The function f(x) = (x - 4)(x + 2) can be expanded to f(x) = x^2 - 4x + 2x - 8, which simplifies to f(x) = x^2 - 2x - 8.

This is a quadratic function, and its graph is a parabola. The minimum value of the function occurs at the vertex of the parabola.

The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.

The x-coordinate of the vertex is given by -b/2a. In this case, a = 1 and b = -2, so the x-coordinate of the vertex is -(-2)/2(1) = 1.

Substituting x = 1 into the function gives the y-coordinate of the vertex, which is the minimum value of the function: f(1) = (1)^2 - 2(1) - 8 = -9.

So the equivalent equation of the function in which the minimum value of f appears as a constant or coefficient is f(x) = (x - 1)^2 - 9, which corresponds to choice (D).

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The function f(x) = (x - 4)(x + 2) can be expanded to f(x) = x^2 - 4x + 2x - 8, which simplifies to f(x) = x^2 - 2x - 8.

This is a quadratic function, and its graph is a parabola. The minimum value of the function occurs at the vertex of the parabola.

The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.

The x-coordinate of the vertex is given by -b/2a. In this case, a = 1 and b = -2, so the x-coordinate of the vertex is -(-2)/2(1) = 1.

Substituting x = 1 into the function gives the y-coordinate of the vertex, which is the minimum value of the function: f(1) = (1)^2 - 2(1) - 8 = -9.

So the equivalent equation of the function in which the minimum value of f appears as a constant or coefficient is f(x) = (x - 1)^2 - 9, which corresponds to choice (D).

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The function f(x) = (x - 4)(x + 2) can be expanded to f(x) = x^2 - 4x + 2x - 8, which simplifies to f(x) = x^2 - 2x - 8.

This is a quadratic function, and its graph is a parabola. The minimum value of the function occurs at the vertex of the parabola.

The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.

The x-coordinate of the vertex is given by -b/2a. In this case, a = 1 and b = -2, so the x-coordinate of the vertex is -(-2)/2(1) = 1.

Substituting x = 1 into the function gives the y-coordinate of the vertex, which is the minimum value of the function: f(1) = (1)^2 - 2(1) - 8 = -9.

So the equivalent equation of the function in which the minimum value of f appears as a constant or coefficient is f(x) = (x - 1)^2 - 9, which corresponds to choice (D).

f(x) = (x - 4)(x + 2)Which of the following is an equivalent equation of the function f(x) above in which the minimum value of f appears as a constant or coefficient?(A) f(x) = -x2 + 4(B) f(x) = x2 - 2x - 8(C) f(x) = x2 + 4(D) f(x) = (x - 1)2 - 9

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