The function you provided is \( H(x) = -\frac{1}{2}x(x - 20) \). This can be simplified and analyzed in a number of ways. Let me break it down:
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Expand the Function: \[ H(x) = -\frac{1}{2}(x^2 - 20x) = -\frac{1}{2}x^2 + 10x \] So, the expanded form is: \[ H(x) = -\frac{1}{2}x^2 + 10x \]
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Identify the Type of Function: This is a quadratic function in the standard form \( ax^2 + bx + c \) where:
- \( a = -\frac{1}{2} \) (which indicates that the parabola opens downward),
- \( b = 10 \),
- \( c = 0 \).
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Vertex of the Parabola: The vertex form of a quadratic function can provide the maximum value since this parabola opens downward. The x-coordinate of the vertex can be found using: \[ x = -\frac{b}{2a} = -\frac{10}{2 \times -\frac{1}{2}} = \frac{10}{1} = 10 \] To find the y-coordinate of the vertex, substitute \( x = 10 \) back into the function: \[ H(10) = -\frac{1}{2}(10)^2 + 10(10) = -\frac{1}{2}(100) + 100 = -50 + 100 = 50 \] Thus, the vertex is at \( (10, 50) \).
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Intercepts:
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X-intercepts: Set \( H(x) = 0 \): \[ -\frac{1}{2}x(x - 20) = 0 \implies x = 0 \quad \text{or} \quad x - 20 = 0 \implies x = 20 \] Therefore, the x-intercepts are at \( (0, 0) \) and \( (20, 0) \).
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Y-intercept: Set \( x = 0 \): \[ H(0) = -\frac{1}{2}(0)(0 - 20) = 0 \] So, the y-intercept is at \( (0, 0) \).
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Summary of Key Points:
- The vertex is at \( (10, 50) \).
- The x-intercepts are \( (0, 0) \) and \( (20, 0) \).
- The y-intercept is \( (0, 0) \).
- Since \( a < 0 \), the parabola opens downward, indicating that the vertex is the maximum point of the function.
If you have any additional requests regarding this function, such as sketching the graph or finding specific values, feel free to ask!