I'm not sure what you want me to do with this. So I'll do everything to it!
Describing a Parabola
First, let's put it in a slightly different form: 3(x-4) 2-6=0
This way of looking at the equation tells us a lot about the shape of the graph. The basics of it are y=a(x-p) 2+q.
The letter 'a' represents the amplitude of the parabola, or how steeply it slopes. Whether 'a' is positive or negative determines whether the parabola opens up or down. In this case, 'a' is 3, meaning the parabola opens up and has an amplitude of 3.
The letters 'p' and 'q' indicate the parabola's position on the graph. The lowest point on this parabola is at (4,-6). This point is called the vertex of the parabola. Because 'a' is positive, the parabola rises up and away from this point on both sides.
Solutions/Zeros
But seeing as 'y' equals zero in your original question, I would imagine you want the solutions, or the points where the parabola crosses the x-axis. Basically, what to make 'x' so your original equation actually works. This is also hidden in 3(x-4) 2-6=0, but it's a little trickier. So now we bring in the dreaded algebra.
3(x-4) 2=6 (add 6 to both sides)
(x-4) 2=2 (divide both sides by 3)
x-4=±√2 (take the square root of both sides)
x=4±√2 (add 4 to both sides)
So there are two solutions, 4+√2 and 4-√2. You can check this by putting them back into your equation, if you have time.
Inversions
This is getting a little more advanced, but I said everything. And I meant it.
To invert a parabola, or any function of x and y, you simply change the variables around. In this case, you will end up with 3(y-4) 2-6=x. It's hard to visualize this, or demonstrate the mechanics of the new function, so let's put y in terms of x again. Oddly, this follows pretty much the exact same steps as finding the solutions. Or perhaps not so oddly - we're still trying to isolate a variable.
3(y-4) 2=x+6
(y-4) 2=(x+6)/3
y-4=±√((x+6)/3)
y=±√((x+6)/3)+4
In terms of the formal definition of a function, this actually isn't one. The inverse of a function is not necessarily a function as well. This graph will essentially look like a sideways parabola - in fact, taking the inverse of a function always mirrors it along the line x=y.
Derivatives
I'm not going to explain this, but I wanted to do everything. Derivatives explain how quickly a variable changes as another variable changes. So here are the first and second derivatives of y=3(x-4) 2-6:
y' = 6x - 24
y'' = 6
Integrals
Nope, still not done. I've been told I take things too far sometimes, can you imagine?
An integral describes the area between a function and the x-axis. The indefinite integral of y=3(x-4) 2-6 is 3/4(x 3) - 12x 2 + 10x.
Okay, NOW I'm done.
