Questions: 1. y=1/2x^2-3 2. y=-2x^2+3 3. y=-2x^2+4x

Transcript text: 1. $y=\frac{1}{2}x^{2}-3$ 2. $y=-2x^{2}+3$ 3. $y=-2x^{2}+4x$

Solution

Solution Steps

Step 1: Identify the function and its properties

The given function is $ y = \frac{1}{2}x^2 - 3 $.

Step 2: Determine the direction of the parabola

Since the coefficient of $ x^2 $ is positive ($ \frac{1}{2} $), the parabola opens upwards.

Step 3: Find the vertex

The vertex form of a parabola $ y = ax^2 + bx + c $ has its vertex at $ x = -\frac{b}{2a} $. Here, $ a = \frac{1}{2} $ and $ b = 0 $, so the vertex is at $ x = 0 $.

Final Answer

The vertex of the parabola $ y = \frac{1}{2}x^2 - 3 $ is at $ (0, -3) $.

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