Question

follow the steps for graphing a rational function to graph the function ( g(x)=\frac{5x}{x^{2}-25} ).
(type an ordered pair using integers or fractions.)
b. there are no holes in the graph.
determine the behavior of the graph on either side of any vertical asymptotes, if any exist. select the correct choice and, if necessary, fill in the answer box(es) to complete your choice.
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 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 each answer only once.)
c. 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.)
d. there is no vertical asymptote.
determine the horizontal asymptote(s), if any exist. select the correct choice and, if necessary, fill in the answer box(ex) to complete your choice.
a. the function has one horizontal asymptote,

Explanation:

Step1: Find vertical asymptotes

For a rational function \(G(x)=\frac{N(x)}{D(x)}\), vertical asymptotes occur where \(D(x) = 0\) (and \(N(x)
eq0\)).
Set \(x^{2}-25=(x + 5)(x - 5)=0\). So \(x=-5\) and \(x = 5\).
Check \(N(-5)=5\times(-5)=-25
eq0\) and \(N(5)=5\times5 = 25
eq0\).

Step2: Analyze behavior near vertical asymptotes

For \(x=-5\):
Let \(x=-5 + h\), where \(h\to0\).
\(G(-5 + h)=\frac{5(-5 + h)}{(-5 + h)^{2}-25}=\frac{-25+5h}{25-10h+h^{2}-25}=\frac{-25 + 5h}{h^{2}-10h}=\frac{h(-25/h + 5)}{h(h - 10)}\)
As \(h\to0^{+}\) (approaching \(-5\) from the right), \(G(-5 + h)\approx\frac{-25}{- 10h}\to\infty\)
As \(h\to0^{-}\) (approaching \(-5\) from the left), \(G(-5 + h)\approx\frac{-25}{-10h}\to-\infty\)
For \(x = 5\):
Let \(x=5 + h\), where \(h\to0\).
\(G(5 + h)=\frac{5(5 + h)}{(5 + h)^{2}-25}=\frac{25+5h}{25 + 10h+h^{2}-25}=\frac{25+5h}{h^{2}+10h}=\frac{h(25/h+5)}{h(h + 10)}\)
As \(h\to0^{+}\) (approaching \(5\) from the right), \(G(5 + h)\approx\frac{25}{10h}\to\infty\)
As \(h\to0^{-}\) (approaching \(5\) from the left), \(G(5 + h)\approx\frac{25}{10h}\to-\infty\)

Step3: Find horizontal asymptote

For \(G(x)=\frac{5x}{x^{2}-25}\), since the degree of the numerator \(n = 1\) and the degree of the denominator \(m=2\) (\(n\lt m\))
By the rule of horizontal asymptotes for rational functions (\(y = 0\) when \(n\lt m\))

Answer:

For vertical asymptotes:

• Option C: It approaches \(\infty\) on one side of the asymptote(s) at \(x=-5\) and \(-\infty\) on the other. It approaches \(\infty\) on one side of the asymptote(s) at \(x = 5\) and \(-\infty\) on the other.

For horizontal asymptote:

• Option A: The function has one horizontal asymptote, \(y = 0\)