2) What is the greatest integer value of b such that -4 is not in the range of y = x^2 + bx + 12?
This parabola turns upward.....therefore...we want to have the vertex lie above ( a, -4) where a is the x coordinate of the vertex....so
a = -b / [ 2 (1) ]
2a =-b
-2a =b
So.....we want to solve this to find the y coordinate of the vertex
(a)^2 + (-2a)(a) + 12 = -4
a^2 - 2a^2 = -16
-a^2 = -16
a^2 =16
a = ± 4
b achieves its greatest value when a = -4 → -2(-4) = b = 8
Note that the graph here : https://www.desmos.com/calculator/6vgqqov8mg
shows that when b = 8, the vertex is at (-4, -4) and when b = 9 that part of the graph lies below y = -4
But when b = 7, the graph lies wholly above y = -4
So....b =7 is the greatst integer value that guarantees that -4 is not in the range of x^2 + bx + 12