How do you find the equation of a line tangent to the function #y=x^3-x^2-4x# at (3,6)?

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Answer 1 · 2016-07-22T04:02:55

#y=17x-45#

To find the slope of the tangent, we must use differentiation. The derivative of the function is

#f'(x) = 3x^2-2x-4#

so that, at the point #(3,6)#, the slope of the tangent is

#f'(3) = 3 xx 3^2-2 xx 3-4=17#.

So, we have to find the equation of a straight line passing through #(3,6)# with a slope of 17:

#{y-6}/{x-3} = 17#

or

#y-6 = 17(x-3) = 17x-51#

or

#y = 17x-45#