Question

follow the steps for graphing a rational function to graph the function ( r(x)=\frac{9}{(x - 5)(x^{2}-81)} ).

a. it approaches ( infty ) on one side of the asymptote(s) at ( x=) and ( -infty ) on the other. it approaches either ( infty ) or ( -infty ) on both sides of the asymptote(s) at ( x=).
(type integers or simplified fractions. use a comma to separate answers as needed. type each answer only once.)
b. it approaches ( infty ) on one side of the asymptote(s) at ( x=) and ( -infty ) on the other.
(type an integer or a simplified fraction. use a comma to separate answers as needed. type each answer only once.)
c. it approaches either ( infty ) or ( -infty ) on both sides of the asymptote(s) at ( x=).
(type an integer or a simplified fraction. use a comma to separate answers as needed. type

Explanation:

Step1: Find the domain

Set the denominator equal to zero: \((x - 5)(x^{2}-81)=(x - 5)(x - 9)(x + 9)=0\). The vertical asymptotes occur at \(x=-9,x = 9\) (since \(x = 5\) is a removable discontinuity as the numerator is non - zero when \(x = 5\)).

Step2: Analyze the behavior near vertical asymptotes

For a vertical asymptote \(x=a\), we consider the sign of the function \(R(x)=\frac{9}{(x - 5)(x - 9)(x + 9)}\) as \(x\to a^{+}\) and \(x\to a^{-}\).

• For \(x = 9\):

Let \(x=9 + h\) where \(h\to0^{+}\). Then \(R(x)=\frac{9}{(9 + h-5)(9 + h-9)(9 + h + 9)}=\frac{9}{(4 + h)(h)(18+h)}\approx\frac{9}{4h\times18}\to+\infty\) as \(h\to0^{+}\).
Let \(x=9 - h\) where \(h\to0^{+}\). Then \(R(x)=\frac{9}{(9 - h-5)(9 - h-9)(9 - h + 9)}=\frac{9}{(4 - h)(-h)(18 - h)}\approx\frac{9}{4\times(-h)\times18}\to-\infty\) as \(h\to0^{+}\).

• For \(x=-9\):

Let \(x=-9 + h\) where \(h\to0^{+}\). Then \(R(x)=\frac{9}{(-9 + h-5)(-9 + h-9)(-9 + h + 9)}=\frac{9}{(-14 + h)(-18+h)(h)}\approx\frac{9}{(-14)\times(-18)h}\to+\infty\) as \(h\to0^{+}\).
Let \(x=-9 - h\) where \(h\to0^{+}\). Then \(R(x)=\frac{9}{(-9 - h-5)(-9 - h-9)(-9 - h + 9)}=\frac{9}{(-14 - h)(-18 - h)(-h)}\approx\frac{9}{(-14)\times(-18)\times(-h)}\to-\infty\) as \(h\to0^{+}\).

Answer:

A. It approaches \(\infty\) on one side of the asymptote(s) at \(x = 9\) and \(-\infty\) on the other. It approaches either \(\infty\) or \(-\infty\) on both sides of the asymptote(s) at \(x=-9\).