Question 76·200 Super-Hard SAT Math Questions·Advanced Math

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In the $xy$ -plane, the graphs of $y=x^2-4x+7$ and $y=kx+1$ intersect at exactly one point. If $k$ is a positive constant, which choice is the value of $k$ ?

A line and a parabola meeting exactly once are tangent: they touch at one point. Set their $y$ -expressions equal to get a quadratic, then use one intersection means a zero discriminant. The discriminant is $b^2-4ac$ for $ax^2+bx+c=0$ . Desmos can make the intersection count visible, but a near-touch on a graph cannot establish an exact value.

Hints

Hint 1
At an intersection, both graphs give the same $y$ -value. Set their right sides equal. How many distinct real $x$ -values can that equation have if the graphs share only one point?
Hint 2
For a quadratic in the form $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . It equals zero when the quadratic has one distinct real solution. First identify $a$ , $b$ , and $c$ .
Hint 3
Undoing a square gives two branches, one positive and one negative. Keep both until you've solved for $k$ ; then use the condition that $k$ is positive.

Step-by-step

Make the intersection quadratic have one root
Step 1
See how the number of intersections changes
Type $k=1$ , $y=x^2-4x+7$ , and $y=kx+1$ in Desmos. The convenient positive test value $1$ gives two intersections; click them to see $(2,3)$ and $(3,4)$ . We need one instead. A line that touches a parabola at one point is tangent to it.
Step 2
Set the two heights equal
At a shared point, both formulas give the same $y$ -value, so set their right sides equal: $x^2-4x+7=kx+1$ . Each distinct real $x$ that solves this equation gives one shared point.
Step 3
Write the intersection equation as a quadratic
Put the equation in standard form, with $0$ on one side. Subtract $kx+1$ from both sides: $x^2-4x+7-kx-1=0$ . Combine the $x$ -terms and constants: $x^2-(k+4)x+6=0$ .
Step 4
Require one real root
For $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ is zero when the two roots merge into one. Here $a=1$ , $b=-(k+4)$ , and $c=6$ . Substitute them into $b^2-4ac=0$ : $[-(k+4)]^2-4(1)(6)=0$ .
Step 5
Isolate the square
Square the negative bracket and multiply: $(k+4)^2-24=0$ . Add $24$ to both sides: $(k+4)^2=24$ .
Step 6
Keep both square-root branches
Take both square-root branches. Since $24=4\cdot6$ , simplify the radical as you write them: $k+4=\pm\sqrt{24}=\pm2\sqrt6$ .
Step 7
Select the positive slope
Subtract $4$ from both sides: $k=-4\pm2\sqrt6$ . Because $\sqrt6>2$ , the plus branch is positive: $2\sqrt6>4$ . The minus branch is negative, so the required value of $k$ is $-4+2\sqrt6$ . Choice C.

Question 76·200 Super-Hard SAT Math Questions·Advanced Math

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