Question

What is the minimum possible value for y=x^2+12x+5? I am not sure where to start.

Answer 1

Assuming you are not allowed to use calculus, complete the square to find the vertex.

x^2 + 12 x = y-5

x^2 + 12 x + 6^2 = y - 5 + 6^2

(x+6)^2 = y + 31

vertex at x = -6, y = -31
so never smaller than -31

Answer 2

differentiate y with respect to x (dy/dx). That gives you the slope at any point on the graph. The slope is always 0 at the max and min points so put dy/dx = 0 and solve for x.

Once you get that value of x put it back into y to find the corresponding y value for the minimum point.