How to write an equation of the line tangent to the curve #(3x^2)/(x^2+x+1# at point p(-1,3) in the form ax+by=c?

Question

How to write an equation of the line tangent to the curve #(3x^2)/(x^2+x+1# at point p(-1,3) in the form ax+by=c? I tried to differentiate it using quotients rule but im not sure if what i got is right. please help,

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Answer 1 · 2018-03-28T09:19:13

Equation of tangent is #3x+y=0#

The slope of tangent is given by first derivative. As #y=(3x^2)/(x^2+x+1)#, using quotient formula

#y'=(dy)/(dx)=(6x(x^2+x+1)-3x^2(2x+1))/(x^2+x+1)^2#

= #(3x^2+6x)/(x^2+x+1)^2#

and at #(-1,3)# we have #((dy)/(dx))_(x=-1)=(3-6)/(1-1+1)^2=-3#

As it passes through #(-1,3)#, equation of tangent is

#y-3=-3(x+1)# i.e. #3x+y=0#

graph{(3x+y)(y-(3x^2)/(x^2+x+1))=0 [-7.08, 2.92, -0.54, 4.46]}