Point of intersection of $f_a(x)=ax^{2}+3x+1$ and $g_b(x)=\frac{b}{x}$ For what values of $a$ and $b$, the two functions $f_a(x)=ax^2+3x+1$ and $g_b(x)=\frac{b}{x}$ are tangent to each other at a point where the $x\text{-coordinate}=1$.
The points of intersection are where:
$f_a(1)=g_b(1)$
which gives 
$$a+4=b\text{ and } b-4=a$$
Now what to do with this information? Or if my approach is right?
 A: Notice, the slope of tangent of $f_{a}(x)=ax^2+3x+1$ at a general point is given as $$\frac{d}{dx}(f_{a}(x))=\frac{d}{dx}(ax^2+3x+1)$$$$\color{red}{f'_{a}(x)=2ax+3}$$ Similarly, the slope of tangent of $g_{b}(x)=\frac{b}{x}$ at a general point is given as $$\frac{d}{dx}(g_{b}(x))=\frac{d}{dx}\left(\frac{b}{x}\right)$$$$\color{red}{g'_{b}(x)=-\frac{b}{x^2}}$$ Since, the curves $f_{a}(x)$ & $g_{b}(x)$ are tangent to each other at a point having $x=1$ Thus the angle between tangents at $x=1$ must be zero i.e. slopes of common tangent at both the curves are equal hence, we have $$$$ 
$$(f'_{a}(x))_{x=1}=(g'_{b}(x))_{x=1}$$ $$\left(2ax+3\right)_{x=1}=\left(-\frac{b}{x^2}\right)_{x=1}$$ $$2a+3=-b$$ $$\bbox[4pt, border: 1px solid blue;] {\color{blue}{2a+b+3=0}}$$ Above is the required relation between $a$ & $b$. Hence there are infinitely many real values of $a$ & $b$ satisfying the above relation.
A: It looks like your approach is right, and you cannot gather any further information about $a$ and $b$.
There are just infinitely many pairs of $a$'s and $b$'s that satisfy the requirements of the problem, and you have shown all the requirements on the $a$'s and $b$'s.
So you are done! So report your answer, and start on another problem.
A: Isn't “to be tangent to each other in $x_0$”  defined by $f(x_0)=g(x_0)$ and $g'(x_0)=f'(x_0)$? know that $f(1)=g(1)$ and $f'(1)=g'(1)$.  In this case we derive two linear equations with solutions $a=-7/3$ and $b=5/3$.
