What is the Inverse function of $f(x)=192x-16x^2$? 
What is the inverse function of: $$f(x)=192x-16x^{2}$$

I have been finding myself  going in a circle in trying to complete this problem which otherwise looks simple, but for some reason I am at a block.  Could someone help explaining how to go about this problem?
 A: The answers of @nayrb and @lab bhatacharjee are correct as far as they go, but as a (retired) teacher, I feel that I must address a fault of the math education system, here in the States as well as, perhaps, elsewhere. To describe a function, you need to mention the domain and the target space, and that is particularly important here. Sketch the graph! You see that there’s a maximum at $(6,576)$, and that on either side of the vertical line $x=6$, there are points on the graph at equal height. So the function fails the “horizontal line test” unless you restrict the domain. Let’s restrict to the interval $\langle-\infty,6]$, i.e. the closed half-line to the left of $6$. But what about our inverse function? It’s not defined for $y>576$, and that means that its domain has to be no bigger than $\langle-\infty,576]$. It’s only now, once we’ve restricted both the domain and the target space of our original function, that we really have an inverse function, and its formula is indeed the one given by @lab, with the minus sign of course.
A: Instead of completing the square, as in lab bhattacharjee’s answer, you can use the quadratic formula. You have $y=192x-16x^{2}$; rewrite it as $16x^2-192x+y=0$, and treat $y$ as the constant term to get
$$x=\frac{192\pm\sqrt{192^2-64y}}{32}=6\pm\frac{\sqrt{64(576-y)}}{32}=6\pm\frac{\sqrt{576-y}}4\;.$$
This gives you two functions, both defined for $y\le 576$:
$$x=6+\frac{\sqrt{576-y}}4\;,$$ and $$x=6-\frac{\sqrt{576-y}}4\;.$$
The first corresponds to the righthand side of the parabola $y=192x-16x^2$, and the second to its lefthand side.
A: Let $f(x)=y=192x-16x^2\implies -y=16(x^2-12x)=16(x-6)^2-16\cdot 36$
$\implies 576-y=16(x-6)^2$
$\implies x=f^{-1}(y)=6±\frac{\sqrt{576-y}}{4}$
Clearly $y ≤576$ to make $f(y)$ invertible in real numbers. 
A: The function will only have an inverse if you restrict it to a region where it is one-to-one.  You can look at this graph:
 
and see that you need to restrict your function to one of the regions where $6\le x\le \infty$ and $-\infty<y\le 576$ or $-\infty<x\le 6$ and the same $y$-region.
The quadratic formula (as indicated above) can give you the proper formula. Wolfram Alpha works it out nicely here:
Wolfram Alpha Knows the Quadratic Formula
You can "show steps" to see how Wolfram Alpha does this; it's essentially completing the square.
The two solutions correspond to the two branches of the inverse function. 
