Example

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

To find the maximum or minimum value of the quadratic equation y=−4x2+16x−11y = -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)=−16−8=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+32−11=5y = -4(2)^2 + 16(2) - 11 = -16 + 32 - 11 = 5. The vertex is (2, 5). Therefore, the maximum value of the quadratic equation is 5, which occurs at x=2x = 2.

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Updated 2026-07-02

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