217k views
0 votes
Find the asymptote of 3x^2/x(x+3) +1

User Henryabra
by
7.9k points

1 Answer

5 votes

Answer:

Vertical asymptote:
x=-3

Horizontal asymptote:
y=4

Explanation:

Given:

The function is given as:


f(x)=(3x^2)/(x(x+3))+1

First, we will simplify the given function.


f(x)=(3x^2)/(x(x+3))+1\\f(x)=(3x)/(x+3)+(x+3)/(x+3)\\f(x)=(3x+x+3)/(x+3)\\f(x)=(4x+3)/(x+3)

Now, vertical asymptotes occur when denominator is 0. So,


x+3=0\\ x=-3

Therefore, the vertical asymptote is:
x=-3

When the degree of numerator is equal to that of the denominator, we have horizontal asymptote equal to
y=(a)/(b)
where, 'a' is the leading coefficient of numerator and 'b' is the leading coefficient of denominator.

Here, degree of numerator and denominator are same and equal to 1. The value of 'a' is 4 and 'b' is 1. So, horizontal asymptote is given as:


y=(a)/(b)\\y=(4)/(1)\\y=4

User MJZ
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories