Question

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

the correct choice and, if necessary, fill in the answer box to complete your choice.

a. the graph of ( r ) intersects the horizontal or oblique asymptote at ( (-2,0) ).
(simplify your answer. type an ordered pair. use a comma to separate answers as needed.)

b. the graph of ( r ) intersects the horizontal or oblique asymptote at infinitely many points.

c. there is no point at which the graph of ( r ) intersects the horizontal or oblique asymptote.

d. 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 evaluating ( r ) there. select the correct choice and fill in the answer box(es) to complete your choice.

a. the graph of ( r ) is above the ( x )-axis on the interval(s) and below the ( x )-axis on the interval(s)
(type your answers in interval notation. use a comma to separate answers as needed.)

b. the graph of ( r ) is above the ( x )-axis on the interval(s)
(type your answer in interval notation. use a comma to separate answers as needed.)

c. the graph of ( r ) is below the ( x )-axis on the interval(s)
(type your answer in interval notation. use a comma to separate answers as needed.)

Explanation:

Step1: Find the horizontal asymptote

For a rational function \(R(x)=\frac{f(x)}{g(x)}\) where \(f(x)=x + 2\) (degree \(n = 1\)) and \(g(x)=x(x + 9)=x^{2}+9x\) (degree \(m=2\)). Since \(n

Step2: Find the intersection of the graph and the horizontal asymptote

Set \(R(x)=0\), so \(\frac{x + 2}{x(x + 9)}=0\). A fraction is zero when the numerator is zero and the denominator is non - zero. Solving \(x+2=0\) gives \(x=-2\). When \(x =-2\), the denominator \(x(x + 9)=(-2)(-2 + 9)=(-2)\times7=-14
eq0\). So the graph intersects the horizontal asymptote \(y = 0\) at the point \((-2,0)\).

Step3: Find the real zeros of the numerator and denominator

The zero of the numerator \(x+2\) is \(x=-2\). The zeros of the denominator \(x(x + 9)\) are \(x = 0\) and \(x=-9\). These points divide the \(x\) - axis into the intervals \((-\infty,-9)\), \((-9,-2)\), \((-2,0)\) and \((0,\infty)\).

Step4: Test the sign of \(R(x)\) in each interval

• For the interval \((-\infty,-9)\), let \(x=-10\). Then \(R(-10)=\frac{-10 + 2}{-10(-10 + 9)}=\frac{-8}{-10\times(-1)}=\frac{-8}{10}<0\).
• For the interval \((-9,-2)\), let \(x=-5\). Then \(R(-5)=\frac{-5 + 2}{-5(-5 + 9)}=\frac{-3}{-5\times4}=\frac{-3}{-20}>0\).
• For the interval \((-2,0)\), let \(x=-1\). Then \(R(-1)=\frac{-1+2}{-1(-1 + 9)}=\frac{1}{-1\times8}<0\).
• For the interval \((0,\infty)\), let \(x = 1\). Then \(R(1)=\frac{1 + 2}{1(1 + 9)}=\frac{3}{10}>0\).

Answer:

A. The graph of \(R\) intersects the horizontal or oblique asymptote at \((-2,0)\).
A. The graph of \(R\) is above the \(x\) - axis on the intervals \((-9,-2),(0,\infty)\) and below the \(x\) - axis on the intervals \((-\infty,-9),(-2,0)\)