Let's analyze the function:
[tex]f(x)=\frac{x^2-4x+3}{x^2-4x-5}[/tex]To do this let's first find the domain of the function, since this is a rational function we know that the denominator can be zero so let's find for which values of x this happens:
[tex]\begin{gathered} x^2-4x-5=0 \\ (x-5)(x+1)=0 \\ \text{then} \\ x-5=0 \\ x=5 \\ \text{ or} \\ x+1=0 \\ x=-1 \end{gathered}[/tex]Hence we conclude that the fucntion is not defined for x=5 or x=-1, which means that the domain of the function is:
[tex]\text{domain}f=(-\infty,-1)\cup(-1,5)\cup(5,\infty)[/tex]This also means that we may have somo vertical asympotes at x=5 and x=-1; to determine if we do we need to calculate the one-sided limits at those points if the result is negative or positive infinity we have vertical asymptotes. Let's calculate this limits:
[tex]\lim _{x\to-1^-}\frac{x^2-4x+3}{x^2-4x-5}=\infty[/tex]and
[tex]\lim _{x\to5^-}\frac{x^2-4x+3}{x^2-4x-5}=-\infty[/tex]To determine this limits we notice that, as we approach both values the denominator increases arbitrarily when they approach the number from the left; for example, notice that:
[tex]f(-1.0001)=\frac{(-1.0001)^2-4(-1.0001)+3}{(-1.0001)^2-4(-1.0001)-5}=13410.98[/tex]as we approach more an more to -1 the function value will increase and for that reason we conclude the results of the limits are what we stated.
Now, let's determine the zeros of the function:
[tex]\begin{gathered} f(x)=0 \\ \frac{x^2-4x+3}{x^2-4x-5}=0 \\ x^2-4x+3=0 \\ (x-3)(x-1)=0 \\ \text{then } \\ x-3=0 \\ x=3 \\ \text{ or} \\ x-1=0 \\ x=1 \end{gathered}[/tex]Hence the function is zero at x=3 and x=1.
Therefore, we conclude that the local behavior of the functions is:
• The function has two vertical asymptotes: x=-1 and x=5
,• The zeros of the function are x=3 and x=1.
Now, let's find the end behavior. To do this we calculate the limits:
[tex]\begin{gathered} \lim _{x\to-\infty}\frac{x^2-4x+3}{x^2-4x-5} \\ \text{and} \\ \lim _{x\to\infty}\frac{x^2-4x+3}{x^2-4x-5} \end{gathered}[/tex]Let's find the first one:
[tex]\begin{gathered} \lim _{x\to-\infty}\frac{x^2-4x+3}{x^2-4x-5}=\lim _{x\to-\infty}\frac{\frac{x^2}{x^2}-\frac{4x}{x^2}+\frac{3}{x^2}}{\frac{x^2}{x^2}-\frac{4x}{x^2}-\frac{5}{x^2}} \\ =\lim _{x\to-\infty}\frac{1-\frac{4}{x}+\frac{3}{x^2}}{1-\frac{4}{x}-\frac{5}{x^2}} \\ =\frac{1}{1} \\ =1 \end{gathered}[/tex]And:
[tex]\begin{gathered} \lim _{x\to\infty}\frac{x^2-4x+3}{x^2-4x-5}=\lim _{x\to\infty}\frac{\frac{x^2}{x^2}-\frac{4x}{x^2}+\frac{3}{x^2}}{\frac{x^2}{x^2}-\frac{4x}{x^2}-\frac{5}{x^2}} \\ =\lim _{x\to\infty}\frac{1-\frac{4}{x}+\frac{3}{x^2}}{1-\frac{4}{x}-\frac{5}{x^2}} \\ =\frac{1}{1} \\ =1 \end{gathered}[/tex]Since both limits are the same we can conclude that we only have one horizontal asymptote, and that it's given as:
[tex]y=1[/tex]Therefore, the horizontal asymptote is y=1 and this described the end behaviour of the function that can be also written as:
[tex]\begin{gathered} \text{ As }x\rightarrow\infty,f(x)\rightarrow1 \\ \text{ As }x\rightarrow-\infty,f(x)\rightarrow1 \end{gathered}[/tex]All the information we found can be seen in the graph of the function:
