Question

the function $f(x) = \frac{3x + 9}{x - 2}$ is a rational function. answer parts (a) - (j).

a. the x-intercept(s) is(are) $x = \square$.
(simplify your answer. use a comma to separate answers as needed.)
b. the function has no x-intercept.

f. find the equation(s) of all vertical asymptotes. select the correct choice and, if necessary, fill in the answer box(es) to complete your
a. the function has one vertical asymptote, $\square$.
(simplify your answer. type an equation. use integers or fractions for any numbers in the equation.)
b. the function has two vertical asymptotes. the leftmost asymptote is $\square$ and the rightmost asymptote is $\square$.
(simplify your answers. type equations. use integers or fractions for any numbers in the equation.)
c. the function has no vertical asymptotes.

g. find the equation(s) of all horizontal asymptotes or slant asymptotes. select the correct choice and, if necessary, fill in the answer
a. the horizontal asymptote is $\square$.
(simplify your answer. type an equation. use integers or fractions for any numbers in the equation.)
b. the function has no horizontal asymptote. it has a slant asymptote of $\square$.
(simplify your answer. type an equation. use integers or fractions for any numbers in the equation. type your answer in slope-
c. the function has no horizontal asymptote and has no slant asymptote.

Explanation:

Part A (x - intercept)

Step 1: Recall x - intercept definition

The x - intercept of a function \(y = f(x)\) is the value of \(x\) when \(y=0\) (i.e., \(f(x) = 0\)). For the rational function \(f(x)=\frac{3x + 9}{x-2}\), we set \(f(x)=0\).

Step 2: Solve for \(x\)

Set \(\frac{3x + 9}{x - 2}=0\). A fraction is zero when its numerator is zero (and the denominator is not zero). So we solve \(3x+9 = 0\).
Subtract 9 from both sides: \(3x=-9\).
Divide both sides by 3: \(x=- 3\). We check the denominator at \(x = - 3\): \(-3-2=-5
eq0\), so it is a valid solution.

Step 1: Recall vertical asymptote definition

For a rational function \(f(x)=\frac{N(x)}{D(x)}\), vertical asymptotes occur where the denominator \(D(x) = 0\) and the numerator \(N(x)
eq0\) at those points.

Step 2: Solve for denominator zero

For \(f(x)=\frac{3x + 9}{x-2}\), set the denominator \(x - 2=0\). Solving for \(x\), we get \(x = 2\).
Check the numerator at \(x = 2\): \(3(2)+9=6 + 9=15
eq0\). So there is one vertical asymptote at \(x = 2\).

Step 1: Recall degree rules

For a rational function \(f(x)=\frac{N(x)}{D(x)}\), let the degree of \(N(x)\) be \(n\) and the degree of \(D(x)\) be \(d\).

• If \(n
• If \(n=d\), horizontal asymptote is \(y=\frac{\text{leading coefficient of }N(x)}{\text{leading coefficient of }D(x)}\).
• If \(n=d + 1\), there is a slant asymptote (found by long division).
• If \(n>d + 1\), no horizontal or slant asymptote.

For \(f(x)=\frac{3x + 9}{x-2}\), the degree of \(N(x)=3x + 9\) is \(n = 1\) and the degree of \(D(x)=x - 2\) is \(d=1\) (so \(n=d\)).

Step 2: Calculate horizontal asymptote

The leading coefficient of \(N(x)\) is 3 and the leading coefficient of \(D(x)\) is 1. So the horizontal asymptote is \(y=\frac{3}{1}=3\).

Answer:

A. The x - intercept(s) is(are) \(x=-3\)

Part f (Vertical Asymptotes)