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Find the equation of the line tangent to the graph of f at (3, 140), where f is given by f(x) = 8x^3 − 9x^2 + 5.

Mathematics
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just differentiate the cubic equation to get f'(x)
now f'(3) will give you slope of tangent
24x^2-18x

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Other answers:

now you have slope of tangent and a point in the tangent , use slope point form to write equation of tangent
f'(3)?
yep substite x=3 in the differential
so slope is 162
so now the mx+b?
slope point form y-y1=m(x-x1)
gt it?
yeah
:)
y-140=162(x-3)?
right
yep
thanks
welcomes

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