Find the minimum/maximum of y=2x^2+12x-22

The minimum value of this function is -40.

Step-by-step explanation:

Recall that the minimum of a quadratic whose graph is a parabola that opens up, as this one does, is the vertex of the graph.  The x-coordinate of the vertex is given by x = -b / (2a), where a is the coefficient of the x^2 term and b is that of the x term.

Here, x = -12 / (2*2), or x = -12/4, or x = -3.

Find the y-value at this x-value:  f(-3) = 2(-3)^2 + 12(-3) - 22, or

                                                      f(-3) = 2(9) - 36 - 22, or -40.

The vertex is at (-3, -40).  The minimum value of this function is -40.

answer: 2/4

step-by-step explanation: