(Calculus) Derivative Thinking Question Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.
The Question:
Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?
Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1. 
Find the required conditions on the parameters a,b,c in terms of a.
My Progress So Far:
So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.
Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.
I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.

If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-\frac{f'(x)}{f(x)^2}
$$
Thus, assuming of course that $f(x)\ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-\frac{f'(x)}{f(x)^2}=f'(x)
$$
which can only happen if $f'(x)=0$. Indeed, if $f'(x)\ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.
You can observe that this is independent of $f$ being a quadratic polynomial.
In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))\ne0$ (or $g(-b/(2a))$ would be undefined.

