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)
$y = 16x - 31$

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