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Consider the parabola y = 4x − x2. (a) Find the slope of the tangent line to the parabola at the point (1, 3).
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do u know derivatives?
i'm just getting into them
i know what they are
i got it
just plug in 1 and get 2
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(b) Find an equation of the tangent line in part (a).
ok u have \[ \large f(x)=4x-x^2 \] find \(f'(x)\)
then the tangent line at (1,3) will be \[ \large y-3=f'(1)\cdot(x-1) \]
how did you get there?
to define a line u need either two points or one point and the slope. in this case u have the point (1,3) and the slope of the tangent which is the derivative.
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ok
did u get \(f'(1)=\)
no
is it 0?
\[ \large f(x)=4x-x^2 \] then \[ \large f'(x)=4-2x \] so \[ \large f'(1)=4-2\cdot1=2 \]
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ohh
this is the slope of the tangent line
ok
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