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**serious331** define $\displaystyle f(x) = x^3+2x+1$ for all x. Find the equation of the tangent line to the graph of $\displaystyle f:R \rightarrow R$ at the point (2,13).

I know that for a function $\displaystyle f:I \rightarrow R$ (I is a neighborhood of the point $\displaystyle x_0$) that is differentiable at $\displaystyle x_0$, the tangent to the graph of the point $\displaystyle ({x_0},(f(x_0))$ is given by the equation:

$\displaystyle y = f(x_0)+f'(x_0)(x-{x_0})$

To find the slope of the line, wehave

$\displaystyle \lim_{x->{x_0}} \frac{f(x)-f(x_0)}{x-{x_0}}$

$\displaystyle = \lim_{x->{x_0}} \frac{(x^3+2x+1)-({x_0}^3+2{x_0}+1)}{x-{x_0}} $

$\displaystyle = \lim_{x->{x_0}} (x^3-{x_0}^3)+2 = 2$

is the slope of this function is 2 then? I cannot figure out if this is correct and how to find the equation of the tangent line (Headbang)