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OpenStudy (anonymous):
Determine all values of x, (if any) at which the graph of the function has a horizontal tangent. Y(x)=x^3+12x^2+8
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OpenStudy (campbell_st):
find the 1st derivative... set to zero and then solve for x. this will give the point(s) where the tangent is horizontal. to find the equation of the tangents, substitute the x values into the original equation to get the ordered pairs... the equations will be the y values from the ordered pairs
OpenStudy (anonymous):
so you get 0 and -8?
OpenStudy (campbell_st):
that's correct... they are the points...
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