If x can be any integer, what is the least possible value of the expression (4x^2)-10?

Hmm, the squaerd will make it positive
if we had x=-10, we get 400-10 or 390
if we had x=0, we get -10
if we had x=10, we get 390
so the minimum value occurs at x=0 because it gets smaller as x appraooches  0 but gets bigger again as it becomes more negative

least value is -10, which is at x=0

Y=4x^2-10  to find the least possible value we need to find when the rate of change is equal to zero.

dy/dx=8x, well dy/dx=0 only when x=0 so the least vale that y can be is:

4*0^2-10

0-10

-10