To graph the piecewise function
\[ f(x) = \begin{cases} -3x^2 & \text{for } x \leq 0 \ \frac{1}{2}x + 3 & \text{for } x > 0 \end{cases} \]
we need to analyze and graph both pieces of the function separately.
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For \(x \leq 0\): The function is defined as \(f(x) = -3x^2\).
- This is a downward-opening parabola.
- The vertex of the parabola is at the origin (0, 0).
- When \(x = 0\), \(f(0) = 0\).
- As \(x\) approaches negative infinity, \(f(x)\) approaches negative infinity.
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For \(x > 0\): The function is defined as \(f(x) = \frac{1}{2}x + 3\).
- This is a linear function with a slope of \(\frac{1}{2}\) and a y-intercept of 3.
- When \(x = 0\), the limit approaches \(f(0) = 3\) (but since it's only for \(x > 0\), the point (0, 3) is not included in the graph).
- When \(x = 2\), \(f(2) = \frac{1}{2}(2) + 3 = 4\).
- When \(x = 4\), \(f(4) = \frac{1}{2}(4) + 3 = 5\).
- The line will continue to rise as \(x\) increases.
Steps to Create the Graph:
- Plot the downward-opening parabola \(f(x) = -3x^2\) for \(x \leq 0\) which includes points like (0,0), (-1, -3), (-2, -12), etc.
- For \(x > 0\), plot the line \(f(x) = \frac{1}{2}x + 3\) which starts just above (0, 3) (open circle at (0, 3)), passing through (2, 4), and (4, 5), etc.
- The graph should show a clear transition at the point \(x = 0\) from the parabola to the line.
Final Selection
Without access to the graphs provided in the question, I can't select an exact option. However, based on the described behavior of the function, look for:
- A parabola that opens downwards on the left side (where \(x \leq 0\)).
- A linear line with a positive slope beginning right above (0, 3) on the right side (where \(x > 0\)) and continuing upwards.
Check the available options for the one that best matches these characteristics.
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