What is the vertex of the quadratic function f(x) = x^2 – 6x?

To find the vertex of the quadratic function f(x) = x^2 – 6x, we can use the vertex formula. The standard form of a quadratic function is given by f(x) = ax^2 + bx + c, where a, b, and c are constants. In this case, we have:

  • a = 1
  • b = -6
  • c = 0

The vertex of a parabola represented by a quadratic function can be found using the formula:

x = -b / (2a)

Substituting the values of a and b, we get:

x = -(-6) / (2 * 1) = 6 / 2 = 3

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting x back into the original function:

f(3) = (3)^2 – 6(3) = 9 – 18 = -9

Thus, the vertex of the quadratic function f(x) = x^2 – 6x is (3, -9). This means the graph of the function reaches its minimum point at this coordinate.

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