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Find the asymptote of 3x^2/x(x+3) +1

User Henryabra
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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
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