After rearranging, you will have a quadratic in $x$. When does a quadratic have exactly one real solution?

Solving the discriminant equation will likely produce two possible values of $k$. Use the fact that $k$ is positive to choose the correct one.

Enter $y=x^2-4x+7$.

(Anik‌o.аі)

Enter $y=kx+1$ and let Desmos create a slider for $k$.

Move the slider until the line appears tangent to the parabola (the graphs touch at exactly one point rather than crossing twice).

When there is exactly one intersection, note the slider value of $k$ (it should be close to $0.9$). Then choose the option that matches that value (the only positive option near $0.9$).

At an intersection point, both $y$-values match:

Ani‌кο​ A​І Тutоr

Rearrange to get a quadratic in $x$:

The line and parabola intersect at exactly one point when the quadratic equation in $x$ has exactly one real solution. That happens when the discriminant is 0.

For $ax^2+bx+c=0$, the discriminant is $b^2-4ac$.

Here, $a=1$, $b=-(k+4)$, and $c=6$.

So $(k+4)^2=24$, which gives

Thus $k=-4\pm 2\sqrt{6}$.

Of the two values $-4\pm 2\sqrt{6}$, only $-4+2\sqrt{6}$ is positive.

Therefore, the value of $k$ is $-4+2\sqrt{6}$.