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Can someone double check my work to see if I'm doing it correctly?

Find the equation of the line tangent to the graph of $(2,1)$ where $f$ is given by $f(x) = 2x^3 - 2x^2 + 1$

1) $f'(x) = 6x^2-4x$ (First I found derivative)

2) $f'(2) = 6(2)^2-4(2) = 16$ (Then found slope by plugging $x$ coordinate into derivative)

3) $y-1 = 16(x-2) =$ (Then I plugged slope, $x$, and $y$ into point slope formula and solved)

$y = 16x - 31$

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  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 8 '15 at 19:20
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Hint:

As noted in the answer of @abel $P=(2,1)$ is not a point of the graph of $y=f(x)$. If you want find a tangent to the graph passing through $P$ ( if it exists) you can proceed in two ways.

1) Take all straight lines through $P$, i.e $y-1=m(x-2)$ and find if there exists some $m$ such that the system: $$ \begin{cases} y-1=m(x-2)\\ y=f(x) \end{cases} $$ has a double solution.

2) Search a point $X=(x,f(x))$ such that

$$ \dfrac{f(x)-y_P}{x-x_P}=f'(x) $$

i.e.

$$ f(x)-1=f'(x)(x-2) $$

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i dont think your answer is correct. i get $f(2) = 9$ so $(2,1)$ is not on the graph of $y = f(x).$

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