How do you find the tangent line of #f(x) = 3sinx-2x^2 # at x=-1?

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Answer 1 · 2017-04-15T19:26:14

#y=5.621x+1.096#

Find the slope at by taking the derivative of #f(x)#.

#f'(x)=3cos(x)-4x#

Now plug in #x=-1#

#f'(-1)=3cos(-1)-4(-1)~~ 5.621#

This is the slope but we need a point to find the tangent line. Plug in #-1# for #f(x)#

#f(-1)=3sin(-1)-2(-1)^2~~-4.524#

So now we have the point #(-1,-4.524)# and the slope #5.621#. Plug into point-slope form:

#y-(-4.524)=5.621(x-(-1))#

#y=5.621x+5.621-4.524#

#y=5.621x+1.096#