Y=X^2 -4x -32. Use this function and identify the minimum.

Thank you very much.

The minimum of a parabola is the y-value of the vertex. See previous response.

To identify the minimum of the function y = x^2 - 4x - 32, we can use the concept of vertex and the properties of quadratic functions.

Step 1: The vertex of a quadratic function in the form y = ax^2 + bx + c is given by the coordinates (h, k), where h = -b/2a and k is the value of y at the vertex.

In this case, we have a = 1, b = -4, and c = -32. So, h = -(-4) / (2*1) = 4/2 = 2.

Step 2: Substitute the value of h back into the equation to find k.

y = (2)^2 - 4(2) - 32
= 4 - 8 - 32
= -36.

Therefore, the coordinates of the vertex are (2, -36), and the minimum value of the function y = x^2 - 4x - 32 is -36.

Note: Another way to find the minimum is by recognizing that the coefficient of x^2 term is positive (a = 1). This indicates that the parabola opens upward and the vertex represents the minimum point.