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Completion requirements
Sketch the graphs of \(y = - 2x + 1\) and \(y = \frac{1}{-2x + 1}\) on
the same grid. Label the invariant points of the two functions. Show any
horizontal and vertical asymptotes of \(y = \frac{1}{-2x + 1}\).
An understanding of where the graphs of the functions \(y = f(x)\) and \(y = \frac{1}{f(x)}\) intersect, and knowing how asymptotes are related to \(y = f(x)\) and \(y = \frac{1}{f(x)}\) makes it possible to sketch one graph using the other. Note that the sketching procedures that follow are used to give the approximate shape of the graph. For a more accurate graph, use a table of values or technology.
The graph of \(y = - 2x + 1\) has a slope of \(–2\) and a \(y\)-intercept of \(1\). A table of values can be used to graph \(y = \frac{1}{-2x + 1}\). There are rapid changes in \(y\) when the denominator is near zero, so it is a good idea to use smaller increments near \(x = \frac{1}{2}\) .
The two functions intersect at \((0, 1)\) and \((1, –1)\). There is a vertical asymptote where \(-2x + 1 = 0\), so the line \(x = \frac{1}{2}\) is a vertical asymptote. The line \(y = 0\) is a horizontal asymptote.
| \(x\) | \(y = \frac{1}{-2x + 1}\) |
| \(-4\) | \(\frac{1}{9}\) |
| \(-3\) | \(\frac{1}{7}\) |
| \(-2\) | \(\frac{1}{5}\) |
| \(-1\) | \(\frac{1}{3}\) |
| \(0\) | \(1\) |
| \(\frac{1}{4}\) | \(2\) |
| \(\frac{1}{2}\) | undefined |
| \(\frac{3}{4}\) | \(-2\) |
| \(1\) | \(-1\) |
| \(2\) | \(-\frac{1}{3}\) |
| \(3\) | \(-\frac{1}{5}\) |
| \(4\) | \(-\frac{1}{7}\) |
| \(5\) | \(-\frac{1}{9}\) |
The two functions intersect at \((0, 1)\) and \((1, –1)\). There is a vertical asymptote where \(-2x + 1 = 0\), so the line \(x = \frac{1}{2}\) is a vertical asymptote. The line \(y = 0\) is a horizontal asymptote.