Question 209·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Try the simple test value $m=1$. Type $m=1$, $y=(x-m)^2+1$, and $y=mx$ in Desmos. Click the two intersections: Desmos shows $(1,1)$ and $(2,2)$. So one value of $m$ can give two solution points. We need the condition that makes the graphs meet only once.

At an intersection, both equations give the same $y$, so set their right sides equal: $(x-m)^2+1=mx$. Each $x$ that solves this equation gives one system solution, because either original equation then gives its $y$.

Expand the square: $x^2-2mx+m^2+1=mx$. Subtract $mx$ from both sides: $x^2-3mx+m^2+1=0$. Now the coefficient of $x$ is $-3m$, not $-2m$; the line's $mx$ must be subtracted too.

For $ax^2+bx+c=0$, the discriminant is $D=b^2-4ac$. Here $a=1$, $b=-3m$, and $c=m^2+1$. Substitute all three coefficients: $D=(-3m)^2-4(1)(m^2+1)$. Simplify: $D=5m^2-4$.

Exactly one shared point means the quadratic has one real $x$-root, so its discriminant is $0$. Set $D=0$: $5m^2-4=0$. Add $4$: $5m^2=4$. Divide by $5$: $m^2=\frac45$.

Take both square roots: $m=\pm\sqrt{\frac45}$. These are two distinct real values of $m$, and each makes the line and parabola meet exactly once. The requested count is two. Choice C.