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The vertical asymptote of the graph of \(y = \frac{1}{-(x - 1)^2}\) passes through the \(x\)-intercept of the graph of \(y = -(x - 1)^2\). The reciprocal function is of the form \(y = \frac{1}{f(x)}\), so \(y = 0\) is the horizontal asymptote. The points \((0, –1)\) and \((2, –1)\) have \(y\)-values of \(–1\), so the graphs of the two functions will intersect at these points.

Use the asymptotes and the points of intersection to sketch the graph of \(y = \frac{1}{-(x - 1)^2}\).

The lines \(x = - 3\) and \(x = 1\) are vertical asymptotes, so the \(x\)-intercepts of the graph of \(y = f(x)\) are \((–3, 0)\) and \((1, 0)\). The points where \(\frac{1}{f(x)} = \pm 1\) are where the graphs of the two functions intersect, so include these points as well.

In the lower section of the graph of \(y = \frac{1}{f(x)}\), the maximum value occurs at \((-1, -\frac{1}{2})\), so the minimum value of \(y = f(x)\) is found to be at \((-1, -2)\) by taking the reciprocal of the \(y\)-value.

Use the \(x\)-intercepts, the invariant points, and the minimum value to sketch the graph of \(y = f(x)\).