How do you find the slope of the line tangent to the curve #3x^2 - 2xy + y = 11# at the point (1,-2)?

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Answer 1 · 2015-10-26T19:36:37

We have that #y=y(x)# hence the tangent is

#y-y(1)=y'(1)(x-1)#

we know that #y(1)=-2# and we need to find #y'(1)#.

The derivative of the given equation is

#d(3x^2-2xy+y-11)/dx=6x-2y-2xdy/dx+dy/dy=0#

For #x=1# , #y=-2# we get

#6*1-2*(-2)-2*1dy/dx+dy/dx=0=>10-dy/dx=0=>dy/dx=10#

Hence the slope is #y'(1)=10# and the tangent is

#y-1=10*(x-1)=>y=10x-9#