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Tell how each graph differs from the graph of \( y=x^{2} \). \( \begin{array}{ll}\text { 1. } y=x^{2}+2 & \text { 4. } y=\frac{1}{5} x^{2} \\ \text { 2. } y=x^{2}-8 & \text { 5. } y=(x+3)^{2} \\ \text { 3. } y=4 x^{2} & \end{array} \)

Ask by Ford Santiago. in the United States
Mar 18,2025

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1. **\(y = x^2 + 2\)**: The graph is shifted upward by 2 units. 2. **\(y = x^2 - 8\)**: The graph is shifted downward by 8 units. 3. **\(y = 4x^2\)**: The graph is vertically stretched by a factor of 4. 4. **\(y = \frac{1}{5} x^2\)**: The graph is vertically compressed by a factor of \(\frac{1}{5}\). 5. **\(y = (x+3)^2\)**: The graph is shifted to the left by 3 units.

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Bonus Knowledge

The graphs differ from \( y=x^{2} \) primarily in their transformations. For instance, \( y=x^{2}+2 \) shifts the vertex of the parabola upward by 2 units, while \( y=x^{2}-8 \) moves it down, placing the vertex at (0, -8). Meanwhile, \( y=(x+3)^2 \) translates the graph horizontally to the left by 3 units, changing the vertex to (-3, 0). In terms of shape, \( y=4x^{2} \) vertically stretches the graph, making it narrower than the standard parabola, whereas \( y=\frac{1}{5}x^{2} \) does the opposite, resulting in a wider parabola. These transformations maintain the parabolic shape while altering position and steepness!

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