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)

$y = 16x - 31$

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


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

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


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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.