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Find the maximum y-value on the grapr f(x)=-x^(2)+10x+1

User OKonyk
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Final answer:

The maximum y-value on the graph of f(x) = -x² + 10x + 1 is 26.

Step-by-step explanation:

To find the maximum y-value, let's determine the vertex of the parabola represented by the function f(x). The vertex of a parabola in the form f(x) = ax² + bx + c is given by the x-coordinate -b/2a. For the given function f(x) = -x² + 10x + 1, the x-coordinate of the vertex can be calculated as -b/2a = -10/(2 * (-1)) = 5. Substitute x = 5 into the function to find the corresponding y-coordinate: f(5) = -(5)² + 10(5) + 1 = -25 + 50 + 1 = 26. Hence, the maximum y-value occurs at x = 5, with a corresponding y-value of 26.

This parabola opens downward since the coefficient of x² is negative (-1 in this case), meaning the vertex represents the maximum point on the graph. Therefore, the highest y-value for this function is 26, located at x = 5. This value represents the peak of the parabola as it opens downward, indicating that as x moves away from the vertex, the y-values decrease indefinitely.

Find the maximum y-value on the grapr f(x)=-x^(2)+10x+1-example-1
User Bujar
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