Supplemental Videos
The main topics of this section are also presented in the following videos:
The main topics of this section are also presented in the following videos:
We have previously seen that a polynomial function is defined for all values of \(x\text{,}\) and its graph is a smooth curve without any breaks or holes. The graph of a rational function, on the other hand, will have breaks or holes at those \(x\)-values where it is undefined.
Investigate the graph of \(f(x) = \displaystyle{\frac{2}{(x - 3)^2}}\) near \(x = 3\text{.}\)
This function is undefined for \(x = 3\text{,}\) so there is no point on the graph with \(x\)-coordinate \(3\text{.}\) However, we can make a table of values for other values of \(x\text{.}\) Plotting the ordered pairs in the table results in the points shown in Figure345.
\(x\) | \(y\) |
\(0\) | \(\dfrac{2}{9}\) |
\(1\) | \(\dfrac{1}{2}\) |
\(2\) | \(2\) |
\(3\) | undefined |
\(4\) | \(2\) |
\(4\) | \(\dfrac{1}{2}\) |
\(6\) | \(\dfrac{2}{9}\) |
Next, we make a table showing \(x\)-values close to \(3\text{,}\) as in Figure346a. As we choose \(x\)-values closer and closer to \(3\text{,}\) \((x - 3)^2\) gets closer to \(0\text{,}\) so the fraction \(\dfrac{2}{(x-3)^2}\) gets very large. This means that the graph approaches, but never touches, the vertical line \(x=3\text{.}\) In other words, the graph has a vertical asymptote at \(x=3\text{.}\) We indicate the vertical asymptote by a dashed line, as shown in Figure346b.
The vertical line \(x=3\) in the example above is an example of a vertical asymptote. In general, we have the following result.
If \(Q(a) = 0\) but \(P(a) \ne 0\text{,}\) then the graph of the rational function \(f(x) = \displaystyle{\frac{P(x)}{Q(x)}}\) has a vertical asymptote at \(x=a\text{.}\)
If \(P(a)\) and \(Q(a)\) are both zero in the rational function \(f(x)=\frac{P(x)}{Q(x)}\text{,}\) but the reduced expression for \(f(x)\) is defined at \(x=a\text{,}\) then the graph of \(f(x)\) has a hole at \(x=a\) rather than an asymptote.
Find the vertical asymptotes of \(G(x) = \displaystyle{\frac{4x^2}{x^2 - 4}}\text{.}\)
To find the vertical asymptotes, we are looking for values of \(a\) such that \(4a^2 \neq 0\) and \(a^2-4=0\text{.}\) The equation \(a^2-4=0\) has solutions \(a=2,-2\text{,}\) and both \(4(\alert{2})^2\neq0\) and \(4(\alert{-2})^2\neq0\text{.}\) This means that there are two vertical asymptotes, namely \(x=2\) and \(x=-2\text{.}\)
Near a vertical asymptote, the graph of a rational function has one of the four characteristic shapes, illustrated in Figure348. Locating the vertical asymptotes can help us make a quick sketch of a rational function.
Locate the vertical asymptotes and sketch the graph of \(g(x) = \displaystyle{\frac{x}{x + 1}}\text{.}\)
The denominator, \(x+1\text{,}\) equals zero when \(x = -1\text{.}\) Because the numerator does not equal zero when \(x = -1\text{,}\) there is a vertical asymptote at \(x = -1\text{.}\) The asymptote separates the graph into two pieces.
We can use the Table feature of a calculator to evaluate \(g(x)\) for several values of \(x\) on either side of the asymptote, as shown in Figure350a. We plot the points found in this way; then connect the points on either side of the asymptote to obtain the graph shown in Figure350b.
Find the vertical asymptotes of \(f(x) = \displaystyle{\frac{1}{x^2 - 4}}\) . Locate any \(x\)-intercepts.
Evaluate the function at \(x = -3\text{,}\) \(-1\text{,}\) \(1\text{,}\) and \(3\text{.}\) Sketch a graph of the function.
The denominator, \(x^2-4\text{,}\) equals zero when \(x=2\) and \(x=-2\text{.}\) Because the numerator does not equal zero for either of these, there are two vertical asymptotes, at \(x=2\) and \(x=-2\text{.}\)
To find the \(x\)-intercepts, we are looking for values of \(x\) such that the function is equal to zero:
Since \(1\) divided by any real number is nonzero, there is no such \(x\text{.}\) This means that there are no \(x\)-intercepts.
\(x\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
\(y\) | \(\frac{1}{5} \) | \(\frac{-1}{3} \) | \(\frac{-1}{3} \) | \(\frac{1}{5} \) |