Why would a 2 degree polynomial with 2 variables on intersecting with a line give either 0, 1 or 2 points? For clarification, the 2nd degree polynomial with two variables is $f(x,y)$, and the intersecting line is also a (first degree) polynomial in $x$ and $y$.

Why would the 2nd degree polynomial with 2 variables intersect with a line in either 0, 1 or 2 points?

I have no clue on how even to begin this, so any basic idea/hint would be helpful. I have tried a couple of straightforward easy examples, but I haven't been able to generalize it.
 A: There is an exception to what the problem describes.  Let $f(x,y)$ be a second degree polynomial and $g(x,y)$ a first degree polynomial.   If $g$ divides $f$ as polynomials, then all (infinitely many) solutions to $g(x, y)= 0$ also solve $f(x,y)=0$.
Otherwise we can show that the equations $f( x,y)=0$ and $g(x,y)=0$ have at most two solutions in common.  To begin note that since $g(x,y)$ is first degree,  at least one of the variables $x,y$ will have a nonzero coefficient.   Assume for the sake of argument that $y$ has a nonzero coefficient (otherwise we exchange the roles of $x$ and $y$ in the steps that follow).
Now rewrite the equation $g(x,y)=0$ as giving $y=mx+b$.
Substitute that for $y$ in the equation $f(x,y)=0$.  The result is an equation only in $x$ of degree (at most) two.
The point to be careful of is that because $g$ doesn't divide $f$, the result cannot simplify to $0=0$.  In other words there will not be more than two solutions for $x$ and corresponding solutions for $y=mx+b$.
