We will have the following:
First, we are given the expression:
[tex]y=\frac{1}{x}[/tex]Now we will have a table, from x = -10 to x = 10, that is:
[tex]\begin{cases}f(-10)=-\frac{1}{10}\Rightarrow f(-10)=-0.1 \\ f(-9)=-\frac{1}{9}\Rightarrow f(-9)=-0.111111111\ldots \\ f(-8)=-\frac{1}{8}\Rightarrow f(-8)=-0.125 \\ f(-7)=-\frac{1}{7}\Rightarrow f(-7)=-0.1428571429\ldots \\ f(-6)=-\frac{1}{6}\Rightarrow f(-6)=-0.166666666\ldots \\ f(-5)=-\frac{1}{5}\Rightarrow f(-5)=-0.2 \\ f(-4)=-\frac{1}{4}\Rightarrow f(-4)=-0.25 \\ f(-3)=-\frac{1}{3}\Rightarrow f(-3)=-0.333333333\ldots \\ f(-2)=-\frac{1}{2}\Rightarrow f(-2)=-0.5 \\ f(-1)=-\frac{1}{1}\Rightarrow f(-1)=-1 \\ f(0)=\infty \\ f(1)=\frac{1}{1}\Rightarrow f(1)=1 \\ f(2)=\frac{1}{2}\Rightarrow f(2)=0.5 \\ f(3)=\frac{1}{3}\Rightarrow f(3)=0.333333333\ldots \\ f(4)=\frac{1}{4}\Rightarrow f(4)=0.25 \\ f(5)=\frac{1}{5}\Rightarrow f(5)=0.2 \\ f(6)=\frac{1}{6}\Rightarrow f(6)=0.166666666\ldots \\ f(7)=\frac{1}{7}\Rightarrow f(7)=1428571429\ldots \\ f(8)=\frac{1}{8}\Rightarrow f(8)=1.125 \\ f(9)=\frac{1}{9}\Rightarrow f(9)=0.111111111\ldots \\ f(10)=\frac{1}{10}\Rightarrow f(10)=0.1\end{cases}[/tex]From this we can see that the function comes from the negative infinite near y = 0 but never 0, at x = 0 the expression grows in the positive axis and negative axis [From the left and right respectively] and goes to infinity, so at x = 0 the expression is undefined; then to the right it comes from an incredibly great number and approaches y = 0 but never touches it, the graph using the points calculated is:
So, there is a vertical asymptote at x = 0, since it is the point in which the expression is undefined:
Now, from this information we can see that the domain and range are:
[tex]\text{Dom}_f=(-\infty,0)\cup(0,\infty)[/tex][tex]\text{Ran}_f=(-\infty,0)\cup(0,\infty)[/tex]
