Hi, so I have the following exercice as a homework, it's in french so I'll do my best to translate it.
We got \(\displaystyle y = f(x) = \frac{3x + ax + b}{x^2+1}\)
where a and b are real numbers.
Determine a and b the curve f(x) is tangent to the line \(\displaystyle y = 4x + 3\) on the point \(\displaystyle I(0;3) \).
Before I explain what I tried so far, I have to say, I haven't really started school yet. We only got some homework by some new teachers, and what this teacher has been sending us is slightly more difficult from what we learned last year (Which is much, mostly because of the pandemic, and some other issues my country had to deal with which caused schools to close a lot).
So I tried some things we would do last year like :
 Combining both equations then doing the quadratic equation in order to get this : \(\displaystyle \delta = x^2 + 4ac \)
Then comparing it to 0 so there's only one possibility for x.
But that doesn't work here since the degree of the equation I get is 3 and I found no way of simplifying it. And we haven't learned how to deal with third degree equations.
Using the following formula : \(\displaystyle y = f'(a) (xa) + f(a) \) .
 Combining both of the methods above in different orders.
Replacing x and y in the curve equation by those given by the I point, also combined with the methods above.
But none of the things I tried worked out.
This is what I got after typing the equations I got on desmo :
While they are intersecting on 0;3, they are nowhere near tangent
We got \(\displaystyle y = f(x) = \frac{3x + ax + b}{x^2+1}\)
where a and b are real numbers.
Determine a and b the curve f(x) is tangent to the line \(\displaystyle y = 4x + 3\) on the point \(\displaystyle I(0;3) \).
Before I explain what I tried so far, I have to say, I haven't really started school yet. We only got some homework by some new teachers, and what this teacher has been sending us is slightly more difficult from what we learned last year (Which is much, mostly because of the pandemic, and some other issues my country had to deal with which caused schools to close a lot).
So I tried some things we would do last year like :
 Combining both equations then doing the quadratic equation in order to get this : \(\displaystyle \delta = x^2 + 4ac \)
Then comparing it to 0 so there's only one possibility for x.
But that doesn't work here since the degree of the equation I get is 3 and I found no way of simplifying it. And we haven't learned how to deal with third degree equations.
Using the following formula : \(\displaystyle y = f'(a) (xa) + f(a) \) .
 Combining both of the methods above in different orders.
Replacing x and y in the curve equation by those given by the I point, also combined with the methods above.
But none of the things I tried worked out.
This is what I got after typing the equations I got on desmo :
While they are intersecting on 0;3, they are nowhere near tangent
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