Sketch a rational function subject to the given conditions. Horizontal asymptote: $$ y=0 $$ Vertical asymptote: $$ x=-1 $$ $$ y $$-intercept: $$ (0,1) $$ No $$ x $$-intercepts Range: $$ (0, \infty) $$

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To sketch a rational function that meets the specified conditions, we need to construct a function that satisfies all the given criteria. Let's break down the conditions step by step:

1. **Horizontal Asymptote: $$ y = 0 $$**:
   - This means that as $$ x $$ approaches infinity (both positive and negative), the function should approach 0. This typically occurs when the degree of the numerator is less than the degree of the denominator.

2. **Vertical Asymptote: $$ x = -1 $$**:
   - This indicates that the function is undefined at $$ x = -1 $$, which usually happens when the denominator of the rational function equals zero at this point.

3. **$$ y $$-intercept: $$ (0, 1) $$**:
   - The function must pass through the point $$ (0, 1) $$. This means that when $$ x = 0 $$, the function should equal 1.

4. **No $$ x $$-intercepts**:
   - This means that the function should never cross the x-axis, which implies that the numerator must not equal zero for any real $$ x $$.

5. **Range: $$ (0, \infty) $$**:
   - The function should only take positive values, which aligns with the requirement of having no $$ x $$-intercepts.

### Constructing the Function

Given these conditions, we can construct a rational function of the form:

$$
f(x) = \frac{a}{(x + 1)^b}
$$

Where:
- $$ a $$ is a positive constant to ensure the function is positive.
- $$ b $$ is a positive integer to ensure the function approaches 0 as $$ x $$ approaches infinity.

### Finding the Constants

1. **To satisfy the $$ y $$-intercept**:
   - We need $$ f(0) = 1 $$:
   $$
   f(0) = \frac{a}{(0 + 1)^b} = \frac{a}{1^b} = a
   $$
   Thus, $$ a = 1 $$.

2. **Choosing $$ b $$**:
   - We can choose $$ b = 1 $$ for simplicity. Therefore, the function becomes:
   $$
   f(x) = \frac{1}{x + 1}
   $$

### Verifying the Conditions

- **Horizontal Asymptote**: As $$ x \to \infty $$, $$ f(x) \to 0 $$.
- **Vertical Asymptote**: At $$ x = -1 $$, $$ f(x) $$ is undefined.
- **$$ y $$-intercept**: $$ f(0) = 1 $$.
- **No $$ x $$-intercepts**: The numerator is always positive (1), so there are no $$ x $$-intercepts.
- **Range**: The function takes values from $$ (0, \infty) $$.

### Final Function

The function that meets all the conditions is:

$$
f(x) = \frac{1}{x + 1}
$$

### Sketching the Function

1. The vertical asymptote at $$ x = -1 $$ means the graph approaches infinity as $$ x $$ approaches -1 from the left and approaches 0 from the right.
2. The horizontal asymptote at $$ y = 0 $$ means the graph approaches the x-axis as $$ x $$ goes to positive or negative infinity.
3. The $$ y $$-intercept at $$ (0, 1) $$ indicates the point where the graph crosses the y-axis.

Now, let's sketch the function based on these characteristics.

- The graph will be in the first quadrant, starting from the point $$ (0, 1) $$ and approaching the horizontal asymptote $$ y = 0 $$ as $$ x $$ increases.
- As $$ x $$ approaches -1 from the left, the function will go to positive infinity, and as $$ x $$ approaches -1 from the right, it will drop down towards 0.

This completes the construction and sketching of the rational function.
The rational function that satisfies all the given conditions is $$ f(x) = \frac{1}{x + 1} $$.

Sketch a rational function subject to the given conditions. Horizontal asymptote: \( y=0 \) Vertical asymptote: \( x=-1 \) \( y \)-intercept: \( (0,1) \) No \( x \)-intercepts Range: \( (0, \infty) \)

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