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The parabola \( y=-x^{2}+12 x-11 \) has an axis of symmetry of \( x=6 \). Which of the following represents its range? \( \begin{array}{ll}\text { (1) } y \geq-11 & \text { (3) } y \leq 6 \\ \text { (2) } y \leq 25 & \text { (4) } y \geq 10\end{array} \)

Ask by Erickson Stephens. in the United States
Mar 13,2025

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The range of the parabola is \( y \leq 25 \).

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Bonus Knowledge

To find the range of the parabola given by the equation \( y = -x^2 + 12x - 11 \), we can complete the square or find the vertex directly. Since the parabola opens downwards (the coefficient of \( x^2 \) is negative), its maximum value occurs at the vertex. Given that the axis of symmetry is \( x = 6 \), we plug \( x = 6 \) back into the equation to find the maximum \( y \): \[ y = -6^2 + 12(6) - 11 = -36 + 72 - 11 = 25. \] Thus, the maximum value of \( y \) is 25, and the parabola opens downwards, indicating that the range of \( y \) is \( y \leq 25 \). Therefore, the answer is (2) \( y \leq 25 \).

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