Question

follow the steps for graphing a rational function to graph the function ( r(x)=\frac{x^{2}+9 x + 18}{x + 6} )

b. the function has two oblique asymptotes. the oblique asymptote with a negative slope is, and the oblique asymptote with a positive slope is
(type equations. use integers or fractions for any numbers in the equations )
c. the function has no oblique asymptote
determine the points, if any, at which the graph of ( r ) intersects the horizontal or oblique asymptote. if one exists. select the correct choice and, if necessary, fill in the answer box within your choice
a. the graph of ( r ) intersects the horizontal or oblique asymptote at
(simplify your answer. type an ordered pair. use a comma to separate answers as needed )
b. there is no point at which the graph of ( r ) intersects the horizontal or oblique asymptote
c. there is no horizontal or oblique asymptote
use the real zeros of the numerator and denominator of ( r ) to divide the ( x )-axis into intervals. determine where the graph of ( r ) is above or below the ( x )-axis by choosing a number in each interval and evaluatin complete your choice
(type your answer in interval notation. use a comma to separate answers as needed )
a. the graph of ( r ) is below the ( x )-axis on the interval(s)
b. the graph of ( r ) is above the ( x )-axis on the interval(s)
c. the graph of ( r ) is above the ( x )-axis on the interval(s) and below the ( x )-axis on the interval(s)
use the results from the previous steps to graph ( r ). choose the correct graph
a.
b.
c.

Explanation:

Step1: Simplify the rational function

First, factor the numerator \(x^{2}+9x + 18=(x + 3)(x+6)\). Then \(R(x)=\frac{x^{2}+9x + 18}{x + 6}=\frac{(x + 3)(x + 6)}{x+6}=x + 3\) for \(x
eq-6\).

Step2: Analyze asymptotes

Since \(R(x)=x + 3\) (a linear function) after simplification (with a hole at \(x=-6\)), there is no oblique asymptote (because the simplified function is linear, not a non - linear polynomial divided by a linear polynomial in the form that would create an oblique asymptote).

Step3: Check intersection with asymptote

Since there is no horizontal or oblique asymptote (the function \(y=x + 3\) is not an asymptote in the traditional sense for a rational function here, as the original rational function simplifies to a linear function with a removable discontinuity), we consider the nature of the function.
The \(x\) - intercept is found by setting \(y = 0\): \(0=x+3\Rightarrow x=-3\), so the point is \((-3,0)\). The \(y\) - intercept is found by setting \(x = 0\): \(y=0 + 3=3\), so the point is \((0,3)\).
The function \(y=x + 3\) is a straight line. The domain is all real numbers except \(x=-6\).

Answer:

C. The function has no oblique asymptote.
C. There is no horizontal or oblique asymptote.
Since \(R(x)=x + 3\) (a straight line with a hole at \(x=-6\)), it is above the \(x\) - axis when \(x+3>0\Rightarrow x>-3\) and below the \(x\) - axis when \(x + 3<0\Rightarrow x<-3\) (but \(x
eq-6\)). So the graph is above the \(x\) - axis on the interval \((-3,\infty)\) and below the \(x\) - axis on the interval \((-\infty,-6)\cup(-6,-3)\)