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 $(0, -3)$ .

{"axisType": 3, "coordSystem": {"xmin": -10, "xmax": 10, "ymin": -10, "ymax": 10}, "commands": ["y = (1/2)x**2 - 3"], "latex_expressions": [" $y = \\frac{1}{2}x^2 - 3$ "]}

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