Chapter 2: Problem 12
\(y=\frac{1}{2}(x-3)^2+2\)
Short Answer
Expert verified
The vertex of the quadratic equation \(y=\frac{1}{2}(x-3)^2+2\) is at point (3, 2). The axis of symmetry is \(x = 3\). And the parabola opens upwards and is wider than \( y = x^2\).
Step by step solution
Key Concepts
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