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I have the following exercise.

$1$. Find the slope of the tangent to the curve $y = x^3 -4x + 1$ at the point where $x = a$.

$2$. Find the equations of the tangent lines in the points $(1, -2)$ and $(2,1)$.

$3$. Graph the curve and the two tangents in a single Cartesian plane.

I understand perfectly how to elaborate the answers for $2$nd and $3$rd part, but I have doubts about how the first part should be answered: Find the slope of the tangent to the curve $y = x^3 -4x + 1$ at the point where $x = a$.

Could you please help me? Many thanks in advance.

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  • $\begingroup$ Take the derivative at $a$? $\endgroup$ – Andrew Li Mar 11 '18 at 0:42
  • $\begingroup$ I understand perfectly how to elaborate the answers for 2nd and 3th part Use one of those to find the tangent at point $(a, a^3-4a+1)$. $\endgroup$ – dxiv Mar 11 '18 at 0:43
  • $\begingroup$ Can you get started with the third part at all? Where are you stuck? $\endgroup$ – saulspatz Mar 11 '18 at 0:51
  • $\begingroup$ Please remember that you can choose an aswer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 12 '18 at 23:55
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HINT

  1. Remember that the slope in a point x is given by $f'(x)$
  2. The line equation in a point $(x_0,y_0)$ is $(y-y_0)=f'(x_0)(x-x_0)$
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  • $\begingroup$ the slope of the tangent to the curve y=x slope of the tangent to the curve y=x3−4x+1 at the point where x=a3−4x+1 at the point where x=a is m=3a^2-4. is this correct? $\endgroup$ – Tulipa Kauffmaniana Mar 11 '18 at 4:28
  • $\begingroup$ Yes exactly! $m=3a^2-4$ is the slope $\endgroup$ – gimusi Mar 11 '18 at 6:57
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(1) Find $\,f'$(x) and then substitute x = a as the derivative of a function at a point is equal to the tangent slope at that point. You're supposed to get 3a$^2$-4.

(2) You can use the point-slope form of a line ie $\,y-y_0 = \frac{dy}{dx}.(x-x_0) $ where $\,(x_o,y_0)$ are the given points.

Here the equations should simplify to $\,y = -(x+1)$ and $\, y= 8x-15$

(3)

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