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For the functio h(x)=1/2x²-7x 7 state the maximum/minimum of h(x)

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Final answer:

The quadratic function h(x) = 1/2x^2 - 7x has a minimum but no maximum. Its minimum occurs at the vertex (7, -49/2), found using the vertex formula for quadratics.

Step-by-step explanation:

To determine the maximum or minimum of the function h(x) = \frac{1}{2}x^2 - 7x, we need to consider its shape and the values of its coefficients. This is a quadratic function, and since the coefficient of the x2 term is positive (\frac{1}{2}), the graph of this function opens upwards, implying that it will have a minimum point but no maximum point.

The vertex of a quadratic function in the form of f(x) = ax2 + bx + c is given by the formula (-b/2a, f(-b/2a)), and it corresponds to the maximum or minimum of the function. For the function h(x), a = \frac{1}{2} and b = -7. Thus, the x-coordinate of the vertex is x = -(-7)/(2 * \frac{1}{2}) = 7. To find the y-coordinate, we substitute x back into the function: h(7) = \frac{1}{2}(7)^2 - 7(7) = \frac{1}{2}(49) - 49 = \frac{49}{2} - 49 = \frac{-49}{2}.

The minimum of h(x) occurs at the vertex, (7, -\frac{49}{2}). There is no maximum value since the function continues indefinitely upwards on both sides as x moves away from the vertex.

User Ovidiu G
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