y=(x-1)^2-2

The equation you provided is in vertex form of a quadratic function, which is given by the formula:

\[ y = a(x - h)^2 + k \]

In this case, the equation you provided is:

\[ y = (x - 1)^2 - 2 \]

Here, the parameters are:

• \( a = 1 \) (this indicates the parabola opens upwards),
• \( h = 1 \) (the x-coordinate of the vertex),
• \( k = -2 \) (the y-coordinate of the vertex).

Thus, the vertex of the parabola is at the point \( (1, -2) \).

Key features of the parabola:

1. Vertex: \( (1, -2) \)

2. Direction: Opens upwards (since \( a > 0 \)).

3. Axis of symmetry: The line \( x = 1 \).

4. Y-intercept: To find the y-intercept, set \( x = 0 \): \[ y = (0 - 1)^2 - 2 = 1 - 2 = -1 \] So, the y-intercept is \( (0, -1) \).

5. X-intercepts: Set \( y = 0 \) and solve: \[ 0 = (x - 1)^2 - 2 \] \[ (x - 1)^2 = 2 \] \[ x - 1 = \pm \sqrt{2} \] \[ x = 1 \pm \sqrt{2} \] Thus, the x-intercepts are \( (1 + \sqrt{2}, 0) \) and \( (1 - \sqrt{2}, 0) \).

Summary

• Vertex: \( (1, -2) \)
• Opens upwards
• Axis of symmetry: \( x = 1 \)
• Y-intercept: \( (0, -1) \)
• X-intercepts: \( (1 + \sqrt{2}, 0) \) and \( (1 - \sqrt{2}, 0) \)

If you need further assistance with this function or related topics, feel free to ask!