After you set $kx - 4 = (x - 1)^2 + 2$, expand and simplify to get a standard quadratic equation in $x$ that depends on $k$.

For a quadratic equation, when does it have exactly one real solution? Think about the discriminant $b^2 - 4ac$.

When you solve the discriminant equation for $k$, you will get two values. Use the condition $k > 0$ to decide which one is valid.

Enter the parabola as y = (x - 1)^2 + 2.

Type y = kx - 4. Desmos will prompt you to add a slider for k; add it so you can adjust the slope of the line.

Move the $k$ slider and watch how the line moves relative to the parabola. Look for the value of $k$ where the line just touches the parabola at a single point (the line is tangent), instead of crossing it at two points.

Optionally, add a new expression: (k + 2)^2 - 28. As you move the $k$ slider, observe when this expression equals $0$; that $k$ value matches the one where the line and parabola touch at exactly one point.

At intersection points, the $y$-values of the line and the parabola are equal, so set the equations equal:

Now expand the right-hand side.

First expand $(x - 1)^2$:

So

Set this equal to the line:

Move everything to one side to get a quadratic in $x$:

So the $x$-coordinates of intersection points satisfy

The line intersects the parabola at exactly one point, so this quadratic in $x$ has exactly one real solution.

For a quadratic $ax^2 + bx + c = 0$, there is exactly one real solution when the discriminant $b^2 - 4ac$ equals $0$.

Here, $a = 1$, $b = -(k + 2)$, and $c = 7$, so the discriminant is

Set this equal to $0$:

Solve

Take the square root of both sides:

So

This gives two possible values: $k = -2 + 2\sqrt{7}$ and $k = -2 - 2\sqrt{7}$. Since the problem states $k > 0$, we choose