Question 53·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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Let $b$ be a real number. Consider the system of equations

y = -x^2 + 4x + 3

y = 2x + b

For which value of $b$ does the system have exactly one real solution?

A tangent line touches a parabola at one point, giving a system exactly one solution. The cue is a line with an unknown intercept, a parabola, and a requested solution count. In Desmos, graph the parabola, add a line for each possible intercept, and count shared points. Don’t mistake two close crossings for one touch.

Hints

Hint 1
An intersection is a point on both graphs, so it solves both equations. Keep the parabola graphed and try each possible value of $b$ in the line. How many shared points does each line have?
Hint 2
A tangent line touches the parabola once without cutting through it. If a line looks like it touches, zoom in: two nearby crossings can look like a single point.

Step-by-step

Approach 1: Graph the possible lines
Step 1
Check the line when $b=0$
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.
Step 2
Check the line when $b=2$
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.
Step 3
Check the line when $b=6$
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.
Step 4
Find the line that touches once
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.

Approach 2: Check the one-solution condition with algebra
Step 1
Set the two outputs equal
At a shared point, both equations give the same $y$ , so set their right sides equal: $-x^2+4x+3=2x+b$ .
Step 2
Put the equation in quadratic form
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.
Step 3
Build the discriminant
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: $D=16-4b$ .
Step 4
Find the constant for one root
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.

Question 53·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

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Strategy

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Step-by-step

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