Question 475425
We could use calculus or a graph (actually, the graph still helps as it gives a good visual), but the easiest way to do it is to treat them as functions of y and set them equal.

*[tex \LARGE y = -\frac{1}{4}x^2 + 5 = 8 - kx]

Since the quadratic is differentiable everywhere, we can say that a tangency point occurs when the equation has only one solution for x. We turn this into a quadratic in terms of x

*[tex \LARGE \frac{1}{4}x^2 - kx + 3 = 0], set the discriminant equal to zero.

*[tex \LARGE k^2 - 4(\frac{1}{4})(3) = 0]

*[tex \LARGE k^2 - 3 = 0 \Rightarrow k = \pm \sqrt{3}]