What is the equation of the line tangent to #f(x)=-3x^2-3x +6 # at #x=1#?

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Answer 1 · 2016-09-08T07:29:06

The equation of tangent at #(1,0)# is #9x+y=9#

At #x=1#, #f(1)=-3xx1^2-3xx1+6=-3-3+6=0#

Hence we need a tangent at #(1,0)#.

The slope of the tangent is given by value of #(dy)/(dx)# at #x=1#

Now as #f(x)=-3x^2-3x+6#, #(dy)/(dx)=-6x-3# and as such slope of tangent is #-6xx1-3=-9#.

Now equation of line at #(x_1,y_1)# with a slope of #m# is given by #(y-y_1)=m(x-x_1)#

Hence, the equation of tangent at #(1,0)# with a slope of #-9# is

#(y-0)=-9(x-1)# or #y=-9x+9# or #9x+y=9#

graph{(y+3x^2+3x-6)(9x+y-9)=0 [-5, 5, -20, 20]}