Given two functions defined on the real numbers:

$$
\begin{aligned}
& f(x)=t x^{2}-x+1 \\
& g(x)=2 t x-1
\end{aligned}
$$

where $t$ is a real parameter. How many common points do the graphs of $f$ and $g$ have?
The graphs of the two functions intersect at points where their values are equal. That is:

$$
t x^{2}-x+1=2 t x-1
$$

Rearranging:

$$
t x^{2}-(1+2 t) x+2=0
$$

For the case where $t \neq 0$, the quadratic equation has a double root if its discriminant is 0. Since the discriminant is

$$
D=(1+2 t)^{2}-8 t=(2 t-1)^{2}
$$

from this, $t=\frac{1}{2}$; in this case, the graphs of the two functions have one common point. In all other cases, the discriminant is positive, meaning the graphs have two common points.

The case where $t=0$ must be examined separately. In this case, $f(x)=-x+1, g(x)=-1$, and the two graphs also have one common point.