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f(x) = X/2 -4/x^2 + 1

thats the line.

dy/dx is 1/2 + 8/x^3 that was first question.

2nd = the normal at point (2/1) on the curve above cuts the x and y axes at A and B respectively. Calculate the length of AB leaving answer in simplified surd form.

ok.

gradient of tangent at that point is 1/2 + 8/16 = 3/2

normal is -2/3.

ok, now from here i'm not sure where to go or what to do =( halp =)