The first function, $$ f(x)=\frac{x^{2}-4}{x-2} $$

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UpStudy AI · Accepted Answer

Certainly! Let's analyze and simplify the function you've provided:

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

### Step 1: Factor the Numerator

The numerator $$ x^{2} - 4 $$ is a **difference of squares**, which can be factored as follows:

$$
x^{2} - 4 = (x - 2)(x + 2)
$$

### Step 2: Simplify the Expression

Now, substitute the factored form of the numerator back into the function:

$$
f(x) = \frac{(x - 2)(x + 2)}{x - 2}
$$

Notice that $$ x - 2 $$ appears in both the numerator and the denominator. For all values of $$ x $$ **except** $$ x = 2 $$ (since division by zero is undefined), you can cancel out $$ x - 2 $$:

$$
f(x) = x + 2 \quad \text{for} \quad x \neq 2
$$

### Step 3: Consider the Domain and Discontinuity

- **Simplified Function:** $$ f(x) = x + 2 $$
  
- **Domain:** All real numbers except $$ x = 2 $$, because at $$ x = 2 $$, the original function is undefined (division by zero).

- **Graphical Interpretation:** The function $$ f(x) = \frac{x^{2} - 4}{x - 2} $$ is equivalent to $$ f(x) = x + 2 $$ for all $$ x \neq 2 $$. However, there is a **hole** in the graph at $$ x = 2 $$. If you were to graph this function, it would look like the line $$ y = x + 2 $$ with a missing point at $$ (2, 4) $$.

### Summary

After simplifying, the function can be expressed as:

$$
f(x) = x + 2 \quad \text{for} \quad x \neq 2
$$

There's a hole at $$ x = 2 $$ because the original function is not defined at that point.

If you have any further questions or need additional clarification on this function, feel free to ask!
The function simplifies to $$ f(x) = x + 2 $$ for all $$ x $$ except $$ x = 2 $$, where there is a hole in the graph.

The first function, \( f(x)=\frac{x^{2}-4}{x-2} \)

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