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.