What is the vertex of the function g(x) = 8x² – 64x + 128?

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The vertex of a quadratic function in the form of g(x) = ax² + bx + c can be found using the vertex formula. For the given function g(x) = 8x² - 64x + 128, we identify the coefficients: a = 8, b = -64, and c = 128.

The x-coordinate of the vertex is calculated using the formula:

x = -b / (2a)

Plugging in the values of b and a:

x = -(-64) / (2 * 8) = 64 / 16 = 4

Next, we substitute this value of x back into the function to find the y-coordinate of the vertex:

g(4) = 8(4)² - 64(4) + 128

Calculating this:

g(4) = 8(16) - 256 + 128 = 128 - 256 + 128 = 0

Thus, the coordinates of the vertex are (4, 0).

In conclusion, the vertex of the quadratic function g(x) = 8x² - 64x + 128 is (4, 0). This point represents the minimum value of the function since the leading coefficient a is positive, indicating that the parabola opens upwards.

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