Question 158·Medium·Advanced Math

Type $y=-x^2+4x+3$ and $y=2x+0$ into Desmos. Click both intersections; Desmos shows $(-1,-2)$ and $(3,6)$. Each intersection solves both equations, so two shared points mean this value of $b$ gives two solutions.

Keep those graphs and add $y=2x+2$. Click where this new line meets the parabola: Desmos shows points near $(-0.414,1.172)$ and $(2.414,6.828)$. It still crosses twice, even though one crossing is near the parabola’s top.

Add $y=2x+6$. Desmos shows this line above the parabola, with no shared point. A line that misses the parabola gives no solution, not one.

Add $y=2x+4$ and click its single shared point with the parabola, $(1,6)$. The line touches rather than crosses, so this system has exactly one real solution when $b=4$. One shared point means one solution; two crossings mean two. Choice C.

At a shared point, both equations give the same $y$, so set their right sides equal: $-x^2+4x+3=2x+b$.

Subtract $2x+b$ from both sides: $-x^2+2x+3-b=0$. Each real value of $x$ that solves this equation gives one intersection of the original graphs.

For $Ax^2+Bx+C=0$, the discriminant $D=B^2-4AC$ is zero for exactly one real root: the plus and minus parts of the quadratic formula then agree. Here $A=-1$, $B=2$, and $C=3-b$. Substitute them: $D=2^2-4(-1)(3-b)$. Square $2$ and multiply $-4(-1)$: $D=4+4(3-b)$. Distribute the $4$: $D=4+12-4b$. Combine the constants: $D=16-4b$.

Set the discriminant to zero: $16-4b=0$. Type $16-4b\sim0$ in Desmos; the tilde asks it to find $b$. Under PARAMETERS, Desmos shows $b=4$, the value that makes the system have exactly one real solution. Choice C.