91Ó°ÊÓ

To expand the function, we need to apply the square formula. The given function is \(f(x) = (x-1)^2 - 2\). We start by expanding \((x-1)^2\):\[(x-1)^2 = x^2 - 2x + 1\]Substitute this back into the function:\[f(x) = x^2 - 2x + 1 - 2\]Simplify the expression to get:\[f(x) = x^2 - 2x - 1\]

To graph the function \(f(x) = x^2 - 2x - 1\), we need to identify its key features. This is a quadratic function of the form \(ax^2 + bx + c\), where \(a = 1\), \(b = -2\), and \(c = -1\).1. **Vertex**: The vertex formula for a quadratic function is given by \(x = -\frac{b}{2a}\). For this function:\[x = -\frac{-2}{2(1)} = 1\]Substitute \(x = 1\) back into the function to find the y-coordinate of the vertex:\[f(1) = (1)^2 - 2(1) - 1 = -2\]So, the vertex is \((1, -2)\).2. **Axis of Symmetry**: The line \(x = 1\).3. **Y-intercept**: Set \(x = 0\) and solve for \(f(x)\):\[f(0) = (0)^2 - 2(0) - 1 = -1\]So, the y-intercept is \((0, -1)\).4. **Plot Points**: We already have the vertex and y-intercept. To better define the graph, calculate a few more points around these, such as \(x = -1\) and \(x = 2\). - For \(x = -1\): \(f(-1) = (-1)^2 - 2(-1) - 1 = 2\). - For \(x = 2\): \(f(2) = (2)^2 - 2(2) - 1 = -1\).5. **Graph**: On a coordinate plane, plot the vertex \((1, -2)\), y-intercept \((0, -1)\), and additional points \((-1, 2)\) and \((2, -1)\). Draw a symmetric parabola opening upwards through these points.