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 #1
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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.
Sep 6, 2013
 #18
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Wow, 18 replies and not a single attempt at an answer... Come on, Internet, we can do better than this.

Anyway, there are a lot of irrationals in that denominator, but there should be a way to make them all go away. First we'll try the conjugate rule.

The conjugate rule works like this: If you have something like 1/(√6+√5), you multiply the top and bottom by (√6 - √5), the conjugate. This works with the FOIL method, or First-Outside-Inside-Last.

Start with (√6+√5)(√6-√5). Take the First term from each set of brackets and multiply them. . . . . √6 x √6 = 6
. . . . . . . . . . . . . . . . . . . Then the Outside term from each set of brackets. . . . . . . . . . . . . . . . √6 x -√5 = -√30 (The negative is important!)
. . . . . . . . . . . . . . . . . . . Then the Inside. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . √5 x √6 = √30
. . . . . . . . . . . . . . . . . . . Then the Last. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . √5 x -√5 = -5

When you add them all together, you get a nice, simple, 1. Just 1. So now the fraction is (√6 - √5)/1.

The complicated part about your problem is that you have three square roots instead of two. We can still use the conjugate rule, it's just going to be a little trickier.

First, we decide where we split our denominator. I'm going to treat it as (√2 + √3) + √5. This is the same thing we had before, just organized differently. So our conjugate is going to be (√2 + √3) - √5.

Now we use the FOIL rule, which gets more complicated again... but it still works. You just have to use it twice, at the same time. I'll show you what I mean:

Multiplying ((√2 + √3) + √5) x ((√2 + √3) - √5)
First - (√2 + √3) x (√2 + √3) This is where we use FOIL again. I'm not going to show you each step, but it works out to 2√6 + 5.
Outside - (√2 + √3) x -√5 = -√10 - √15
Inside - (√2 + √3) x √5 = √10 + √15
Last - √5 x -√5 = -5

So, adding all that together, the fraction becomes (√2 + √3 - √5) / (2√6). Usually, this will leave you with another number still on the bottom, but then you can use the conjugate rule (I know, again?) to get rid of it. In this case, though, simply multiply the top and the bottom by √6.

I'll skip a few steps and tell you that the answer is (√12 + √18 - √30) / 12.

Now, always remember how to use FOIL. It's not just for square roots: the basic form applies to everything you could possibly multiply together. Happy homeworking!
Sep 6, 2013
 #1
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775/1212 is as far reduced as it ever will be.

Reducing fractions is based on dividing the top and bottom of your fraction by the same number. This works because a fraction is a ratio, meaning it compares two numbers by relative size. Another example of a ratio is this: If I have four apples and you have two, I still have twice as many apples as you do, even if we divide both our numbers by two. You will still have one apple, and I will still have two - twice as many.

You can break 775 down into smaller numbers by trying to divide it by other numbers. If you start with 2 (because everything is divisible by 1, and it's not very helpful), and keep going up, you'll see that 5 is the smallest number that divides 775 evenly. Doing it, you see that 775=5x155. You can divide 155 by 5 as well, which gives you 775=5x5x31. 31 isn't divisible by anything other than 1 and itself, so it's a prime number. 5 is as well, so we call 5x5x31 the prime factors of 775. If you know the prime numbers, you can skip all the numbers in between them when you're testing numbers because if the other ones are factors, you would have noticed their prime factors first.

1212 can also be broken down into 1212=2x2x3x101. 101 is a prime number, even though there are a lot of smaller numbers that might divide it. Because none of the prime factors of 775 and 1212 are the same, the fraction can't be reduced anymore.

If you want a decimal answer, you can just put it in a calculator. You'll get about 0.639494389... and no matter how big your calculator is, it will keep going forever.
Sep 6, 2013
 #1
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I'm not sure what exactly the function will look like, because it depends on where the brackets are. There are three possibilities, so I'll explain each of them.

Option 1: f(x) = √(x 2) - 1

This function looks like it should be the same as f(x) = x-1. After all, taking a square root should just undo the square, right? But when we see this, we have to remember that this implies only the positive square root! For example, 1^2 and (-1)^2 are both 1, but √1 is always assumed to mean 1. So both √(5^2) and √(-5^2) are assumed to be 5. This means that the graph of this function will actually look like a wide V, with the corner at (0,-1).

Option 2: f(x) = (√x) 2 - 1

Depending on how good your calculator is, this may result in half of what you'd expect. It is commonly assumed that the square root of a negative number doesn't exist, so this function might look like f(x) = x-1 for all positive values of x (and zero). However, this graph should technically look exactly like f(x) = x-1, because while the square root of a negative number certainly wouldn't show up on a graph, it does technically exist. These are called imaginary numbers, and while I won't get into too much detail, in this case, the square and square root actually would simply reverse each other. Basically, this demonstrates that with certain functions, order matters, and while a square root doesn't just undo a square, a square does undo a square root.

Option 3: f(x) = √(x 2 - 1)

I think this is the least likely option of the three, but for completeness' sake, I'm throwing it in anyway. It's also the most interesting looking graph. What happens is an odd combination of the quirks of the first two options. If you set x as any number in between 1 and -1, then x 2-1 is negative, and so its square root doesn't exist. This leaves the rest of the graph, which still looks a lot like the first option, but curves downward to approach f(x) = 0 a lot more quickly towards the end. It looks a bit like a funnel, and if you take a step back and look at a big enough area on the graph that the gap at the bottom is too small to see, it looks exactly the same as the first one.

Anyway, pick a graph, and always remember to include your brackets. They're important!
Sep 6, 2013

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