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Calculus and differentiating

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The first step is to find the general form of the slope of f(x). We can find this by differentiating f(x):
 
f'(x) = -(x-2)^2*(2x+1)^3*(14x-13) 
 
Note: At any given x, f'(x) will give us the slope of the tangent line of f(x) at that particular point.
 
Let y = f(x) and m = f'(x). At x=-1 then,
 
y = f(-1) = (2-(-1))3(2(-1)+1)= 27
m = f'(-1) = -((-1)-2)^2*(2(-1)+1)^3*(14(-1)-13) = -243
 
A line takes the general form:
 
y = mx + b
 
plugging in the values...
 
(27) = (-243)(-1) + b
b = -216
 
Hence, the equation of the line tangent to the graph of f(x) is 
 
y = -243x - 216