To analyze the function \( A(w) = w^2 + 5w \), we can start by rewriting it in a more familiar form. This is a quadratic function that opens upwards (since the coefficient of \( w^2 \) is positive).
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Finding the Vertex: The vertex of a quadratic function in the form \( A(w) = aw^2 + bw + c \) occurs at \( w = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 5 \).
\[ w = -\frac{5}{2(1)} = -\frac{5}{2} \]
Since this vertex is negative, the function will increase as \( w \) increases from 0 to infinity.
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Behavior of the Function:
- As \( w \) increases, \( A(w) \) will also keep increasing for \( w > 0 \).
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Options Evaluation:
- A(w) has a maximum value at w=5: This is incorrect; it has a minimum value at the vertex (which we don't reach because the vertex is negative).
- A times w has a maximum value at w=5: This seems based on a misunderstanding; not applicable.
- A(w) increases as w increases for 0<w<5: While it is true that \( A(w) \) is increasing for this range, it doesn’t stop increasing as \( w \) goes beyond 5.
- A times w increases as w increases for 0<w<5: Same as previous point but wrongly rephrased.
- A(w) has a minimum value at w=5: This is incorrect for the same reason as the first point.
- A(w) increases as w increases when w>0: This statement is true.
The most accurate choice is:
A(w) increases as w increases when w>0.