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I want to calculate the tangent line of $f(x) = Ax^3 + Bx^2 + Cx + D$. But I am not sure what are the required steps.

I remember that I should first get the derivative, which is $3Ax^2 + 2Bx + C$ I believe. Then something like $Y-Y_1 = M(x-x_1)$ or something like that (sorry I did this long time ago).

Would someone please put me in the right way?

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(+1), and welcome back to math! – The Chaz 2.0 Mar 19 '12 at 19:15
up vote 7 down vote accepted

You have quoted the result essentially correctly. We want the tangent line at the point on the curve $y=f(x)$ which has $x$-coordinate say $x_1$. Calculate the $y$ coordinate $y_1$, using $y_1=f(x_1)$. Then the equation of the tangent line can be written as $$y-y_1=f'(x_1)(x-x_1).$$ In our situation, we have $f(x)=Ax^3+Bx^2+Cx+D$, so as you wrote we have $f'(x)=3Ax^2+2Bx+C$. Thus $$y_1=Ax_1^3+Bx_1^2+Cx_1+D\qquad\text{and}\quad f'(x_1)=3Ax_1^2+2Bx_1+C.$$

Remark: The number $f'(x_1)$ gives the slope of the tangent line at the point on the curve which has $x$-coordinate equal to $x_1$. Now if you take a "general" point $(x,y)$ on the tangent line, then the slope of the line that joins this point $(x,y)$ to $(x_1,y_1)$ is equal to $$\frac{y-y_1}{x-x_1}$$ (the change in $y$ divided by the change in $x$). Since this slope is supposed to be $f'(x_1)$, we get $$\frac{y-y_1}{x-x_1}=f'(x_1).$$ Now multiply both sides by $x-x_1$ to get the equation of the tangent line that you quoted.

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Thanks! What does is name of function (with power of 3) in english? Quadratic is the one with power of 2 I guess...what goes for this one?! – Sean87 Mar 19 '12 at 20:25
@Sean87: Cubic polynomial, often abbreviated as "cubic." – André Nicolas Mar 19 '12 at 20:28

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