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

User Broofa
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2 Answers

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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


User Arunachalam K
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7.8k points
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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
User Thomas Easo
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