Question

follow the steps for graphing a rational function to graph the function ( r(x)=\frac{x^{2}+5 x - 24}{x^{2}-x - 6} ).
a. the graph of ( r ) intersects the horizontal or oblique asymptote at
(simplify your answer. type an ordered pair, using integers or fractions. 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 below the ( x )-axis on the interval(s)
(type your answer 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 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.)

Explanation:

Step1: Simplify the rational function

Factor the numerator \(x^{2}+5x - 24=(x + 8)(x-3)\) and the denominator \(x^{2}-x - 6=(x - 3)(x + 2)\).
So \(R(x)=\frac{(x + 8)(x - 3)}{(x - 3)(x + 2)}=\frac{x + 8}{x + 2}\), \(x
eq3\)

Step2: Find the horizontal asymptote

Since the degree of the numerator and the denominator is the same (degree \(n = 1\)), the horizontal asymptote is \(y=\frac{a_{n}}{b_{n}}\), where \(a_{n}\) is the leading coefficient of the numerator and \(b_{n}\) is the leading coefficient of the denominator. Here \(a_{n}=1\) and \(b_{n}=1\), so \(y = 1\)

Step3: Find the intersection of the graph and the horizontal asymptote

Set \(R(x)=1\), \(\frac{x + 8}{x+2}=1\). Cross - multiply: \(x + 8=x + 2\), \(8=2\) (no solution)

Step4: Find the zeros of the simplified function

Set \(y=\frac{x + 8}{x + 2}=0\), then \(x=-8\) (zero of the function)
The vertical asymptote is \(x=-2\)

Step5: Test intervals

The intervals are \((-\infty,-8)\), \((-8,-2)\), \((-2,\infty)\)

• For \(x=-9\) (in \((-\infty,-8)\)): \(R(-9)=\frac{-9 + 8}{-9+2}=\frac{-1}{-7}=\frac{1}{7}>0\)
• For \(x=-5\) (in \((-8,-2)\)): \(R(-5)=\frac{-5 + 8}{-5+2}=\frac{3}{-3}=-1<0\)
• For \(x=0\) (in \((-2,\infty)\)): \(R(0)=\frac{0 + 8}{0+2}=4>0\)

Answer:

For the first part:

• Option C: There is no point at which the graph of \(R\) intersects the horizontal or oblique asymptote.

For the second part:

• The graph of \(R\) is above the \(x\) - axis on the intervals \((-\infty,-8)\cup(-2,\infty)\) and below the \(x\) - axis on the interval \((-8,-2)\). So we choose option C, with the graph of \(R\) above the \(x\) - axis on \((-\infty,-8)\cup(-2,\infty)\) and below the \(x\) - axis on \((-8,-2)\)