Ldaniel
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.



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AravindG
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just differentiate the cubic equation to get f'(x)

AravindG
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now f'(3) will give you slope of tangent

Ldaniel
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24x^218x

AravindG
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now you have slope of tangent and a point in the tangent , use slope point form to write equation of tangent

Ldaniel
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f'(3)?

AravindG
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yep substite x=3 in the differential

Ldaniel
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so slope is 162

Ldaniel
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so now the mx+b?

AravindG
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slope point form
yy1=m(xx1)

AravindG
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gt it?

Ldaniel
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yeah

AravindG
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:)

Ldaniel
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y140=162(x3)?

Ldaniel
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right

AravindG
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yep

Ldaniel
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thanks

AravindG
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welcomes