Finding a tangent line to a parabole/hyperbole

[Sap67]

First of all, excuse me if this is the wrong category. I learn math in french and we just refer to those as functions and derivatives, I just saw calculus had those so I decided to post it here. The series of pictures I'm about to send is in french but I'll explain what I did and what I need.

Maths-1

Image Maths-1 hosted in ImgBB

ibb.co

I had this function: x^2 -3x -3 (P)
I simply had to study its variations and then draw it

Maths-2

Image Maths-2 hosted in ImgBB

ibb.co

I had this function: (2x-3) / (x-1) (H)
I simply had to study its variations and then draw it

Maths-3

Image Maths-3 hosted in ImgBB

ibb.co

Representation of both functions, with the tangent that I was able to draw thanks to a graphical calculator but still can't figure out how to get yet.

Maths-4

Image Maths-4 hosted in ImgBB

ibb.co

I had to get the coordinates of the intersections between the parabola (P) and the hyperbola (H)

Now the third part of the question is to determine the equation of the tangent to (H) (which is also tangent to (P) since the tangent is at their intersection point), but instead of having the absciss point like we usually do, now I only have the coordinates at which the tangent intersects (Which is at (0 , -3) ), which I've tried working with but after several hours of trial and error, I give up.
I do know the equation is y = -3x - 3 and that it also intersects with the hyperbola (H) at (-2.66 , 5) but I only got that by luck on a graphical calculator. Plus I wouldn't know that if I hadn't cheated so I can't use those (-2.66 , 5) coordinates.

 

[Dr.Peterson]

The tangent you drew is tangent to P, not to H; it intersects both curves at C.

Please state what the third part of the problem actually said. Are you to find the equation of one particular tangent to one particular curve at one particular point, or many? Are you to find where it intersects each curve?

If it was to find the tangent to P at C, then you just have to find the derivative for P, y' = 2x - 3, and evaluate it at x=0 to find the slope of the line (-3); then use the slope-intercept or point-slope form to write the equation of this line. Since its y-intercept is at y=-3 (point C), this equation is y = -3x - 3, as you say.

 

[Sap67]

The exercice is in french so I'll translate it in english :
"We have (P) a parabola with an equation f(x) = x² - 3x - 3 and (H) a hyperbola with an equation g(x) = (2x-3) / (x+1) in an orthonormal benchmark (o, vector i, vector j)
A) Study the variations of each function and trace their curves (P) and (H)
B) Determine the coordinates of the intersection points of (P) and (H)
C) Determine the equation of the tangent to (H) led from the point E(0;3)"


Thank you for pointing that out, I was working on the tangent to (P) when I needed the tangent to (H). I don't know how I could just glance over that and didn't notice my mistake after so many attempts.

Well that means I need to get the derivative of (H) which is 5/(x+1)². If I replace x with 0 in the normal function I should get -3. If I replace x with 0 in the derivative, I should get 5.
With that, I can use that formula : y = f'(a) (x-a) + f(a)
( "a" refers to the absciss point which is 0 here) and I get y = 5x-3 which tangent to (H) when I look at the graphical calculator. So thank you for your help sir or madam.

 

[Dr.Peterson]

Good!

C) Determine the equation of the tangent to (H) led from the point E(0;3)"

I assume it said (0;-3)!