How do you find the equation of the line tangent to f(x) = 6x^2 - 1 at x = 3?

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Answer 1 · 2018-05-28T21:03:31

#y=36x-55#

#f(x)=6x^2-1# , #color(white)(aa)# #x##in##RR#

#f'(x)=12x#

#f(3)=53#

#f'(3)=36#

The equation of the tangent line at #A(3,f(3))# will be

#y-f(3)=f'(3)(x-3)# #<=>#

#y-53=36(x-3)# #<=>#

#y=36x-55#

graph{(y-6x^2+1)(y-36x+55)=0 [-41.1, 41.1, -20.55, 20.55]}