Question 202007: Hi there, could someone please help me with the following Advanced Functions problem worth 10 marks?
a)Determine the slope of the tangent to y=1/(1-x) at the point (2,-1) without constructing a table. Explain what you are doing.
b)Determine the general equation of the secant from any point (A,f(A)) to the point on f(x)=4x where x=2. Describe how you would use this to find the equation of the tangent at the point (2,f(2)).
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! 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
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