Find the equation of the tangent to the curve y = (2x -3)^3 at the point (1, - 1), giving your answer in the form y = mx + c.
y = (2x -3)^3
y = (2x)^3 + 3.((2x)^2)(-3) + 3.(2x).(-3)^2 + (-3)^2 using Pascal's Triangle.
y = 8x^3 - 36x^2 + 54x - 27
dy/dx = 24x^2 - 72x + 54
at point (1,-1); dy/dx = 24 -72 + 54 = 6
Therefore tangent line is of the form y=6x + c
at point (1,-1); -1=6.1 + c
Therefore c = -7 and tangent line is y = 6x - 7.
RS
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