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the graph of y = x^3 -9x^2 - 16x +1 has a slope of 5 at two points. Find the coordinates of the points
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y'=3x^2-18x-16=5
solve for x
I know that y'= 3x^2-18x-16 0=3x^2-18-21 I am stuck
you have forgotten how to solve the quadratic\[3x^2-18x-21=0\]this is definately a skill you need to have down for calculus first divide out the common factor of 3, then see if you can factor the equation if not, use the quadratic formula
(3x-21)(x+1)..Is this correct
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yeah, but you could factor out the 3\[3x^2-18x-21=0\implies x^2-6x-7=0\]\[\implies(x-7)(x+1)=0\]
x=7 x=-1 and i plugged it in the org equation..Thanks
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