# Table of Contents
- [Rational Functions](#rational-functions)
- [The $x$-Intercepts of a Rational Function](#the-x-intercepts-of-a-rational-function)
- [Example 1: $k(x)$](#example-1-kx)
- [Example 2: $g(x)$](#example-2-gx)
- [Example 3: $h(x)$](#example-3-hx)
- [Example 4: $f(x)$ (No Real Solutions)](#example-4-fx-no-real-solutions)
- [Example 5: $f(x)$ (No Intercepts)](#example-5-fx-no-intercepts)
- [Vertical Asymptotes](#vertical-asymptotes)
- [Example 6: $k(x)$](#example-6-kx)
- [Example 7: $k(x)$](#example-7-kx)
- [Example 8: $f(x)$](#example-8-fx)
- [Example 9: $f(x)$ (No Vertical Asymptotes)](#example-9-fx-no-vertical-asymptotes)
- [Example 10: $f(x)$](#example-10-fx)
- [Horizontal Asymptotes](#horizontal-asymptotes)
- [Part 1: Larger Denominator Degree](#part-1-larger-denominator-degree)
- [Example 1: $f(x)$](#example-1-fx)
- [Example 2: $f(x)$](#example-2-fx)
- [Example 3: $g(x)$](#example-3-gx)
- [Part 2: Equal Degrees](#part-2-equal-degrees)
- [Example 1: $f(x)$](#example-1-fx-1)
- [Example 2: $f(x)$](#example-2-fx-1)
- [Example 3: $f(x)$](#example-3-fx-1)
- [Part 3: Larger Numerator Degree](#part-3-larger-numerator-degree)
- [Example 1: $f(x)$ (No Horizontal Asymptote)](#example-1-fx-no-horizontal-asymptote)
- [Example 2: $f(x)$ (No Horizontal Asymptote)](#example-2-fx-no-horizontal-asymptote)
- [Graphing Rational Functions](#graphing-rational-functions)
- [Example: $f(x)$](#example-fx)
- [Example: $g(x)$](#example-gx)
- [Example: $h(x)$](#example-hx)

# 4.5 - Rational Functions

- Rational functions have the form $$f(x)=\frac{polynomial}{polynomial}.$$

Examples: $$f(x)=\frac{2x^2-x-5}{x+2}$$

$$g(x)=\frac{7x-5}{x+2}$$

$$h(x)=\frac{2x^2-x-5}{x^3-x+2}$$

$$k(x)=\frac{(x-2)(x+5)}{(x+4)(x-8)}$$

- Recall: $$\frac{0}{a}=0$$

if $a\neq 0$. BUT:

$$\frac{a}{0}\neq 0 \text{ is not defined.}$$

## The $x$-intercepts of a rational function.

- An $x$ value which makes the numerator zero produces $x$-intercepts
for the function.
---

### Example 1. For $$k(x)=\frac{(x-2)(x+5)}{(x+4)(x-8)},$$
the numerator is zero when $x=2,-5$, so $(2,0)$ and $(-5,0)$ are $x$-intercepts.
(That is $y=k(2)=k(-5)=0$.)

---

### Example 2. Find the $x$ intercepts of $$g(x)=\frac{(2x+7)(3x-5)}{2x^2+5}.$$

To find the $x$-intercepts, set $g(x) = 0$ and solve for $x$. Since $g(x)$ is a fraction, the numerator must equal zero for the function to equal zero.

The numerator of $g(x)$ is:
$$
(2x+7)(3x-5)
$$
Set the numerator equal to zero:
$$
(2x+7)(3x-5) = 0
$$

#### Solving for $x$:
Using the zero product property, solve each factor:

1. For $2x+7 = 0$:
$$
2x = -7 \quad \Rightarrow \quad x = -\frac{7}{2}
$$

2. For $3x-5 = 0$:
$$
3x = 5 \quad \Rightarrow \quad x = \frac{5}{3}
$$

#### $x$-Intercepts:
The $x$-intercepts are:
$$
x = -\frac{7}{2} \quad \text{and} \quad x = \frac{5}{3}
$$

---

### Example 3. Find the $x$ intercepts of $$h(x)=\frac{3x^2+5x}{2x-1}.$$

To find the $x$-intercepts, set $h(x) = 0$ and solve for $x$. The numerator must equal zero for the function to equal zero.

The numerator of $h(x)$ is:
$$
3x^2 + 5x
$$
Set the numerator equal to zero:
$$
3x^2 + 5x = 0
$$

#### Solving for $x$:
Factor the quadratic equation:
$$
x(3x + 5) = 0
$$

Using the zero product property:
1. $x = 0$
2. $3x + 5 = 0 \quad \Rightarrow \quad 3x = -5 \quad \Rightarrow \quad x = -\frac{5}{3}$

#### $x$-Intercepts:
The $x$-intercepts are:
$$
x = 0 \quad \text{and} \quad x = -\frac{5}{3}
$$

---

### Example 4. Find the $x$ intercepts of $$f(x)=\frac{x^2+7}{x-5}.$$

To find the $x$-intercepts, set $f(x) = 0$ and solve for $x$. The numerator must equal zero for the function to equal zero.

The numerator of $f(x)$ is:
$$
x^2 + 7
$$
Set the numerator equal to zero:
$$
x^2 + 7 = 0
$$

#### Solving for $x$:
Rearrange the equation:
$$
x^2 = -7
$$
This equation has no real solutions because the square of a real number cannot be negative.

#### $x$-Intercepts:
The function $f(x)$ has no $x$-intercepts because $x^2 + 7 = 0$ has no real solutions.

---

### Example 5. Find the $x$ intercepts of $$f(x)=\frac{13}{x-9}.$$

To find the $x$-intercepts, set $f(x) = 0$ and solve for $x$. The numerator must equal zero for the function to equal zero.

The numerator of $f(x)$ is:
$$
13
$$
Since the numerator is a constant and does not equal zero, the function $f(x)$ can never equal zero.

#### $x$-Intercepts:
The function $f(x)$ has no $x$-intercepts because the numerator $13 \neq 0$.

---

Vertical asymptotes.

- An $x$ value of $a$ which makes the denominator zero is not in the domain of the function because you can't divide by zero.

- If the numerator is not also zero (which it will be for most examples
we look at) the vertical line $x=a$ is a **vertical asymptote** for the function, with graph of one of four possibilities:





- The graph becomes nearly vertical as it approaches the line $x=a$.

- The graph does not cross $x=a$.

- $f(a)$ is not defined.

## Example 6. Find the $x$ intercepts and vertical asymptotes of $$k(x)=\frac{(x+6)(x-2)}{(x+4)(x-8)}$$

- The denominator is zero when $(x+4)=0$ or $x-8=0$, so $x=-4$ and $x=8$ are the vertical asymptotes.

- The numerator is zero when $x+6=0$ or $x-2=0$, i.e., when $x=2,-6$. The $x$ intercepts are $(2,0)$ and
$(-6,0)$.

## Example 7. Find the $x$ intercepts and vertical asymptotes of $$k(x)=\frac{(x-2)(x+5)}{(3x+4)(x-8)}$$

#### $x$-Intercepts:
To find the $x$-intercepts, set $k(x) = 0$. This occurs when the numerator equals zero.

The numerator of $k(x)$ is:
$$
(x-2)(x+5)
$$
Set the numerator equal to zero:
$$
(x-2)(x+5) = 0
$$
Using the zero product property:
1. $x - 2 = 0 \quad \Rightarrow \quad x = 2$
2. $x + 5 = 0 \quad \Rightarrow \quad x = -5$

Thus, the $x$-intercepts are:
$$
x = 2 \quad \text{and} \quad x = -5
$$

---

#### Vertical Asymptotes:
Vertical asymptotes occur when the denominator equals zero, provided the numerator does not also equal zero at those points.

The denominator of $k(x)$ is:
$$
(3x+4)(x-8)
$$
Set the denominator equal to zero:
$$
(3x+4)(x-8) = 0
$$
Using the zero product property:
1. $3x + 4 = 0 \quad \Rightarrow \quad 3x = -4 \quad \Rightarrow \quad x = -\frac{4}{3}$
2. $x - 8 = 0 \quad \Rightarrow \quad x = 8$

Thus, the vertical asymptotes are:
$$
x = -\frac{4}{3} \quad \text{and} \quad x = 8
$$

---

### Summary:
- $x$-Intercepts: $x = 2$ and $x = -5$
- Vertical Asymptotes: $x = -\frac{4}{3}$ and $x = 8$

---

### Example 8: Vertical Asymptotes of $f(x) = \frac{(x-2)(x+5)}{x^2 + 4x - 5}$

Vertical asymptotes occur where the denominator is zero, provided the numerator does not also equal zero at those points.

#### Step 1: Factor the denominator
The denominator is:
$$
x^2 + 4x - 5 = (x + 5)(x - 1)
$$

#### Step 2: Set the denominator equal to zero
$$
(x + 5)(x - 1) = 0
$$
This gives:
$$
x = -5 \quad \text{and} \quad x = 1
$$

#### Step 3: Check for cancellation
The numerator is $(x-2)(x+5)$. Since $x + 5$ appears in both the numerator and denominator, it cancels out, so there is no vertical asymptote at $x = -5$.

Thus, the only vertical asymptote is:
$$
x = 1
$$

---

### Example 9: Vertical Asymptotes of $f(x) = \frac{(x-2)(x+5)}{2x^2 + 7}$

#### Step 1: Analyze the denominator
The denominator is $2x^2 + 7$.

#### Step 2: Check if the denominator can be zero
$$
2x^2 + 7 = 0 \quad \Rightarrow \quad x^2 = -\frac{7}{2}
$$
This has no real solutions because the square of a real number cannot be negative.

#### Step 3: Conclusion
Since the denominator is never zero, there are no vertical asymptotes.

---

### Example 10: Vertical Asymptotes of $f(x) = \frac{(x-2)(x+5)}{(x-3)^2}$

Vertical asymptotes occur where the denominator is zero.

#### Step 1: Set the denominator equal to zero
$$
(x-3)^2 = 0
$$
This gives:
$$
x = 3
$$

#### Step 2: Conclusion
The vertical asymptote is:
$$
x = 3
$$

# Horizontal asymptotes. Part 1.

- If the degree of the denominator is larger than the numerator, then the graph approaches the $x$-axis as it goes to the far right and left. This makes the $x$-axis ($y=0$) a horizontal asymptote:





### Examples: Horizontal Asymptote $y = 0$

The horizontal asymptote of a rational function depends on the degree of the numerator and denominator:

- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$.

#### Example 1: $f(x) = \dfrac{x^4 - 3x - 1}{x^7 + 5x - 5}$

- The degree of the numerator is $4$ (highest power in $x^4 - 3x - 1$).
- The degree of the denominator is $7$ (highest power in $x^7 + 5x - 5$).

Since the degree of the numerator ($4$) is less than the degree of the denominator ($7$), the horizontal asymptote is:
$$
y = 0
$$

---

#### Example 2: $f(x) = \dfrac{(x-2)(x+5)}{x^3 + 4x - 5}$

- The numerator expands to a polynomial of degree $2$ because $(x-2)(x+5)$ gives $x^2 + 3x - 10$.
- The degree of the denominator is $3$ (highest power in $x^3 + 4x - 5$).

Since the degree of the numerator ($2$) is less than the degree of the denominator ($3$), the horizontal asymptote is:
$$
y = 0
$$

---

#### Example 3: $g(x) = \dfrac{2 - 3x}{x^2 + 4x - 5}$

- The degree of the numerator is $1$ (highest power in $-3x$).
- The degree of the denominator is $2$ (highest power in $x^2 + 4x - 5$).

Since the degree of the numerator ($1$) is less than the degree of the denominator ($2$), the horizontal asymptote is:
$$
y = 0
$$

---

### Conclusion:
For all three functions, the degree of the denominator is greater than the degree of the numerator, so the horizontal asymptote for each is:
$$
y = 0
$$

# Horizontal asymptotes. Part 2.

- If the degrees of the denominator and numerator are the same then the graph has a horizontal asymptote of the form $y=a$ where $a$ is the quotient of the leading terms.


### Example 1: Horizontal Asymptote of $f(x) = \frac{6x^4 - 3x - 1}{2x^4 + 5x - 5}$

- The degree of the numerator is $4$ (highest power in $6x^4 - 3x - 1$).
- The degree of the denominator is $4$ (highest power in $2x^4 + 5x - 5$).

When the degrees of the numerator and denominator are equal, the horizontal asymptote is determined by the ratio of the leading coefficients:
$$
y = \frac{6x^4}{2x^4} = \frac{6}{2} = 3
$$

Thus, the horizontal asymptote is:
$$
y = 3
$$

---

### Example 2: Horizontal Asymptote of $f(x) = \frac{-2x^2 - 3x - 1}{5 + x + 5x^2}$

- The degree of the numerator is $2$ (highest power in $-2x^2 - 3x - 1$).
- The degree of the denominator is $2$ (highest power in $5 + x + 5x^2$).

Since the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients:
$$
y = \frac{-2x^2}{5x^2} = \frac{-2}{5}
$$

Thus, the horizontal asymptote is:
$$
y = -\frac{2}{5}
$$

---

### Example 3: Horizontal Asymptote of $f(x) = \frac{2x(x-2)}{(3x+5)(2x-9)}$

- Expand the numerator: $2x(x-2) = 2x^2 - 4x$. The degree of the numerator is $2$.
- Expand the denominator: $(3x+5)(2x-9) = 6x^2 - 27x + 10x - 45 = 6x^2 - 17x - 45$. The degree of the denominator is $2$.

Since the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients:
$$
y = \frac{2x^2}{6x^2} = \frac{2}{6} = \frac{1}{3}
$$

Thus, the horizontal asymptote is:
$$
y = \frac{1}{3}
$$

# Horizontal asymptotes. Part 3.

### Horizontal Asymptotes When the Degree of the Numerator is Larger

If the degree of the numerator is larger than the degree of the denominator, the graph does not have a horizontal asymptote. Instead, the function may have a slant (oblique) asymptote or grow without bound.

---

### Example 1: $f(x) = \dfrac{-2x^3 - 3x - 1}{5x^2 + x + 5}$

- The degree of the numerator is $3$ (from $-2x^3$).
- The degree of the denominator is $2$ (from $5x^2$).

Since the degree of the numerator ($3$) is greater than the degree of the denominator ($2$), the function **does not have a horizontal asymptote**. The graph will grow or decay faster in the numerator's leading term direction.

---

### Example 2: $f(x) = \dfrac{x(x+2)}{x-1}$

- The numerator expands to $x^2 + 2x$, so its degree is $2$.
- The degree of the denominator is $1$ (from $x-1$).

Since the degree of the numerator ($2$) is greater than the degree of the denominator ($1$), the function **does not have a horizontal asymptote**. The graph grows quadratically as $x \to \pm \infty$.

### Horizontal Asymptotes by Comparing Degrees

The horizontal asymptote of a rational function depends on the degrees of the numerator and denominator:

1. **If the degrees are equal**, the horizontal asymptote is the ratio of the leading coefficients.
2. **If the degree of the numerator is less than the degree of the denominator**, the horizontal asymptote is $y = 0$.
3. **If the degree of the numerator is greater than the degree of the denominator**, there is no horizontal asymptote.

---

### Example 1: $f(x) = \dfrac{3x^2 - 3x - 1}{2x^2 + 5x - 5}$

- The degree of the numerator is $2$ (highest power in $3x^2 - 3x - 1$).
- The degree of the denominator is $2$ (highest power in $2x^2 + 5x - 5$).

Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients:
$$
y = \frac{3}{2}
$$

---

### Example 2: $f(x) = \dfrac{x}{2x^2 + x - 5}$

- The degree of the numerator is $1$ (highest power in $x$).
- The degree of the denominator is $2$ (highest power in $2x^2 + x - 5$).

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is:
$$
y = 0
$$

---

### Example 3: $f(x) = \dfrac{x^4}{2x^2 + x - 5}$

- The degree of the numerator is $4$ (highest power in $x^4$).
- The degree of the denominator is $2$ (highest power in $2x^2 + x - 5$).

Since the degree of the numerator is greater than the degree of the denominator, the function has **no horizontal asymptote**.

---

### Summary:

1. $f(x) = \frac{3x^2 - 3x - 1}{2x^2 + 5x - 5} \quad \Rightarrow \quad y = \frac{3}{2}$
2. $f(x) = \frac{x}{2x^2 + x - 5} \quad \Rightarrow \quad y = 0$
3. $f(x) = \frac{x^4}{2x^2 + x - 5} \quad \Rightarrow \quad \text{no horizontal asymptote}$

# Graphing Rational Functions.

## Steps:

1. Find $x$ intercepts (when is numerator zero?)

2. Find vertical asymptotes (when is denominator zero?)

3. Find horizontal asymptotes (is the degree larger in numerator or denominator?)

4. Plot some points to get started. (E.g.: let $x=0$ to get $y$ intercept.)

## Example.

$$f(x)=\dfrac{x-4}{2-x}$$

1. $x$ intercepts: $x-4=0$, so $x=4$. $x$ intercept is $(4,0)$

2. Vertical asymptotes: $2-x=0$, so $x=2$ is vertical asymptote (this is a line, not a point.)

3. Horizontal asymptotes: degree on top and bottom is the same (both $1$), so horizontal asymptote is
$$y=\frac{x}{-x}$$
$$y=-1$$

4. If $x=0$, $$y=\frac{0-4}{2-0}=-\frac{4}{2}=-2.$$ So $y$ intercept: $(0,-2)$.

# Graph.

- The intercepts and asymptotes are plotted.

- From each intercept, we draw the graph toward the asymptote. It's unlikely that the graph will cross
the asymptote and approach from the other side - just plot a few points to see.

- The dotted sections are the last question: does the graph go up or down toward the vertical asymptotes?

- Can be answered by plotting points and/or noting that the graph cannot cross the $x$ axis there because $(4,0)$ is the only $x$ intercept.


| $x$ | $y=\dfrac{x-4}{2-x}$ Substitution and Calculation | $y=f(x)=\dfrac{x-4}{2-x}$ |
|-------|---------------------------------------------|---------|
| -5 | $(-5-4)/(2-(-5)) = -1.3$ | -1.3 |
| -4 | $(-4-4)/(2-(-4)) = -1.3$ | -1.3 |
| -3 | $(-3-4)/(2-(-3)) = -1.4$ | -1.4 |
| -2 | $(-2-4)/(2-(-2))= -1.5$ | -1.5 |
| -1 | $(-1-4)/(2-(-1)) = -1.7$ | -1.7 |
| 0 | $(0-4)/(2-0) = -2.0$ | -2.0 |
| 1 | $(1-4)/(2-1) = -3.0$ | -3.0 |
| 1.5 | $(1.5-4)/(2-1.5) = -5.0$ | -5.0 |
| 2.5 | $(2.5-4)/(2-2.5) = 3.0$ | 3.0 |
| 3 | $(3-4)/(2-3) = 1.0$ | 1.0 |
| 4 | $(4-4)/(2-4) = -0.0$ | -0.0 |
| 5 | $(5-4)/(2-5) = -0.3$ | -0.3 |

# Final graph.


- At both $x=2$ (denominator zero) and $x=4$ (numerator zero), the $y$ value changes sign. The graph switches from
above to below the $x$ axis.

## Example: $f(x) = \dfrac{x-2}{1+x}$

### 1. $x$-Intercepts:
To find the $x$-intercepts, set $f(x) = 0$:
$$
\frac{x-2}{1+x} = 0 \quad \Rightarrow \quad x-2 = 0 \quad \Rightarrow \quad x = 2
$$
The $x$-intercept is:
$$
(2, 0)
$$

---

### 2. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero, provided the numerator is non-zero:
$$
1+x = 0 \quad \Rightarrow \quad x = -1
$$
The vertical asymptote is:
$$
x = -1
$$

---

### 3. Horizontal Asymptotes:
The degrees of the numerator and denominator are equal, so the horizontal asymptote is the ratio of the leading coefficients:
$$
y = \frac{1}{1} = 1
$$
The horizontal asymptote is:
$$
y = 1
$$

---

### 4. $y$-Intercept:
To find the $y$-intercept, substitute $x = 0$ into $f(x)$:
$$
f(0) = \frac{0-2}{1+0} = \frac{-2}{1} = -2
$$
The $y$-intercept is:
$$
(0, -2)
$$

---

### Table of Values for $f(x) = \frac{x-2}{1+x}$

| $x$ | Substitution and Calculation | $f(x)$ |
|-------|---------------------------------------------|---------|
| -3 | $\frac{-3-2}{1+(-3)} = \frac{-5}{-2} = 2.5$ | 2.5 |
| -2 | $\frac{-2-2}{1+(-2)} = \frac{-4}{-1} = 4.0$ | 4.0 |
| -0.5 | $\frac{-0.5-2}{1+(-0.5)} = \frac{-2.5}{0.5} = -5.0$ | -5.0 |
| 0 | $\frac{0-2}{1+0} = \frac{-2}{1} = -2.0$ | -2.0 |
| 1 | $\frac{1-2}{1+1} = \frac{-1}{2} = -0.5$ | -0.5 |
| 2 | $\frac{2-2}{1+2} = \frac{0}{3} = 0.0$ | 0.0 |
| 5 | $\frac{5-2}{1+5} = \frac{3}{6} = 0.5$ | 0.5 |


## Example: $g(x) = \dfrac{-2}{(x-1)^2}$

### 1. $x$-Intercepts:
The numerator \(-2 = 0\) has no solution, so there are no $x$-intercepts.

---

### 2. Vertical Asymptotes:
The denominator $(x-1)^2 = 0$ gives:
$$
x-1 = 0 \quad \Rightarrow \quad x = 1
$$
Thus, the vertical asymptote is:
$$
x = 1
$$

---

### 3. Horizontal Asymptotes:
The degree of the numerator is $0$ (constant term \(-2\)), and the degree of the denominator is $2$ (highest power of $x$ in $(x-1)^2$). Since the numerator's degree is less than the denominator's, the horizontal asymptote is:
$$
y = 0
$$

---

### 4. $y$-Intercept:
To find the $y$-intercept, substitute $x = 0$ into $g(x)$:
$$
g(0) = \frac{-2}{(0-1)^2} = \frac{-2}{1} = -2
$$
The $y$-intercept is:
$$
(0, -2)
$$

---

### Additional Points:
- From the $y$-intercept $(0, -2)$, the graph goes **down** as $x$ approaches the vertical asymptote at $x=1$ from the left.
- At $x=2$:
$$
g(2) = \frac{-2}{(2-1)^2} = \frac{-2}{1} = -2
$$
The point is:
$$
(2, -2)
$$
- From $(2, -2)$, the graph also goes **down** as $x$ approaches $x=1$ from the right.

---

### Table of Values for $g(x) = \frac{-2}{(x-1)^2}$

| $x$ | Substitution and Calculation | $g(x)$ |
|-------|---------------------------------------------|---------|
| -1 | $\frac{-2}{(-1-1)^2} = \frac{-2}{4} = -0.5$ | -0.5 |
| 0 | $\frac{-2}{(0-1)^2} = \frac{-2}{1} = -2.0$ | -2.0 |
| 0.5 | $\frac{-2}{(0.5-1)^2} = \frac{-2}{0.25} = -8.0$ | -8.0 |
| 1.5 | $\frac{-2}{(1.5-1)^2} = \frac{-2}{0.25} = -8.0$ | -8.0 |
| 2 | $\frac{-2}{(2-1)^2} = \frac{-2}{1} = -2.0$ | -2.0 |
| 3 | $\frac{-2}{(3-1)^2} = \frac{-2}{4} = -0.5$ | -0.5 |


### Example: $g(x) = \dfrac{2}{(x+1)^2}$

### 1. $x$-Intercepts:
The numerator $2 = 0$ has no solution, so there are no $x$-intercepts.

---

### 2. Vertical Asymptotes:
The denominator $(x+1)^2 = 0$ gives:
$$
x+1 = 0 \quad \Rightarrow \quad x = -1
$$
Thus, the vertical asymptote is:
$$
x = -1
$$

---

### 3. Horizontal Asymptotes:
The degree of the numerator is $0$ (constant term $2$), and the degree of the denominator is $2$ (highest power of $x$ in $(x+1)^2$). Since the numerator's degree is less than the denominator's, the horizontal asymptote is:
$$
y = 0
$$

---

### 4. $y$-Intercept:
To find the $y$-intercept, substitute $x = 0$ into $g(x)$:
$$
g(0) = \frac{2}{(0+1)^2} = \frac{2}{1} = 2
$$
The $y$-intercept is:
$$
(0, 2)
$$

---

### Additional Points:
- From the $y$-intercept $(0, 2)$, the graph goes **up** as $x$ approaches the vertical asymptote at $x=-1$ from both sides due to the squared term.
- At $x=1$:
$$
g(1) = \frac{2}{(1+1)^2} = \frac{2}{4} = 0.5
$$
The point is:
$$
(1, 0.5)
$$

---

### Table of Values for $g(x) = \frac{2}{(x+1)^2}$

| $x$ | Substitution and Calculation | $g(x)$ |
|-------|---------------------------------------------|---------|
| -3 | $\frac{2}{(-3+1)^2} = \frac{2}{4} = 0.5$ | 0.5 |
| -2 | $\frac{2}{(-2+1)^2} = \frac{2}{1} = 2.0$ | 2.0 |
| -1.5 | $\frac{2}{(-1.5+1)^2} = \frac{2}{0.25} = 8.0$ | 8.0 |
| -0.5 | $\frac{2}{(-0.5+1)^2} = \frac{2}{0.25} = 8.0$ | 8.0 |
| 0 | $\frac{2}{(0+1)^2} = \frac{2}{1} = 2.0$ | 2.0 |
| 1 | $\frac{2}{(1+1)^2} = \frac{2}{4} = 0.5$ | 0.5 |


## Example: $h(x) = \frac{2x}{x^2 + 1}$

### 1. $x$-Intercepts:
The numerator $2x = 0$ gives:
$$
x = 0
$$
The $x$-intercept is:
$$
(0, 0)
$$

---

### 2. Vertical Asymptotes:
The denominator $x^2 + 1 = 0$ gives:
$$
x^2 = -1
$$
This has no real solutions. Thus, there are no vertical asymptotes.

---

### 3. Horizontal Asymptotes:
The degree of the numerator is $1$, and the degree of the denominator is $2$. Since the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is:
$$
y = 0
$$

---

### 4. $y$-Intercept:
To find the $y$-intercept, substitute $x = 0$ into $h(x)$:
$$
h(0) = \frac{2(0)}{0^2 + 1} = 0
$$
The $y$-intercept is:
$$
(0, 0)
$$

---

### Additional Points:
- If $x = 1$:
$$
h(1) = \frac{2(1)}{1^2 + 1} = \frac{2}{2} = 1
$$
The point is:
$$
(1, 1)
$$
- If $x = -1$:
$$
h(-1) = \frac{2(-1)}{(-1)^2 + 1} = \frac{-2}{2} = -1
$$
The point is:
$$
(-1, -1)
$$

---

### Table of Values for $h(x) = \frac{2x}{x^2 + 1}$

| $x$ | Substitution and Calculation | $h(x)$ |
|-------|--------------------------------------------|---------|
| -2 | $\frac{2(-2)}{(-2)^2 + 1} = \frac{-4}{5}$ | -0.8 |
| -1 | $\frac{2(-1)}{(-1)^2 + 1} = \frac{-2}{2}$ | -1.0 |
| 0 | $\frac{2(0)}{0^2 + 1} = 0$ | 0.0 |
| 1 | $\frac{2(1)}{1^2 + 1} = \frac{2}{2}$ | 1.0 |
| 2 | $\frac{2(2)}{2^2 + 1} = \frac{4}{5}$ | 0.8 |