Write a function with vertical asymptote x=4, horizontal asymptote y=1, y intercept at (0,2).
A possible function can be express as:
[tex]f(x)=\frac{x-8}{x-4}[/tex]Let's prove that this function fulfils our conditions. Let's start with the y-intercept, we know that this happens when x=0, then we have:
[tex]f(0)=\frac{0-8}{0-4}=2[/tex]Hence the y-intercept is at (0,2).
Now, we know that a rational function has horizontal asymptote y=b if:
[tex]\begin{gathered} \lim_{x\to\infty}f(x)=b \\ \text{ or } \\ \lim_{x\to-\infty}f(x)=b \end{gathered}[/tex]Let's find these limits:
[tex]\begin{gathered} \lim_{x\to\infty}\frac{x-8}{x-4}=\lim_{x\to\infty}\frac{\frac{x}{x}-\frac{8}{x}}{\frac{x}{x}-\frac{4}{x}} \\ =\lim_{x\to\infty}\frac{1-\frac{8}{x}}{1-\frac{4}{x}} \\ =\frac{1-0}{1-0} \\ =1 \end{gathered}[/tex]and:
[tex]\begin{gathered} \lim_{x\to-\infty}\frac{x-8}{x-4}=\lim_{x\to-\infty}\frac{\frac{x}{x}-\frac{8}{x}}{\frac{x}{x}-\frac{4}{x}} \\ =\lim_{x\to-\infty}\frac{1-\frac{8}{x}}{1-\frac{4}{x}} \\ =\frac{1-0}{1-0} \\ =1 \end{gathered}[/tex]This means that we have a horizontal asymptote y=1 as we wanted.
Now, a rational function has vertical asymptote at x=a if:
[tex]\begin{gathered} \lim_{x\to a^-}f(x)=\pm\infty \\ \text{ or } \\ \lim_{x\to a^+}f(x)=\pm\infty \end{gathered}[/tex]to determine the value of a we need to look where the function is not defined, that is, the values which make the denominator zero, in this case we have:
[tex]\begin{gathered} x-4=0 \\ x=4 \end{gathered}[/tex]Then we need to find the limits:
[tex]\begin{gathered} \lim_{x\to4^-}\frac{x-8}{x-4} \\ \text{ and } \\ \lim_{x\to4^+}\frac{x-8}{x-4} \end{gathered}[/tex]Now, if we approach the value x=4 from the left we notice that as x gets closer to 4 the function gets bigger and bigger, for example:
[tex]f(3.9999)=\frac{3.9999-8}{3.9999-4}=400001[/tex]if we follow this procedure, we conclude that:
[tex]\lim_{x\to4^-}\frac{x-8}{x-4}=\infty[/tex]Similarly, if we approach x=4 from the right the function gets smaller and smaller, for example:
[tex]f(4.0001)=\frac{4.0001-8}{4.0001-4}=-39999[/tex]Then we can conclude that:
[tex]\lim_{x\to4^+}\frac{x-8}{x-4}=-\infty[/tex]Hence, we conclude that the function we proposed has a vertical asymptote x=4 like we wanted.
the properties we gave can be seen in the following graph:
