N 28
we have the rational function
[tex]f(x)=\frac{x^2-4x+3}{x^2-4x-5}[/tex]
Simplify the numerator
[tex]x^2-4x+3=(x-1)(x-3)[/tex]
Simplify the denominator
[tex]x^2-4x-5=(x+1)(x-5)[/tex]
substitute in the given function
[tex]f(x)=\frac{(x-1)(x-3)}{(x+1)(x-5)}[/tex]
Remember that the denominator cannot be equal to zero
The domain of the rational function is all real numbers except for x=-1 and x=5
we have a vertical asymptote at x=-1 and at x=5
Find out horizontal asymptotes
Degree on Top is Equal to the Bottom
In this case, the graph has a horizontal asymptote along y=1/1=1
at y=1
see the attached figure to better understand the problem
End behavior
as x→−∞ ------> f(x)→1
as x→+∞ ------> f(x)→1
as x→-1 to the left ------> f(x) →+∞
as x→-1 to the right -----> f(x)→−∞
as x→5 to the left -----> f(x) →−∞
as x→5 to the right ----> f(x)→+∞
