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Finding the Minimum Value of y=x2+2x8y = x^2 + 2x - 8

To find the minimum value of the quadratic equation y=x2+2x8y = x^2 + 2x - 8, determine the coordinates of its vertex.

Since the coefficient a=1a = 1 is positive, the parabola opens upward, meaning the equation has a minimum value. Find the axis of symmetry: Use the formula x=b2ax = -\frac{b}{2a}. With b=2b = 2 and a=1a = 1, x=22(1)=1x = -\frac{2}{2(1)} = -1. The axis of symmetry is the line x=1x = -1. Find the vertex: Substitute x=1x = -1 into the equation to find yy: y=(1)2+2(1)8=128=9y = (-1)^2 + 2(-1) - 8 = 1 - 2 - 8 = -9. The vertex is (1,9)(-1, -9).

Because the parabola has a minimum, the yy-coordinate of the vertex is the minimum yy-value of the quadratic equation. The minimum value is 9-9, and it occurs when x=1x = -1.

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Updated 2026-04-21

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