First, we make a table. With these kinds of functions, fractions will be a necessity to understand their general shape! Also, since there are two pieces to the graph, we want to make sure to use enough points to capture both pieces!

\(\begin{array} {|c|c|}\hline x & f(x) & (x, f(x)) \\ \hline -3 & \frac{1}{-3} + 4 = -\frac{1}{3} + 4 = \frac{11}{3}& (-3, \frac{11}{3})\\ \hline -2 & \frac{1}{-2} + 4 = -\frac{1}{2} + 4 = \frac{7}{2}& (-2, \frac{7}{2})\\ \hline -1 & \frac{1}{-1} + 4 = -1 + 4 = 3& (-1, 3)\\ \hline -\frac{1}{2} & \frac{1}{-1/2} + 4 = -2 + 4 = 2& (-\frac{1}{2}, 2) \\ \hline -\frac{1}{3} & \frac{1}{-1/3} + 4 = -3 + 4 = 1& (-\frac{1}{3}, 1)\\ \hline 0 & ??? + 4 =???& \text{ draw vertical gap line at \(x = 0\)}\\ \hline \frac{1}{3} & \frac{1}{1/3} + 4 = 3 + 4 = 7& (\frac{1}{3}, 7)\\ \hline \frac{1}{2} & \frac{1}{1/2} + 4 = 2 + 4 = 6& (\frac{1}{2}, 6) \\ \hline 1 & \frac{1}{1} + 4 = 1 + 4 = 3& (1, 5)\\ \hline 2 & \frac{1}{2} + 4 = \frac{1}{2} + 4 = \frac{9}{2}& (2, \frac{9}{2})\\ \hline 3 & \frac{1}{3} + 4 = \frac{1}{3} + 4 = \frac{13}{3}& (3, \frac{13}{3})\\ \hline \end{array}\)

The vertical asymptote goes at the \(x\) value that make the denominator zero. In this example, the denomator is zero when \(x = 0\), so we can add a dashed line at \(x = 0\) to remind ourselves we don't want to graph there.

The horizontal asymptote goes at the \(y\) value that each part of the graph are approaching. In this example, each half of the graph will approach \(y = 4\), so we can add a dashed line at \(y = 4\) to remind ourselves that we don't want to graph there.

Now we plot the points and asymptotes.

When we add the lines, we have the two curves, where the curves follow the points AND asymptotes.