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: \( x - 3 = 0 \).
3. Solve the equation \( x - 3 = 0 \) to find the value of \( x \) that is not in the domain. This gives \( x = 3 \). Therefore, the domain is all real numbers except \( x = 3 \).
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 \( x - 3 \). In this case, \( x^2 + 9 \) does not factor further, so the function is already in its lowest terms.
5. Conclude that the domain of the function is \( \{ x \mid x \neq 3 \} \) and the function in its lowest terms remains \( f(x) = \frac{x^2 + 9}{x - 3} \).