tex]\( \infty \)[/tex] from the right side of [tex]\( x = 3 \)[/tex].
- Use the key points and the fact that the function approaches the oblique asymptote [tex]\( y = x + 1 \)[/tex] as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex].
The final graph should show a clear vertical asymptote at [tex]\( x = 3 \)[/tex], an oblique asymptote with the equation [tex]\( y = x + 1 \)[/tex], and the function’s behavior across key points and general trend. Here's what it should roughly look like:
```
|
10 |
|
5 |
|
0 |---- (vertical asymptote at x=3)
|
-5 |
|
-10 |______________________________
-5 0 5 10
```
Note: Graph is not to scale and drawn with limited tools. For precise visualization, graphing tools or plotting in software would be more accurate.
1. Identify the domain:
The denominator cannot be zero, so we set [tex]\( x - 3 = 0 \)[/tex]. This gives [tex]\( x = 3 \)[/tex] as a point of discontinuity. The function is not defined at [tex]\( x = 3 \)[/tex].
2. Find the asymptotes:
- Vertical Asymptote: Since the function is undefined at [tex]\( x = 3 \)[/tex], there is a vertical asymptote at [tex]\( x = 3 \)[/tex].
- Oblique Asymptote: Because the degree of the numerator (2) is one more than the degree of the denominator (1), there is an oblique asymptote. To find it, we perform polynomial long division:
[tex]\( x^2 - 2x + 6 \div (x - 3) \)[/tex]:
[tex]\[ x^2 - 2x + 6 = (x - 3)(x + 1) + 9 \][/tex]
This simplifies to:
[tex]\[ \frac{x^2 - 2x + 6}{x - 3} = x + 1 + \frac{9}{x - 3} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the term [tex]\( \frac{9}{x - 3} \)[/tex] approaches zero. Hence, the oblique asymptote is:
[tex]\[ y = x + 1 \][/tex]
3. Plot key points:
Evaluate the function at various points to understand its behavior away from [tex]\( x = 3 \)[/tex]:
- [tex]\( f(0) = \frac{0^2 - 2 \cdot 0 + 6}{0 - 3} = -2 \)[/tex]
- [tex]\( f(1) = \frac{1^2 - 2 \cdot 1 + 6}{1 - 3} = -\frac{2.5}{2} \)[/tex]
- [tex]\( f(2) = \frac{2^2 - 2 \cdot 2 + 6}{2 - 3} = -2 \)[/tex]
- [tex]\( f(4) = \frac{4^2 - 2 \cdot 4 + 6}{4 - 3} = 6 \)[/tex]
- [tex]\( f(5) = \frac{5^2 - 2 \cdot 5 + 6}{5 - 3} = 8 \)[/tex]
4. Sketch the asymptotes and key points:
Draw the vertical asymptote at [tex]\( x = 3 \)[/tex] and the oblique asymptote [tex]\( y = x + 1 \)[/tex].
5. Plot the function:
- As [tex]\( x \)[/tex] approaches 3 from the left ([tex]\( x \to 3^- \)[/tex]), [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches 3 from the right ([tex]\( x \to 3^+ \)[/tex]), [tex]\( f(x) \to \infty \)[/tex].
6. Combine information for the graph:
- Draw the curve approaching [tex]\(-\infty\)[/tex] as [tex]\( x \)[/tex] approaches 3 from the left, and approaching [tex]\( \infty \)[/tex] from the right side of [tex]\( x = 3 \)[/tex].
- Use the key points and the fact that the function approaches the oblique asymptote [tex]\( y = x + 1 \)[/tex] as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex].
The final graph should show a clear vertical asymptote at [tex]\( x = 3 \)[/tex], an oblique asymptote with the equation [tex]\( y = x + 1 \)[/tex], and the function’s behavior across key points and general trend. Here's what it should roughly look like:
```
|
10 |
|
5 |
|
0 |---- (vertical asymptote at x=3)
|
-5 |
|
-10 |______________________________
-5 0 5 10
```
Note: Graph is not to scale and drawn with limited tools. For precise visualization, graphing tools or plotting in software would be more accurate.
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