Answer :
To sketch the graph of the curve \( y = \frac{x(x+3)}{1 - x^2} \), we will systematically find all the relevant features including intercepts, asymptotes, and the behavior of the function.
### Step-by-Step Solution:
1. Find the \( x \)-intercepts:
The \( x \)-intercepts occur when \( y = 0 \):
[tex]\[ \frac{x(x + 3)}{1 - x^2} = 0 \implies x(x + 3) = 0 \][/tex]
Solving this, we get:
[tex]\[ x = 0 \quad \text{or} \quad x = -3 \][/tex]
So, the \( x \)-intercepts are at the points \( (0, 0) \) and \( (-3, 0) \).
2. Find the \( y \)-intercept:
The \( y \)-intercept occurs when \( x = 0 \):
[tex]\[ y = \frac{0 \cdot (0+3)}{1 - 0^2} = 0 \][/tex]
So, the \( y \)-intercept is at the point \( (0, 0) \).
3. Determine vertical asymptotes:
Vertical asymptotes occur where the denominator is zero but the numerator is not:
[tex]\[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \][/tex]
So, there are vertical asymptotes at \( x = 1 \) and \( x = -1 \).
4. Determine the behavior around the asymptotes and endpoints:
Near \( x = 1 \) and \( x = -1 \), the function will tend to \(\pm \infty\) since the denominator approaches zero while the numerator does not.
5. Analyze the behavior of the function for large positive and negative \( x \):
As \( x \to \infty \) or \( x \to -\infty \), the dominant terms in the numerator and the denominator are \( x^2 \). Simplifying, we get:
[tex]\[ y \approx -x \quad (\text{since } \frac{x^2}{-x^2} = -1) \][/tex]
This indicates that as \( x \) becomes very large (positively or negatively), the function \( y \) grows linearly but negatively. So, the curve will tend downward as \( x \) moves away from the origin.
6. Sketch the Graph:
- Plot the \( x \)-intercepts at \( (0, 0) \) and \( (-3, 0) \).
- Plot the \( y \)-intercept at \( (0, 0) \).
- Draw vertical asymptotes at \( x = 1 \) and \( x = -1 \).
- Note the behavior as \( x \to \pm \infty \) will indicate a negative linear trend.
Given these features, we can now sketch the graph:
- The curve passes through \( (0, 0) \) and \( (-3, 0) \).
- There are vertical asymptotes at \( x = -1 \) and \( x = 1 \).
- For large \(|x|\), the curve will resemble the line \( y \approx -x\).
Here is a rough sketch to illustrate the points (without the actual plotting of all \( x \) and \( y \) values):
```
y | /
| /
| /
| /
| /
|-----------------------------o------ (-3,0)
| \
| \
| \
| \
|
+------------------------------------------------------------------- x
-1 0 1
```
- It shows the curve moving through the intercepts \( (0, 0) \) and \( (-3, 0) \).
- It goes to infinity near \( x = 1 \) and \( x = -1 \).
- The behavior is linear as it moves far from the origin.
By marking the relevant intercepts and asymptotes on the graph as described, we accurately outline the principal features of the curve defined by [tex]\( y = \frac{x(x+3)}{1-x^2} \)[/tex].
### Step-by-Step Solution:
1. Find the \( x \)-intercepts:
The \( x \)-intercepts occur when \( y = 0 \):
[tex]\[ \frac{x(x + 3)}{1 - x^2} = 0 \implies x(x + 3) = 0 \][/tex]
Solving this, we get:
[tex]\[ x = 0 \quad \text{or} \quad x = -3 \][/tex]
So, the \( x \)-intercepts are at the points \( (0, 0) \) and \( (-3, 0) \).
2. Find the \( y \)-intercept:
The \( y \)-intercept occurs when \( x = 0 \):
[tex]\[ y = \frac{0 \cdot (0+3)}{1 - 0^2} = 0 \][/tex]
So, the \( y \)-intercept is at the point \( (0, 0) \).
3. Determine vertical asymptotes:
Vertical asymptotes occur where the denominator is zero but the numerator is not:
[tex]\[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \][/tex]
So, there are vertical asymptotes at \( x = 1 \) and \( x = -1 \).
4. Determine the behavior around the asymptotes and endpoints:
Near \( x = 1 \) and \( x = -1 \), the function will tend to \(\pm \infty\) since the denominator approaches zero while the numerator does not.
5. Analyze the behavior of the function for large positive and negative \( x \):
As \( x \to \infty \) or \( x \to -\infty \), the dominant terms in the numerator and the denominator are \( x^2 \). Simplifying, we get:
[tex]\[ y \approx -x \quad (\text{since } \frac{x^2}{-x^2} = -1) \][/tex]
This indicates that as \( x \) becomes very large (positively or negatively), the function \( y \) grows linearly but negatively. So, the curve will tend downward as \( x \) moves away from the origin.
6. Sketch the Graph:
- Plot the \( x \)-intercepts at \( (0, 0) \) and \( (-3, 0) \).
- Plot the \( y \)-intercept at \( (0, 0) \).
- Draw vertical asymptotes at \( x = 1 \) and \( x = -1 \).
- Note the behavior as \( x \to \pm \infty \) will indicate a negative linear trend.
Given these features, we can now sketch the graph:
- The curve passes through \( (0, 0) \) and \( (-3, 0) \).
- There are vertical asymptotes at \( x = -1 \) and \( x = 1 \).
- For large \(|x|\), the curve will resemble the line \( y \approx -x\).
Here is a rough sketch to illustrate the points (without the actual plotting of all \( x \) and \( y \) values):
```
y | /
| /
| /
| /
| /
|-----------------------------o------ (-3,0)
| \
| \
| \
| \
|
+------------------------------------------------------------------- x
-1 0 1
```
- It shows the curve moving through the intercepts \( (0, 0) \) and \( (-3, 0) \).
- It goes to infinity near \( x = 1 \) and \( x = -1 \).
- The behavior is linear as it moves far from the origin.
By marking the relevant intercepts and asymptotes on the graph as described, we accurately outline the principal features of the curve defined by [tex]\( y = \frac{x(x+3)}{1-x^2} \)[/tex].
