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the line through the origin with the slopr of -1 is tagnet to the curve at p find the x-and y- coordinates of point p . the equation is 2y^3+6x^2y-12x^2+6y=1
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equation of the straight line is y=-x, but the eqn of st.line into the curve eqn. ; -2x^3+6x^2(-x)-12x^2+6(-x)=1, by remainder thm, f(-1/2) =0. therefore (2x+1) is a factor. therefore -(2x+1)^3 = 0; x= -1/2, y=1/2
thank you, can you please expand on it more.
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