Answer
The function \( y = \frac{x^2}{1+x} \) has a domain of \( x \in \mathbb{R} \setminus \{-1\} \), a y-intercept at \( (0, 0) \), a vertical asymptote at \( x = -1 \), and local maxima and minima at \( x = -2 \) and \( x = 0 \), respectively. The function approaches infinity as \( x \) approaches positive or negative infinity.
Solution
To analyze the function \( y = \frac{x^2}{1+x} \), we can explore several aspects such as its domain, intercepts, asymptotes, and behavior.
### 1. Domain
The function is defined for all \( x \) except where the denominator is zero. Thus, we set the denominator equal to zero:
\[
1 + x = 0 \implies x = -1
\]
So, the domain of the function is \( x \in \mathbb{R} \setminus \{-1\} \).
### 2. Intercepts
- **Y-intercept**: Set \( x = 0 \):
\[
y = \frac{0^2}{1+0} = 0 \implies (0, 0)
\]
- **X-intercepts**: Set \( y = 0 \):
\[
\frac{x^2}{1+x} = 0 \implies x^2 = 0 \implies x = 0 \implies (0, 0)
\]
### 3. Asymptotes
- **Vertical Asymptote**: The function has a vertical asymptote at \( x = -1 \) since the function approaches infinity as \( x \) approaches -1 from the left or right.
- **Horizontal Asymptote**: To find the horizontal asymptote, we analyze the behavior of \( y \) as \( x \to \infty \) or \( x \to -\infty \):
\[
\lim_{x \to \infty} y = \lim_{x \to \infty} \frac{x^2}{1+x} = \lim_{x \to \infty} \frac{x^2/x}{1/x + 1} = \lim_{x \to \infty} \frac{x}{1/x + 1} = \infty
\]
\[
\lim_{x \to -\infty} y = \lim_{x \to -\infty} \frac{x^2}{1+x} = \lim_{x \to -\infty} \frac{x^2/x}{1/x + 1} = \lim_{x \to -\infty} \frac{x}{1/x + 1} = -\infty
\]
### 4. Behavior and Critical Points
To find critical points, we can take the derivative of \( y \):
Using the quotient rule:
\[
y' = \frac{(1+x)(2x) - x^2(1)}{(1+x)^2} = \frac{2x + 2x^2 - x^2}{(1+x)^2} = \frac{x^2 + 2x}{(1+x)^2} = \frac{x(x + 2)}{(1+x)^2}
\]
Setting the derivative equal to zero to find critical points:
\[
x(x + 2) = 0 \implies x = 0 \text{ or } x = -2
\]
### 5. Second Derivative Test
To determine the nature of the critical points, we can use the second derivative test. We can differentiate \( y' \) again, but we can also analyze the sign of \( y' \):
- For \( x < -2 \), \( y' > 0 \) (increasing)
- For \( -2 < x < 0 \), \( y' < 0 \) (decreasing)
- For \( x > 0 \), \( y' > 0 \) (increasing)
Thus, \( x = -2 \) is a local maximum, and \( x = 0 \) is a local minimum.
### Summary
- **Domain**: \( x \in \mathbb{R} \setminus \{-1\} \)
- **Intercepts**: \( (0, 0) \)
- **Vertical Asymptote**: \( x = -1 \)
- **Behavior**: Local maximum at \( x = -2 \), local minimum at \( x = 0 \)
- **End behavior**: \( y \to \infty \) as \( x \to \infty \) and \( y \to -\infty \) as \( x \to -\infty \)
This function has interesting behavior around its critical points and asymptotes, making it a good candidate for further analysis or graphing.
Reviewed and approved by the UpStudy tutoring team
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