This involves derivatives.
For example, if we had f(x) = 3x^2 + 2x + 3...
The 1st derivative would be f ' (x) = 6x + 2.
Note, when 6x + 2 = 0 --> 6x = -2 --> x = -2/6 = -1/3, the slope is 0 at that point on the graph of f(x).
If we had x = 1, our f ' (x) > 0, meaning we have a positive slope at point x = 1 on f(x).
If we had x = -1, our f ' (x) < 0, meaning we have a negative slope at point x= -1 on f(x).
In any case, anytime x < 0, f ' (x) will always be negative, meaning the slope will always be negative on the original f(x) graph.
When $$x \ge 0$$, f ' (x) will always be positive, meaning the slope will always be positive on the original f(x) graph.
Also, if we take the 2nd derivative, we would get f '' (x) = 6.
This means that no matter if the x value is positive, negative, or 0, the original f(x) graph will always be concave up.
See graph: https://www.desmos.com/calculator/8fypoy1rbh