Question

Find the value of x that minimizes or maximizes y = x² + 10x - 45.

Answer 1

x = -5

Step-by-step Solution:

We will take the derivative of the equation and set f' = 0 to find its Max/Min

`y = {x}^(2) + 10x - 49 \\ (dy)/(dx) = 2x + 10 \\ 0 = 2x + 10 \\ 2x = - 10 \\ x = - 5`

Therefore, at x = -5 is where the equation is minimized the most

Answer 2

To find the value of x that minimizes or maximizes y = x² + 10x - 45, use the vertex formula: x = -b/2a. Plugging in the values of a = 1, b = 10, and c = -45, x is found to be -5.

Step-by-step explanation:

To find the value of x that minimizes or maximizes y = x² + 10x - 45, we can use the vertex formula. The vertex formula states that the x-coordinate of the vertex of a quadratic function in the form y = ax² + bx + c is given by x = -b/2a. In this case, a = 1, b = 10, and c = -45. Plugging these values into the formula, we get x = -10/2(1) = -5.