Suppose \(r\) is a rational function. You can graph the function \(r\) using the following six-step procedure:

  1. Find the domain of \(r\).
  2. Reduce \(r(x)\) to lowest terms, if applicable.
  3. Determine the location of any vertical asymptotes or holes in the graph, if they exist.
  4. Find the axis intercepts, if they exist.
  5. Analyze the end behavior of \(r\). Find the horizontal or slant asymptote, if one exists.
  6. Use a sign diagram and plot additional points, as needed, to sketch the graph.
In Exercises sixstepfirst - sixsteplast, use the six-step procedure to graph the rational function. Be sure to draw any asymptotes as dashed lines.
\(f(x) = \dfrac {4}{x + 2}\)

\(f(x) = \dfrac {4}{x + 2}\)
Domain: \((-\infty , -2) \cup (-2, \infty )\)
No \(x\)-intercepts
\(y\)-intercept: \((0, 2)\)
Vertical asymptote: \(x = -2\)
\(\lim _{x \rightarrow -2^{-}} f(x) = -\infty \), \(\lim _{x \rightarrow -2^{+}} f(x) = \infty \)
Horizontal asymptote: \(y = 0\)
\(\lim _{x \rightarrow - \infty } f(x) = 0\)
More specifically, as \(x \rightarrow -\infty , \; f(x) \rightarrow 0^{-}\)
\(\lim _{x \rightarrow \infty } f(x) = 0\)
More specifically, as \(x \rightarrow \infty , \; f(x) \rightarrow 0^{+}\)

\(f(x) = 5x(6-2x)^{-1}\)

\(f(x) = 5x(6-2x)^{-1} = \dfrac {5x}{6 - 2x}\)
Domain: \((-\infty , 3) \cup (3, \infty )\)
\(x\)-intercept: \((0, 0)\)
\(y\)-intercept: \((0, 0)\)
Vertical asymptote: \(x = 3\)
\(\lim _{x \rightarrow 3^{-}} f(x) = \infty \), \(\lim _{x \rightarrow 3^{+}} f(x) = -\infty \)
Horizontal asymptote: \(y = -\frac {5}{2}\)
\(\lim _{x \rightarrow - \infty } f(x) =\) \(-\frac {5}{2}\)
More specifically, as \(x \rightarrow -\infty , \; f(x) \rightarrow -\frac {5}{2}^{+}\)
\(\lim _{x \rightarrow -\infty } f(x) =\) \(-\frac {5}{2}\)
More specifically, as \(x \rightarrow \infty , \; f(x) \rightarrow -\frac {5}{2}^{-}\)

\(g(t) = t^{-2}\)

\(g(t) = t^{-2} = \dfrac {1}{t^{2}}\)
Domain: \((-\infty , 0) \cup (0, \infty )\)
No \(t\)-intercepts
No \(y\)-intercepts
Vertical asymptote: \(t = 0\)
\(\lim _{t \rightarrow 0} g(t) = \infty \)
Horizontal asymptote: \(y = 0\)
\(\lim _{t \rightarrow -\infty } g(t) = 0\)
More specifically, as \(t \rightarrow -\infty , \; g(t) \rightarrow 0^{+}\)
\(\lim _{t \rightarrow \infty } g(t) = 0\)
More specifically, as \(t \rightarrow \infty , \; g(t) \rightarrow 0^{+}\)

\(g(t) = \dfrac {1}{t^{2} + t - 12}\)

\(g(t) = \frac {1}{t^{2} + t - 12} = \frac {1}{(t - 3)(t + 4)}\)
Domain: \((-\infty , -4) \cup (-4, 3) \cup (3, \infty )\)
No \(t\)-intercepts
\(y\)-intercept: \((0, -\frac {1}{12})\)
Vertical asymptotes: \(t = -4\) and \(t = 3\)
\(\lim _{t \rightarrow -4^{-}} g(t) = \infty \), \(\lim _{t \rightarrow -4^{+}} g(t) = -\infty \)
\(\lim _{t \rightarrow 3^{-}} g(t) = -\infty \), \(\lim _{t \rightarrow 3^{+}} g(t) = \infty \)
Horizontal asymptote: \(y = 0\)
\(\lim _{t \rightarrow -\infty } g(t) = 0\)
More specifically, as \(t \rightarrow -\infty , \; g(t) \rightarrow 0^{+}\)
\(\lim _{t \rightarrow \infty } g(t) = 0\)
More specifically, as \(t \rightarrow \infty , \; g(t) \rightarrow 0^{+}\)

\(r(z) = \dfrac {2z - 1}{-2z^{2} - 5z + 3}\)

\(r(z) = \frac {2z - 1}{-2z^{2} - 5z + 3} = -\frac {2z - 1}{(2z - 1)(z + 3)}\)
Domain: \((-\infty , -3) \cup (-3, \frac {1}{2}) \cup (\frac {1}{2}, \infty )\)
No \(z\)-intercepts
\(y\)-intercept: \((0, -\frac {1}{3})\)
\(r(z) = \frac {-1}{z + 3}, \; z \neq \frac {1}{2}\)
Hole in the graph at \((\frac {1}{2}, -\frac {2}{7})\)
Vertical asymptote: \(z = -3\)
\(\lim _{z \rightarrow -3^{-}} r(z) = \infty \), \(\lim _{z \rightarrow -3^{+}} r(z) = -\infty \)
Horizontal asymptote: \(y = 0\)
\(\lim _{z \rightarrow -\infty } r(z) = 0\)
More specifically, as \(z \rightarrow -\infty , \; r(z) \rightarrow 0^{+}\)
\(\lim _{z \rightarrow \infty } r(z) = 0\)
More specifically, as \(z \rightarrow \infty , \; r(z) \rightarrow 0^{-}\)

\(r(z) = \dfrac {z}{z^{2} + z - 12}\)

\(r(z) = \frac {z}{z^{2} + z - 12} = \frac {z}{(z - 3)(z + 4)}\)
Domain: \((-\infty , -4) \cup (-4, 3) \cup (3, \infty )\)
\(z\)-intercept: \((0, 0)\)
\(y\)-intercept: \((0, 0)\)
Vertical asymptotes: \(z = -4\) and \(z = 3\)
\(\lim _{z \rightarrow -4^{-}} r(z) = -\infty \), \(\lim _{z \rightarrow -4^{+}} r(z) = \infty \)
\(\lim _{z \rightarrow 3^{-}} r(z) = -\infty \), \(\lim _{z \rightarrow 3^{+}} r(z) = \infty \)
Horizontal asymptote: \(y = 0\)
\(\lim _{z \rightarrow -\infty } r(z) = 0\)
More specifically, as \(z \rightarrow -\infty , \; r(z) \rightarrow 0^{-}\)
\(\lim _{z \rightarrow \infty } r(z) = 0\)
More specifically, as \(z \rightarrow \infty , \; r(z) \rightarrow 0^{+}\)

\(f(x) = 4x(x^2+4)^{-1}\)

\(f(x) = 4x(x^2+4)^{-1} = \frac {4x}{x^{2} + 4}\)
Domain: \((-\infty , \infty )\)
\(x\)-intercept: \((0,0)\)
\(y\)-intercept: \((0,0)\)
No vertical asymptotes
No holes in the graph
Horizontal asymptote: \(y = 0\)
\(\lim _{x \rightarrow -\infty } f(x) = 0\)
More specifically, as \(x \rightarrow -\infty , f(x) \rightarrow 0^{-}\)
\(\lim _{x \rightarrow \infty } f(x) = 0\)
More specifically, as \(x \rightarrow \infty , f(x) \rightarrow 0^{+}\)

\(f(x) = 4x(x^2-4)^{-1}\)

\(f(x) = 4x(x^2-4)^{-1} = \frac {4x}{x^{2} -4} = \frac {4x}{(x + 2)(x - 2)}\)
Domain: \((-\infty , -2) \cup (-2, 2) \cup (2, \infty )\)
\(x\)-intercept: \((0,0)\)
\(y\)-intercept: \((0,0)\)
Vertical asymptotes: \(x = -2, x = 2\)
\(\lim _{x \rightarrow -2^{-}} f(x) = -\infty \), \(\lim _{x \rightarrow -2^{+}} f(x) = \infty \)
\(\lim _{x \rightarrow 2^{-}} f(x) = -\infty \), \(\lim _{x \rightarrow 2^{+}} f(x) = \infty \)
No holes in the graph
Horizontal asymptote: \(y = 0\)
\(\lim _{x \rightarrow -\infty } f(x) = 0\)
More specifically, as \(x \rightarrow -\infty , f(x) \rightarrow 0^{-}\)
\(\lim _{x \rightarrow \infty } f(x) = 0\)
More specifically, as \(x \rightarrow \infty , f(x) \rightarrow 0^{+}\)

\(g(t) = \dfrac {t^2-t-12}{t^2+t-6}\)

\(g(t) = \frac {t^2-t-12}{t^{2} +t - 6} = \frac {t-4}{t - 2}, \, t \neq -3\)
Domain: \((-\infty , -3) \cup (-3, 2) \cup (2, \infty )\)
\(t\)-intercept: \((4,0)\)
\(y\)-intercept: \((0,2)\)
Vertical asymptote: \(t = 2\)
\(\lim _{t \rightarrow 2^{-}} g(t) = \infty \), \(\lim _{t \rightarrow 2^{+}} g(t) = -\infty \)
Hole at \(\left (-3, \frac {7}{5} \right )\)
Horizontal asymptote: \(y = 1\)
\(\lim _{t \rightarrow -\infty } g(t) = 1\)
More specifically, as \(t \rightarrow -\infty , g(t) \rightarrow 1^{+}\)
\(\lim _{t \rightarrow \infty } g(t) = 1\)
More specifically, as \(t \rightarrow \infty , g(t) \rightarrow 1^{-}\)

\(g(t) = 3- \dfrac {5t-25}{t^2-9}\)

\(g(t) = 3- \frac {5t-25}{t^2-9} = \frac {3t^2-5t-2}{t^{2} -9}\)
\(\phantom {g(t)}= \frac {(3t+1)(t-2)}{(t + 3)(t - 3)}\)
Domain: \((-\infty , -3) \cup (-3, 3) \cup (3, \infty )\)
\(t\)-intercepts: \(\left (-\frac {1}{3}, 0 \right )\), \((2,0)\)
\(y\)-intercept: \(\left (0, \frac {2}{9} \right )\)
Vertical asymptotes: \(t = -3, t = 3\)
\(\lim _{t \rightarrow -3^{-}} g(t) = \infty \), \(\lim _{t \rightarrow -3^{+}} g(t) = -\infty \)
\(\lim _{t \rightarrow 3^{-}} g(t) = -\infty \), \(\lim _{t \rightarrow 3^{+}} g(t) = \infty \)
Horizontal asymptote: \(y = 3\)
\(\lim _{t \rightarrow -\infty } g(t) = 3\)
More specifically, as \(t \rightarrow -\infty , g(t) \rightarrow 3^{+}\)
\(\lim _{t \rightarrow \infty } g(t) = 3\)
More specifically, as \(t \rightarrow \infty , g(t) \rightarrow 3^{-}\)

\(r(z) = \dfrac {z^2-z-6}{z+1}\)

\(r(z) = \frac {z^2-z-6}{z+1} = \frac {(z-3)(z+2)}{z+1}\)
Domain: \((-\infty , -1) \cup (-1, \infty )\)
\(z\)-intercepts: \((-2,0)\), \((3,0)\)
\(y\)-intercept: \((0,-6)\)
Vertical asymptote: \(z = -1\)
\(\lim _{z \rightarrow -1^{-}} r(z) = \infty \), \(\lim _{z \rightarrow -1^{+}} r(z) = -\infty \)
Slant asymptote: \(y = z-2\)
\(\lim _{z \rightarrow -\infty } r(z) = -\infty \)
As \(z \rightarrow -\infty \), the graph is above \(y=z-2\)
\(\lim _{z \rightarrow \infty } r(z) = \infty \)
As \(z \rightarrow \infty \), the graph is below \(y=z-2\)

\(r(z) =-z-2+\dfrac {6}{3-z}\)

\(r(z) = -z-2+\frac {6}{3-z} = \frac {z^2-z}{3-z}\)
Domain: \((-\infty , 3) \cup (3, \infty )\)
\(z\)-intercepts: \((0,0), (1,0)\)
\(y\)-intercept: \((0,0)\)
Vertical asymptote: \(z = 3\)
\(\lim _{z \rightarrow 3^{-}} r(z) = \infty \), \(\lim _{z \rightarrow 3^{+}} r(z) = -\infty \)
Slant asymptote: \(y = -z-2\)
\(\lim _{z \rightarrow -\infty } r(z) = \infty \)
As \(z \rightarrow -\infty \), the graph is above \(y=-z-2\)
\(\lim _{z \rightarrow \infty } r(z) = -\infty \)
As \(z \rightarrow \infty \), the graph is below \(y=-z-2\)

\(f(x) = \dfrac {x^3+2x^2+x}{x^2-x-2}\)

\(f(x) = \frac {x^3+2x^2+x}{x^{2} -x-2} = \frac {x(x+1)}{x - 2}, \, x \neq -1\)
Domain: \((-\infty , -1) \cup (-1, 2) \cup (2, \infty )\)
\(x\)-intercept: \((0,0)\)
\(y\)-intercept: \((0,0)\)
Vertical asymptote: \(x = 2\)
\(\lim _{x \rightarrow 2^{-}} f(x) = -\infty \), \(\lim _{x \rightarrow 2^{+}} f(x) = \infty \)
Hole at \((-1,0)\)
Slant asymptote: \(y = x+3\)
\(\lim _{x \rightarrow -\infty } f(x) = -\infty \)
As \(x \rightarrow -\infty \), the graph is below \(y=x+3\)
\(\lim _{x \rightarrow \infty } f(x) = \infty \)
As \(x \rightarrow \infty \), the graph is above \(y=x+3\)

\(f(x) = \dfrac {5x}{9-x^2} - x\)

\(f(x) = \frac {5x}{9-x^2} - x = \frac {x^{3} - 4x}{9-x^{2}}\)
\(\phantom {f(x)} = \frac {x(x-2)(x+2)}{-(x-3)(x+3)}\)
Domain: \((-\infty , -3) \cup (-3, 3) \cup (3, \infty )\)
\(x\)-intercepts: \((-2, 0), (0, 0), (2, 0)\)
\(y\)-intercept: \((0, 0)\)
Vertical asymptotes: \(x = -3, x = 3\)
\(\lim _{x \rightarrow -3^{-}} f(x) = \infty \), \(\lim _{x \rightarrow -3^{+}} f(x) = -\infty \)
\(\lim _{x \rightarrow 3^{-}} f(x) = \infty \), \(\lim _{x \rightarrow 3^{+}} f(x) = -\infty \)
Slant asymptote: \(y = -x\)
\(\lim _{x \rightarrow -\infty } f(x) = \infty \)
As \(x \rightarrow -\infty \), the graph is above \(y=-x\)
\(\lim _{x \rightarrow \infty } f(x) = -\infty \)
As \(x \rightarrow \infty \), the graph is below \(y=-x\)

\(g(t) =\dfrac {1}{2}t-1 + \dfrac {t+1}{t^2+1}\)

\(g(t) = \frac {1}{2}t-1 + \frac {t+1}{t^2+1} = \frac {t(t^2-2t+3)}{2t^2+2}\)
Domain: \((-\infty ,\infty )\)
\(t\)-intercept: \((0,0)\)
\(y\)-intercept: \((0,0)\)
Slant asymptote: \(y = \frac {1}{2}t-1\)
\(\lim _{t \rightarrow -\infty } g(t) = -\infty \)
As \(t \rightarrow -\infty \), the graph is below \(y = \frac {1}{2}t-1\)
\(\lim _{t \rightarrow \infty } g(t) = \infty \)
As \(t \rightarrow \infty \), the graph is above \(y = \frac {1}{2}t-1\)

\(g(t) = \dfrac {t^{2} - 2t + 1}{t^{3} + t^{2} - 2t}\)

\(g(t) = \frac {t^{2} - 2t + 1}{t^{3} + t^{2} - 2t}=\frac {t - 1}{t(t + 2)}, \; t \neq 1\)
Domain: \((-\infty , -2) \cup (-2, 0) \cup (0, 1) \cup (1, \infty )\)
No \(t\)-intercepts
No \(y\)-intercepts
Vertical asymptotes: \(t = -2\) and \(t = 0\)
\(\lim _{t \rightarrow -2^{-}} g(t) = -\infty \), \(\lim _{t \rightarrow -2^{+}} g(t) = \infty \)
\(\lim _{t \rightarrow 0^{-}} g(t) = \infty \), \(\lim _{t \rightarrow 0^{+}} g(t) = -\infty \)
Hole in the graph at \((1, 0)\)
Horizontal asymptote: \(y = 0\)
\(\lim _{t \rightarrow -\infty } g(t) = 0\)
More specifically, as \(t \rightarrow -\infty , \; g(t) \rightarrow 0^{-}\)
\(\lim _{t \rightarrow \infty } g(t) = 0\)
More specifically, as \(t \rightarrow \infty , \; g(t) \rightarrow 0^{+}\)

Find a possible formula for the function whose graph is given. \(y = f(x)\)

[Picture]

\(f(x) = \dfrac {1}{x - 2}\)
Find a possible formula for the function whose graph is given. \(y = F(x)\)

[Picture]

\(F(x) = \dfrac {x-3}{(x-2)(x-3)} = \dfrac {x-3}{x^2-5x+6}\)
Find a possible formula for the function whose graph is given. \(y = g(t)\)

[Picture]

\(g(t) =\dfrac {t^2-1}{t}\)
Find a possible formula for the function whose graph is given. \(y = G(t)\)

[Picture]

\(G(t) = \dfrac {(t^2-1)(t+1)}{t(t+1)} = \dfrac {t^3+t^2-t-1}{t^2+t}\)
Let \(g(x) = \displaystyle \frac {x^{4} - 8x^{3} + 24x^{2} - 72x + 135}{x^{3} - 9x^{2} + 15x - 7}.\;\) With the help of your classmates:
  • find the \(x\)- and \(y\)- intercepts of the graph of \(g\).
  • find all of the asymptotes of the graph of \(g\) and any holes in the graph, if they exist.
  • find the intervals on which the function is increasing, the intervals on which it is decreasing and the local maximums and minimums, if any exist.
  • sketch the graph of \(g\), using more than one picture if necessary to show all of the important features of the graph.
Example carefulanalysisneeded showed us that the six-step procedure cannot tell us everything of importance about the graph of a rational function and that sometimes there are things that are easy to miss. Without Calculus, we may need to use graphing utilities to reveal the hidden behavior of rational functions. Working with your classmates, use a graphing utility to examine the graphs the rational function given. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it?

\(f(x) = \dfrac {1}{x^{2} + 1}\)

Example carefulanalysisneeded showed us that the six-step procedure cannot tell us everything of importance about the graph of a rational function and that sometimes there are things that are easy to miss. Without Calculus, we may need to use graphing utilities to reveal the hidden behavior of rational functions. Working with your classmates, use a graphing utility to examine the graphs the rational function given. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it?

\(f(x) = \dfrac {x}{x^{2} + 1}\)

Example carefulanalysisneeded showed us that the six-step procedure cannot tell us everything of importance about the graph of a rational function and that sometimes there are things that are easy to miss. Without Calculus, we may need to use graphing utilities to reveal the hidden behavior of rational functions. Working with your classmates, use a graphing utility to examine the graphs the rational function given. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it?

\(f(x) = \dfrac {x^{2}}{x^{2} + 1}\)

Example carefulanalysisneeded showed us that the six-step procedure cannot tell us everything of importance about the graph of a rational function and that sometimes there are things that are easy to miss. Without Calculus, we may need to use graphing utilities to reveal the hidden behavior of rational functions. Working with your classmates, use a graphing utility to examine the graphs the rational function given. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it?

\(f(x) = \dfrac {x^{3}}{x^{2} + 1}\)