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Finding the Maximum or Minimum Value of y=x28x+12y = x^2 - 8x + 12

To find the maximum or minimum value of the quadratic equation y=x28x+12y = x^2 - 8x + 12, find the vertex of its parabola.

Since the coefficient a=1a = 1 is positive, the parabola opens upward, so it has a minimum value. The axis of symmetry is x=b2a=82(1)=4x = -\frac{b}{2a} = -\frac{-8}{2(1)} = 4. Substitute x=4x = 4 into the equation to find the yy-coordinate: y=(4)28(4)+12=1632+12=4y = (4)^2 - 8(4) + 12 = 16 - 32 + 12 = -4. The vertex is (4,4)(4, -4).

Therefore, the minimum value of the quadratic equation is 4-4, which occurs at x=4x = 4.

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Updated 2026-04-21

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