Question 202007
y= 1/(1-x)
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to  find  slope ,,,find  differential  of  above (y')
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form  is  d/dx (C/u),,,where  c=1, u=(1-x),, du/dx = -1,, u^2 = (x^2 -2x +1)
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d/dx(C/u) = (-1/u^2 )d/dx(u) = -1/(x^2-2x+1) * (-1) = 1/(x^2 -2x+1)=y'
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at (2,-1) slope  is y' = 1/(x^2 -2x+1) = 1/{(2)^2 -2(2) +1} = 1
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for  y=4x,,,,y' = 4  (always  as  y=4x  is  a  straight  line.)
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secant  to  straight  line  is  that  straight  line,,,y=4x.
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tan  at  (2,8) is  y=mx+b,  m=-1/4,,y=(-1/4)x +b,,,
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subst(2,8),,,8=(-1/4)2 +b,,,b=8.5,,,,eqn  is  ,,,y=(-1/4)x +8.5
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or  4y = -x + 34