Question

the function $f(x) = \frac{3x + 9}{x - 2}$ is a rational function. answer parts (a) - (i).
a. determine the domain.
the domain of $f$ is \boxed{}.
(type your answer in interval notation. use integers or fractions for any numbers in the expression.)
b. find the coordinates of any removable discontinuities (if any exist). select the correct choice and, if necess
\bigcirc a. there is a removable discontinuity at \boxed{}.
(type an ordered pair, using integers or fractions.)
the function simplifies to $f(x) = \boxed{}$.
(type your answer in factored form.)
\bigcirc b. there are no removable discontinuities.
c. check for symmetry. does the graph have y - axis symmetry, symmetry about the origin, or no symmetry? c
\bigcirc a. the graph of $f$ is symmetric about the y - axis.
\bigcirc b. the graph of $f$ is symmetric about the origin.
\bigcirc c. the graph has no symmetry.
d. find the y - intercept or state that the function does not have a y - intercept. select the correct choice and, if n
\bigcirc a. the y - intercept is $y = \boxed{}$.
(simplify your answer. type an integer or a simplified fraction.)
\bigcirc b. the function has no y - intercept.

Explanation:

Part (a)

Step1: Identify denominator restriction

For \( f(x)=\frac{3x + 9}{x - 2} \), denominator \( x - 2
eq0 \), so \( x
eq2 \).

Step2: Write domain in interval notation

All real numbers except \( x = 2 \), so domain is \( (-\infty, 2)\cup(2, \infty) \).

Step1: Factor numerator and denominator

Numerator: \( 3x + 9 = 3(x + 3) \), denominator: \( x - 2 \). No common factors, so no removable discontinuities.

Step2: Choose correct option

Since no common factors, there are no removable discontinuities.

Step1: Check y - axis symmetry (\( f(-x)=f(x) \))

\( f(-x)=\frac{3(-x)+9}{-x - 2}=\frac{-3x + 9}{-x - 2}=\frac{3x - 9}{x + 2}
eq f(x) \).

Step2: Check origin symmetry (\( f(-x)=-f(x) \))

\( -f(x)=-\frac{3x + 9}{x - 2}=\frac{-3x - 9}{x - 2}
eq f(-x) \). So no symmetry.

Step3: Choose correct option

The graph has no symmetry.

Answer:

\( (-\infty, 2)\cup(2, \infty) \)

Part (b)