Find the domain of the rational function. Then, write it in lowest terms.
$f\left(x\right)=\frac{x^2+9}{x-3}$

• A. {x∣x≠0}, $f\left(x\right)=\frac{1}{x-3}$
• B. {x∣x≠3}, $f\left(x\right)=\frac{x^2+9}{x-3}$
• C. {x∣x≠−3}, $f\left(x\right)=\frac{x^2+9}{x-3}$
• D. {x∣x≠3}, $f\left(x\right)=x+3$

Answer: (B)

Verified step by step guidance

1. Identify the rational function given: \( f(x) = \frac{x^2 + 9}{x - 3} \).
2. Determine the domain of the function by identifying values of \( x \) that make the denominator zero. Set the denominator equal to zero: \( x - 3 = 0 \). Solve for \( x \) to find the value that is not in the domain.
3. The solution to \( x - 3 = 0 \) is \( x = 3 \). Therefore, the domain of the function is all real numbers except \( x = 3 \). In set notation, this is \( \{ x \mid x \neq 3 \} \).
4. Simplify the rational function if possible. Check if the numerator \( x^2 + 9 \) can be factored and if any common factors exist with the denominator \( x - 3 \).
5. Since \( x^2 + 9 \) cannot be factored further and has no common factors with \( x - 3 \), the function is already in its lowest terms. Thus, the simplified form of the function remains \( f(x) = \frac{x^2 + 9}{x - 3} \).