Finding Equation of tangent line 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)
$y = 16x - 31$

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