Slope of the tangent

Question

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.

Answer 1

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 $(y-y_0)=f'(x_0)(x-x_0)$

Answer 2

(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)