How do you find the equation of the line tangent to the graph of #y=3x^2-6x+2#, at the point (3,11)?

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Answer 1 · 2018-04-11T13:31:06

#y-12x+25=0#

#y=3x^2-6x+2#

Apply differentiation (power rule )

#y'=6x-6#

substitute with #x# value to get the slope #m#

#y'=12#

substitute in the equation of the tangent with the point and slope

#y-y_1=m(x-x_1)#

where
#(x_1,y_1)=(3,11)#

#y-11=12(x-3)#

Simplify

#y-12x+25=0#

I hope I helped.