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Finding the Maximum or Minimum Value of y=4x2+16x11y = -4x^2 + 16x - 11

To find the maximum or minimum value of the quadratic equation y=4x2+16x11y = -4x^2 + 16x - 11, determine the vertex of its parabola.

Since the coefficient a=4a = -4 is negative, the parabola opens downward, meaning it has a maximum value. The axis of symmetry is x=b2a=162(4)=168=2x = -\frac{b}{2a} = -\frac{16}{2(-4)} = -\frac{16}{-8} = 2. Substitute x=2x = 2 into the equation to find the yy-coordinate: y=4(2)2+16(2)11=16+3211=5y = -4(2)^2 + 16(2) - 11 = -16 + 32 - 11 = 5. The vertex is (2,5)(2, 5).

Therefore, the maximum value of the quadratic equation is 55, which occurs at x=2x = 2.

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Updated 2026-04-21

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