Question 135·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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In the $xy$ -plane, the line with equation $y = kx - 4$ intersects the parabola with equation

y = (x - 1)^2 + 2

at exactly one point. If $k > 0$ , what is the value of $k$ ?

The phrase exactly one point signals tangency: the line touches the parabola rather than crossing it twice. With radical answer choices and a positive slope, rule out negative values first. Then graph each remaining line in Desmos and count where it meets the parabola. Zoom out before trusting the count, since a second intersection could be offscreen.

Hints

Hint 1
The slope is the number multiplying $x$ in $y=kx-4$ . Since that number must be positive, which radical expressions can you rule out before graphing?
Hint 2
An intersection is a point shared by two graphs. Test each remaining slope by keeping the parabola and the line’s $-4$ intercept unchanged, then count where those two graphs meet.
Hint 3
A tangent line touches a parabola at one point. Zoom out when you test a slope: a line that appears to touch might have another intersection farther away.

Step-by-step

Approach 1: Test the positive slopes in Desmos
Step 1
Rule out negative slopes
Type $2-2\sqrt7$ and $-2-2\sqrt7$ in Desmos. They display about $-3.292$ and $-7.292$ . Both fail the positive-slope condition $k>0$ , so only two choices need graphing.
Step 2
Test one remaining slope
Type the given parabola and $y=(2+2\sqrt7)x-4$ . Desmos shows two intersections: one near $x=1$ and another beyond $x=8$ . This slope fails exactly one point; the higher crossing can be missed in a smaller window.
Step 3
Find the line that touches once
Add $y=(-2+2\sqrt7)x-4$ and keep the parabola visible. Desmos shows the new line touching the parabola once, near $(2.646,4.708)$ . Count that meeting point, not where the two lines meet. The required positive slope is $k=-2+2\sqrt7$ . Choice D.

Approach 2: Prove the exact slope with the discriminant
Step 1
Set the two heights equal
At a shared intersection, both equations give the same $y$ , so set their right sides equal: $(x-1)^2+2=kx-4$ .
Step 2
Expand the square
Use $(x-1)^2=x^2-2x+1$ ; the middle term is negative: $x^2-2x+1+2=kx-4$ .
Step 3
Combine the constants
Add $1$ and $2$ : $x^2-2x+3=kx-4$ . Keep the $-2x$ from expanding the square.
Step 4
Move the line’s right side
Subtract the whole expression $kx-4$ from both sides: $x^2-2x+3-(kx-4)=0$ .
Step 5
Write a quadratic equal to zero
Distribute the minus sign, then combine $3+4$ :
$x^2-2x+3-kx+4=0.$
$x^2-2x-kx+7=0$ . The $-4$ becomes $+4$ when subtracted.
Step 6
Use the one-root condition
One shared point means this quadratic has one real root. For $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ must then be $0$ : the plus and minus versions of the quadratic formula give the same root. Here $a=1$ , $b=-2-k$ , and $c=7$ , so: $(-2-k)^2=4(1)(7)$ .
Step 7
Multiply the coefficients
Since $a=1$ , multiply $4$ by $7$ : $(-2-k)^2=28$ . The square applies to the entire $-2-k$ .
Step 8
Choose the positive square root
Because $k>0$ , $k+2$ is positive. Take the positive square root of the squared expression: $k+2=\sqrt{28}$ . The negative branch would make $k$ negative.
Step 9
Simplify the radical
Since $28=4\cdot7$ and $\sqrt4=2$ , simplify the square root: $k+2=2\sqrt7$ .
Step 10
Isolate the slope
Subtract $2$ from both sides: $k=-2+2\sqrt7$ . This is the line’s positive slope. Choice D.

Question 135·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this