This is a didactic question.
Given a differentiable function $y=f(x) \;, x,y \in \mathbb{R}$, I want to construct an exercise in which we have to find a straight line that passes through a point $P=(x_P,y_P)=(x_P, f(x_P))$ of its graph and is tangent in some other point $X=(x,y)=(x,f(x))$.
This problem reduced to solve the equation: $$ (1) \qquad f(x)-f(x_P)=f'(x)(x-x_P) $$ that has the obvious solution $x=x_P$, but the other solutions (if they exists) are not, in general, ''easy'' solutions , i.e. solutions that can be expressed in closed form with elementary functions.
My problem is to find some class of functions such that equation $(1)$ can have such ''easy'' solutions.
I've find that functions of the form $$ f(x)=ax+\dfrac{b}{x^2}+c $$ work well, since equation $(1)$ in this case become a second degree equation that can be solved in elementary way. But I'm searching for some other class of functions and since I don't have a method to look for them, my imagination failed to produce any interesting result.
So my question is: someone can find a method ( but I doubt that exists) or has enough imagination to find some other class of functions tha work well for this kind of exercises?
Let's consider a cubic function $f(x)=a_0x^3+a_1x^2+a_2x+a_3$. Assume that you know $(p,q)$ a point on the graph of the function. Then, you know that $q=a_0p^3+a_1p^2+a_2p+a_3$. Using this, you can find a relation for $a_3=q-a_0p^3-a_1p^2-a_2p$. So, we write $a_3$ as a polynomial in $a_0$, $a_1$, and $a_2$.
Now, consider the line $y-q=m(x-p)$. This is a line passing through the point $(p,q)$. We would like this line to be tangent to the curve at another point. Then, consider the line as $y=mx-mp+q$ and by substitution, we consider $$ mx-mp+q=a_0x^3+a_1x^2+a_2x+a_3. $$ A solution to this equation is both on the graph of the cubic function and on the line above. Consider $$ a_0x^3+a_1x^2+(a_2-m)x+(a_3+mp-q)=0. $$ This is a cubic polynomial in $x$ and we are interested in its roots. We know that $x=p$ is a root by the definition of $a_3$. So, we can rewrite this as $$ (x-p)(a_0p^2+pa_1+pa_0x-m+a_2+xa_1+a_0x^2)=0. $$ (I used maple to get this factorization and to eliminate $a_3$).
Now, to be a tangent point to the curve, the remaining factor must have a double root. You can observe that this has a double root when the discriminant vanishes. Therefore, you want $$ (a_0p+a_1)^2-4a_0(a_0p^2+pa_1-m+a_2)=0 $$
Therefore, you should choose $a_0$, $a_1$, $a_2$, and $p$ so that this equation holds. This will generate the type of curve that you're looking for.