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Mathematics 21 Online
OpenStudy (anonymous):

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.

OpenStudy (ash2326):

@Headdesk What's the slope of horizontal tangents or horizontal lines?

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

y' = 3ax^2 + 2bx + c Now substitute the x and y for each point. Solve for a, b, and c

OpenStudy (anonymous):

Okay, that makes more sense then how I was attempting to do it before. Thanks!

OpenStudy (anonymous):

Good luck in all of your studies and thx for the recognition! And you're welcome! @Headdesk

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