Let's analyze the function step by step:

Given function:
$f(x) = \frac{x^2 + 1}{2x - 4}$

1. Domain

The domain of a function includes all possible values of $x$ for which the function is defined. The function will be undefined if the denominator is equal to zero.

$2x - 4 = 0$ $2x = 4 \implies x = 2$

Thus, $x = 2$ is excluded from the domain.

Therefore, the domain is: $x \in \mathbb{R}, x \neq 2$ Domain: $x \in \mathbb{R}, x \neq 2$

2. Range

To find the range, we first check for any asymptotes or restrictions. The function may approach infinity as $x$ approaches 2 (vertical asymptote) or as $x$ approaches infinity. We will analyze the horizontal asymptote later.

For now, the function takes all values except at points where asymptotes exist. We'll determine the horizontal asymptote for more clarity on the range.

3. X-Intercept

To find the x-intercept(s), we set the numerator equal to zero and solve for $x$:

$x^2 + 1 = 0$ $x^2 = -1$

Since this equation has no real solutions (the square of a real number cannot be negative), there are no x-intercepts.

X-Intercept: None

4. Y-Intercept

To find the y-intercept, set $x = 0$:

$f(0) = \frac{0^2 + 1}{2(0) - 4} = \frac{1}{-4} = -\frac{1}{4}$

Y-Intercept: $\left( 0, -\frac{1}{4} \right)$

5. Zeros

Since we already saw that $x^2 + 1 = 0$ has no real solutions, the function has no real zeros.

Zeros: None

6. Vertical Asymptote

The function has a vertical asymptote where the denominator is zero:

$2x - 4 = 0 \implies x = 2$

Vertical Asymptote: $x = 2$

7. Horizontal Asymptote

To find the horizontal asymptote, we examine the behavior of the function as $x$ approaches infinity. We divide both the numerator and denominator by $x$:

$f(x) = \frac{x^2 + 1}{2x - 4} \sim \frac{x^2}{2x} = \frac{x}{2}$

Since the degree of the numerator (2) is greater than the degree of the denominator (1), the function has no horizontal asymptote. Instead, it has an oblique asymptote, which can be found through polynomial long division:

Performing the division of $x^2 + 1$ by $2x - 4$ yields the quotient $\frac{1}{2}x + 1$. This indicates that the graph approaches the line $y = \frac{1}{2}x + 1$ as $x \to \infty$.

Horizontal Asymptote: None
Oblique Asymptote: $y = \frac{1}{2}x + 1$

8. Graph

The graph of this function would have:

• A vertical asymptote at $x = 2$
• An oblique asymptote following the line $y = \frac{1}{2}x + 1$
• A y-intercept at $\left( 0, -\frac{1}{4} \right)$

Would you like me to generate a graph of this function?

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Summary:

• Domain: $x \in \mathbb{R}, x \neq 2$
• Range: The range is determined by the behavior of the function near asymptotes.
• X-Intercept: None
• Y-Intercept: $\left( 0, -\frac{1}{4} \right)$
• Zeros: None
• Vertical Asymptote: $x = 2$
• Oblique Asymptote: $y = \frac{1}{2}x + 1$

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Do you have any questions or would you like more details on a specific point?

Here are 5 related questions to explore further:

1. How do you find the oblique asymptote of a rational function?
2. What is the significance of vertical asymptotes in a graph?
3. How can you determine the behavior of a function near its asymptotes?
4. What is the difference between horizontal and oblique asymptotes?
5. Why does $f(x)$ have no x-intercepts in this case?

Tip: When analyzing rational functions, always check for vertical and horizontal (or oblique) asymptotes to understand the behavior of the graph at large values of $x$.