Answer:
Vertical asymptote: [tex]x=-3[/tex]
Horizontal asymptote: [tex]y=4[/tex]
Step-by-step explanation:
Given:
The function is given as:
[tex]f(x)=\frac{3x^2}{x(x+3)}+1[/tex]
First, we will simplify the given function.
[tex]f(x)=\frac{3x^2}{x(x+3)}+1\\f(x)=\frac{3x}{x+3}+\frac{x+3}{x+3}\\f(x)=\frac{3x+x+3}{x+3}\\f(x)=\frac{4x+3}{x+3}[/tex]
Now, vertical asymptotes occur when denominator is 0. So,
[tex]x+3=0\\ x=-3[/tex]
Therefore, the vertical asymptote is: [tex]x=-3[/tex]
When the degree of numerator is equal to that of the denominator, we have horizontal asymptote equal to [tex]y=\frac{a}{b}[/tex] 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:
[tex]y=\frac{a}{b}\\y=\frac{4}{1}\\y=4[/tex]
