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Using differentiation rules, find a cubic function y=ax^3+bx^2+cx+d whose graph has horizontal tangents at (-2,6) and (2,0) Do not graph on a calculator.
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@Headdesk What's the slope of horizontal tangents or horizontal lines?
Try differentiating your original function and setting that to 0 to start with. Once you do that, then fill in the"x" and "y" and solve for a, b, and c.
y' = 3ax^2 + 2bx + c Now substitute the x and y for each point. Solve for a, b, and c
Okay, that makes more sense then how I was attempting to do it before. Thanks!
Good luck in all of your studies and thx for the recognition! And you're welcome! @Headdesk
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