Hello, Ranger SVO!

You're curious . . . good for you!

While playing with parabolas in college (centuries ago),

. . I "discovered" some facts that might help you.

The equation I'm using is $\displaystyle y \:= \:x^2 - 4x - 2$

and I'm looking for the equation of the tangent line at point $\displaystyle (4,\,-2)$

Using calculus I found the equation to be: $\displaystyle y \:= \:4x -18$

How can I do this without calculus? Code:

* | */
| /
* | *P(x,y)
* | */
---------***---o----------
| /↑
| /(½x,0)
| /
|/
o ←(0,-y)
/|

Given the parabola: .$\displaystyle y \:=\:ax^2$, the tangent at point $\displaystyle P(x,y)$

. . has an x-intercept of $\displaystyle \left(\frac{1}{2}x,\,0\right)$ and a y-intercept of $\displaystyle (0,\,-y)$

The tangent at $\displaystyle P(x,y)$ always has a point directly below the vertex

. . with the same vertical displacement (only downward).

Your parabola $\displaystyle y \:=\:x^2-4x-2$ has its vertex at $\displaystyle (2,-6).$

Your point $\displaystyle P(4,-2)$ is 4 units *above* the level of the vertex.

Hence, it has a point 4 units directly *below* the vertex: $\displaystyle Q(2,-10).$

Now you can write the equation through $\displaystyle P$ and $\displaystyle Q.$