linear divided by square root of quadratic I want to find the $x\in\mathbb{R}$ which maximizes $$f(x)=\frac{a+bx}{\sqrt{(x-c)^2+d^2}}$$ for given real numbers $a,b,c,d\neq0$.
Since this is a differentiable function, I do this by solving for $f'(x)=0$, which yields a single extremum $\displaystyle x_0=c+\frac{bd^2}{a+bc}$ (or none if $a+bc=0$). And $\displaystyle f''(x_0)=-\frac{d^2((a+bc)^2+b^2d^2)}{(a+bc)((x-c)^2+d^2)^{5/2}}$, so whether it is a maximum or minimum depends solely on whether $a+bc>0$ or $a+bc<0$.
While the computational part of this is pretty clear to me (and I've verified the computations in Maple), I'm struggling with the intuition. What is the interpretation of $a+bc$ here, and why is it so important that its sign determines the direction of the curve, or even makes $f$ monotonous exactly when $a+bc=0$? I feel like there must be some geometric/parabola interpretation here, but I fail to see it.
 A: Rewrite as
$$
f(x)=\frac{b(x-c)+(a+bc)}{\sqrt{(x-c)^2+d^2}}=\frac{(x-c,d)}{\sqrt{(x-c)^2+d^2}}\cdot\left(b,\frac{a+bc}d\right)
$$
So geometrically, $f$ is the (signed magnitude of the) projection of $(b,\frac{a+bc}{d})$ onto an oriented line, and $a+bc\neq 0$ is the necessary and sufficient condition for the existence of vector $(x-c,d)$ parallel to $(b,\frac{a+bc}{d})$ (which yields $x=c+b\cdot\frac{d^2}{a+bc}$).  Whether that is a maximum or minimum depends on whether they point in the same direction, which is of course what the sign of $a+bc$ tells you.
A: If you substitute X = x-c then you get
$$F(x)=\frac{a+bc + bX}{\sqrt{X^2+d^2}} = \frac{a+bc}{\sqrt{X^2+d^2}} + \frac{bX}{\sqrt{X^2+d^2}}$$
This function has the same maximum/minimum (though the location is shifted by $c$, of course). The left term always has the sign of $a+bc$ and is maximal (or minimal, depending on the sign of $(a+bc)$) when $X=0$. The right term is odd at $X=0$, perturbing the max/min to the side, according to the sign of $b$.
