Question 159·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The line $y = mx + n$ is tangent to the parabola $y = x^2 + bx + c$ . Which of the following expresses all possible values of $b$ in terms of $m$ , $n$ , and $c$ ?

A line tangent to a parabola touches it at exactly one point. Set their $y$ -values equal to get a quadratic equation in $x$ . One touching point means the discriminant is zero. That gives an exact equation for $b$ ; a graph with unknown constants can't reliably supply the exact expression. Watch the order of the constants when you move terms.

Hints

Hint 1
A tangent line and a parabola share exactly one point. At that point, both formulas give the same height. What equation do you get by setting those formulas equal?
Hint 2
The discriminant of a quadratic $Ax^2+Bx+C=0$ is $B^2-4AC$ . It equals zero when the quadratic has exactly one real solution. After moving everything to one side, what are $B$ and $C$ ?
Hint 3
Taking the square root of a squared expression gives two possible signs. Check whether the number under the radical is a perfect square times another expression before simplifying.

Step-by-step

Use the zero-discriminant condition
Step 1
Set the two heights equal
No Desmos needed. The exact expression comes from the one-point tangency condition, not from estimating a graph with unknown constants. At their shared point, the line and parabola have the same $y$ -value, so set their formulas equal:
$x^2+bx+c=mx+n.$
Step 2
Make one side zero
The discriminant applies when a quadratic equation has zero on one side. Subtract $mx+n$ from both sides: $x^2+(b-m)x+(c-n)=0$ . The $x$ -coefficient is $b-m$ , and the constant term is $c-n$ , not $c+n$ .
Step 3
Use tangency to set the discriminant to zero
Each solution for $x$ gives a point where the graphs meet. Tangency means one such point, so the quadratic has one real solution. For $Ax^2+Bx+C=0$ , the discriminant $B^2-4AC$ is zero in that case. Here $A=1$ , $B=b-m$ , and $C=c-n$ , so:
$(b-m)^2-4(c-n)=0.$
Step 4
Isolate the square
Add $4(c-n)$ to both sides:
$(b-m)^2=4(c-n).$
Step 5
Keep both square-root possibilities
Take square roots, keeping both signs because either a positive or a negative number can have the same square:
$b-m=\pm\sqrt{4(c-n)}.$
Step 6
Simplify the radical
Simplify the square root of $4$ : $b-m=\pm2\sqrt{c-n}$ . For real values of $b$ , $c-n$ must be nonnegative.
Step 7
Solve for the requested coefficient
Add $m$ to both sides: $b=m\pm2\sqrt{c-n}$ . These are all possible real values of the parabola's coefficient $b$ that make the line tangent. Choice A.

Question 159·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

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