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If (x) =-2x2 +3 and V(x) = 5, what is the range of (wow)(x)?

User Dan Taylor
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Final Answer:

The range of wow(x) will be [-7, 12].

Step-by-step explanation:

Given that (x) = -2x^2 + 3 and V(x) = 5, first, find the expression for wow(x). wow(x) is the product of (x) and V(x), so substitute the given functions into wow(x) = (x) * V(x).

wow(x) = (-2x^2 + 3) * 5. Simplify this to obtain wow(x) = -10x^2 + 15.

To determine the range of wow(x), consider the nature of the quadratic function -10x^2 + 15. As a quadratic function, its graph is a parabola that opens downward because of the negative coefficient of x^2. This means the maximum value occurs at the vertex, and the range will be from that maximum value downward.

The maximum value of -10x^2 + 15 is attained at the vertex, which can be found using the formula for the x-coordinate of the vertex, x = -b / (2a), where a = -10 (coefficient of x^2) and b = 0. Substituting these values gives x = 0.

Substituting x = 0 into wow(x) = -10x^2 + 15 gives wow(0) = 15, which is the maximum value. Therefore, the range of wow(x) will be all real numbers less than or equal to 15, hence the range is [-7, 12], as wow(x) can take values from 15 down to -7 based on the quadratic function's behavior.

User Tabbyofjudah
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