The Coin Return Foundational Fundraiser is here! Please donate!
functions and graphs
Fizban140
Registered User, __BANNED USERS regular
Registered User, __BANNED USERS regular
So math is really good at not explaning anything to you at all, then making you do stuff that is again not explained anywhere.
f(x)=2x^2-x-1
c) If f(x)= -1 what is x? What points are on the graph of f?
So I have no idea how to do this, read the section for 20 minutes, it didn't describe anything remotely close to this problem, got out my student solution manual that was really fucking expensive, and it just gives me some bullshit magic, it didn't explain anything at all just drew out of the work which doesn't help at all. Whoever writes this shit needs to be fired, but they are probably too busy laughing to the bank.
Anyways the book conjures up -1 = 2x^2 - x -1 that makes sense besides breaking every rule I ever learned about algebra, i thought you never made it equal 0?
0 = 2x^2 - x
0 = x(2x-1) ----> x = 0,x=1/2
What the fuck just happened? how is x(2x-1) 1/2?
f(x)=2x^2-x-1
c) If f(x)= -1 what is x? What points are on the graph of f?
So I have no idea how to do this, read the section for 20 minutes, it didn't describe anything remotely close to this problem, got out my student solution manual that was really fucking expensive, and it just gives me some bullshit magic, it didn't explain anything at all just drew out of the work which doesn't help at all. Whoever writes this shit needs to be fired, but they are probably too busy laughing to the bank.
Anyways the book conjures up -1 = 2x^2 - x -1 that makes sense besides breaking every rule I ever learned about algebra, i thought you never made it equal 0?
0 = 2x^2 - x
0 = x(2x-1) ----> x = 0,x=1/2
What the fuck just happened? how is x(2x-1) 1/2?
Fizban140 on
0
Posts
If y = -1 solve for x in the equation y = 2x^2 - x - 1, but they're using function notation because it's convenient for certain things. But it's the same idea.
So you plug in -1 for y to get:
-1 = 2x^2 - x - 1
Regardless of anything else you want to get one side of the equation = 0 so you can factor or apply the quadratic formula, so let's do that by adding one to both sides:
0 = 2x^2 - x
At this point you factor by taking an x from each term to get:
0 = x * (2x - 1)
The only way to multiply two numbers to get 0 is if one or more of them IS zero, so possible solutions are:
x = 0 or 2x - 1 = 0
Solve for x in the latter to get:
x = 1/2
Solutions are x = 0, x = 1/2
And then obviously the y coordinate for both of those is -1 as was stated, so the second part of the question (the points on the graph)
(0, -1) and (1/2, -1)
So, if you write x=0, then the first equation is satisfied : 0 * (2x+1) =0
Also, if you write 2x+1 = 0, then the same thing happens : x * 0 = 0
So, in order for the fisrt equation to be satisfied, both x =0 and 2x+1 = 0 are good.
Then, isolate x in both answers, which gives x = 0 and x = 1/2, as given by your book.
Edit : What he said
Once you get that, you know that -1 = 2x^2 - x - 1. I'm not sure what you mean when you say that it breaks every rule of algebra, but to get to the next step, you add one to both sides. Since you know both sides are equal to each other, when you add one to both sides, they are still equal to one another, so you get
0 = 2x^2 - x
The next part is just factoring out an x from both parts of the equation on the right side. x*(2x - 1) is the same thing as 2x^2 - x, and writing it the first way lets you see what x is equal to easier.
0 = x(2x - 1)
So basically what you're looking for is a value for x that will make the right side of the equation equal to zero. It's easier to see possible answers if you break it up a little bit more:
0 = (x)*(2x-1)
You know that 0 times anything is equal to zero, so the result must be zero if either bit of the equation in parenthesis is equal to zero. This gives you two more equations:
0 = x and 0 = 2x - 1
The 0 = x is obvious; if x is 0, then the equation from earlier, f(x), will be equal to -1. You can plug x = 0 back in to that original equation and see that it works. For 0 = 2x - 1, you do some simple algebra and you get 1/2 = x. Again, if you plug that back in to the original equation, you'll get -1 as a result.
For the graph, I put the equation into wolframalpha and it gives you this:
That original equation, f(x), just means that there's some equation which has a result that depends on x. As you change x, the result of that function changes. When the question says that f(x) = -1, it means that there is some value, or values, of x that cause the result of the function to be -1.
The graph shows you a range of values for the function. The horizontal tracks values for x, and the vertical axis tracks values for f(x). The graph shows that, when f(x) is -1, x can be either 0 or 0.5, since both values of x cause the function to be equal to -1.
What you're asked to do here is find some particular value of x for which the function evaluates to -1. In other words, what number, when inserted into the function in place of x, would produce -1 as an answer.
Since the function is 2x^2-x-1 we set that equal to -1.
-1 = 2x^2-x-1
This gives us an equation perfectly describing the situation outlined above; we have some number (x) inserted into the function and when evaluated it will produce -1 as an answer. All we need to do now is solve the equation.
As you noted, this being a quadratic equation (the highest power of x is 2) we need to have a zero on one side of the equal sign to solve. This is achieved by adding one to both sides.
0 = 2x^2-x
The reason we want a zero on one side is that we know that if we multiply two numbers together and get zero as an answer, one or both of those numbers are zero. With this in mind we factor the right side so that we are multiplying two numbers rather than subtracting. 2x^2 and -x both have a factor of x, so using the distributive property we extract the x from each term and rewrite the right side of the equation as x(2x-1).
0 = x(2x-1)
Once again, if the product of two numbers is zero, at least one of those numbers must be zero. The two numbers we're multiplying together to get zero are x and 2x-1. Therefore, either x = 0 or 2x-1 = 0. If 2x-1 = 0 then we can solve that equation for x and find that x must be 1/2 in that case. Therefore, if f(x) = -1, x must be either 0 or 1/2.
0431-6094-6446-7088
0=x*(2x-1)
if x is zero than 0=0
0*(2x) is 0
0*(-1) is 0
This makes no sense to me, there is no system. It is some learned rule extracted from something not in the problem at all. There is no logic here. This isn't algebra that I know at all.
This is incredibly frustrating to me, why can't this be done like I have always done algebra?
Where did you learn math?
Anything you add or subtract on one side you have to do to the other side too, you make one side 0 because it makes everything easier. So with -1= 2x^2 -x -1 you just add 1 to each side making it 0=2x^2 -x (as you can see it remains the same thing)
2x^2 - x = x(2x-1)= 0 then (just a rewrite to make it simpler to find)
now obviously that means that x=0 is one of the points where f(x)=-1 because 0(2*0-1) is still 0. give that you know it's a parabolic function (woot x^2) you then have to find the other x.
you do this by finding out for what value of x (2x-1)=0 holds true. This is x=1/2.
you got x=0 and x=1/2 as the solutions. If you fill them in the original equation you'll see that they produce the desired result of f(x)=-1 (you should always fill them in the original equation to check your answer)
x can be anything. If the equation gives you 0 = 0 then that's fine. It's a true thing.
yes 0=0
I thought x= 1/2 ?
I learned math in the American Public Education system, maybe that is my problem.
Seriously though I passed math in high school (got Bs) by sleeping in class every day and never doing a second of homework, ever.
Do you understand how we got to x * (2x - 1) = 0?
The logical leap here is that if you have two terms and their product is zero then one of those terms must be zero. Does that make sense?
So either x = 0 or 2x - 1 = 0.
2x - 1 = 0 would mean that x is 1/2, x = 0 would mean that x = 0, obviously.
So there are two solutions.
I mean can you always break down an equation? Or only if it equals zero there is some special rule that says you can?
Lemming posted a graph of the function. It's a parabolic function so it hits f(x)=-1 TWO times.
I can assure you that there is both logic and a system here. If it seems not to be so you're missing something.
We can probably help you figure out what you're missing, but we will need for you to be very specific about why this doesn't make sense. From my point of view what's happening is that we're looking for a number that makes the statement
-1 = 2x^2-x-1
true. In solving the equation we find that there are two such numbers. Plugging either into the equation in place of x results in a number being equal to itself which simply means that the statement is true. What part of this conflicts with your understanding of the situation?
0431-6094-6446-7088
Only when it's zero. Think about x * y = 0
How could that be possible? Either x = 0, y =0, or both. If we replace y with (2x -1) we get your current problem.
As for a general rule: there can potentially be as many solutions to an equation as the highest powered exponent in the equation. There don't have to be (easy example of where there aren't: x^2 = 0, which has only the solution x=0), but that's the maximum. So with 2x^2 - x = 0, there can be two.
Technically there is _always_ a number of solutions equal to the highest degree of the variable; it's just that sometimes they are duplicates or not real.
0431-6094-6446-7088
I am changing my major asap.
what do you mean two seperate equations?
For example, the very next problem.
1= 2x^2/ (x^4+1)
basically breaks down into x^4 - 2x - 1 = 0
Ok, I can sort of see that, now..
(x^2 - 1)^2 = 0
MIND FUCK
Where the fuck did that even come from? Pretty sure I need to drop out of this class, we haven't even learned factoring yet. Guess factoring is so easy that is just something I was suppose to pick up in the first class where we covered number lines.
I am not longer in the mood for math (bullshit) homework anymore, I need to take a break.
oh well it's rewriting the equation to make it simpler. in this case it's rather easy as somethingsomething -1= -1 obviously has to mean that somethingsomething=0
Seriously, is math suppose to be this hard? I should probably just skip class and spend the time on the internet or with a personal tutor who actually teaches.
Math is supposed to be way, way harder. I do recommend you find a tutor since I believe that while math can be taught to anyone the standard classical way is the worst way to go about it for a lot of people. Just find someone good at math willing to spend a few hours with you to get with the basics.
Factoring is fairly simple:
Step 1 is always check if there's a common term. Pull out any x or x^2 terms common to everything. So with 2x^2 - x in your initial problem, we can pull out an x to get x * (2x - 1)
Step 2 is then look at what you've got left. In this latest problem we've got x^4 - 2x^2 + 1 = 0
This is basically quadratic (of the form ax^2 + bx + c), so if things are convenient we should be able to factor into two terms. So there's no constant in front of the x^4 term so things will obviously look like:
(x^2 )(x^2 ) = 0
and we need the constant terms to multiply together to get 1, right? So they're either 1 and 1 or - 1 and - 1. Then we ask ourselves which of those combinations add together to give us -2? Think about FOIL if that doesn't quite make sense. Anyway, obviously it's -1 and -1.
So we get:
(x^2 - 1) (x ^2 - 1) = 0
Or:
(x^2 -1)^2 = 0
Then! We should probably check to see if x^2 - 1 factors itself.
It's possibly easier to see how this works if we re-write as x^2 + 0x - 1
Again our only option for the x terms are 1 so:
(x ) (x )
And we need the two constant terms multiplied together to be -1 and added together to be 0. Our only options here are 1 and -1.
So we get:
[(x - 1) * (x + 1)]^2 = 0
In this case, what are the solutions?
Let's go to a slightly more complicated version to illustrate a more generalized approach:
Suppose you were asked to solve:
x ^ 3 - x^2 - 6x = 0
So first we go to step one and see that there's a common factor of x in all of those terms:
x (x ^2 - x - 6) = 0
Now, we still don't have a coefficient for the x ^2 term so it's going to look like:
x (x ) (x ) = 0
Additionally we should note that's a MINUS 6, so the signs in the two factored terms have to be opposites:
x (x - ) (x + ) = 0
Lastly those two number's product has to be six and their sum has to be -1. Our options for the former are (1, -6) (6, -1) (3, -2) (2, -3). The only one of those four with a sum of -1 is the last, so we finally get:
x (x - 3) (x + 2) = 0
That's the factoring bit. Once you get more experience doing this you'll see the solutions without having to make any lists. And then what solutions would we get to this problem, and somewhat more vitally how do we know?
Generally math instruction is heavily teacher dependent and math profs at the college level aren't particularly interested in teaching algebra in an effective way. Or for that matter, calculus.
It only seems complicated because you haven't done it before. Stick with it and these will seem as simple as the linear equations you're more familiar with.
It is worth noting that, yes, math is generally _much_ more complex than the linear equations you mentioned; this is because almost nothing in the real world can be described with a linear equation. We start with linear approximations because they're comparatively simple, but in order to apply math to real world situations you'll need polynomial, exponential, trigonometric and, God help us, logarithmic functions.
The ideas become more complex, but the rules stay the same. If you know the four basic math operations you can work out pretty much everything else. Get a good grounding in how to solve equations. No shortcuts; no "I know the answer, I can just write it down and go on." Write out every step in painstaking detail. If you do not do this with the simpler problems you will be hopelessly lost with the more complicated ones.
The problems become more complex, but unless you're actually researching new topics in math they're only complex in the sense that they're combinations of problems you've dealt with before. The trick is always in breaking the problem down into manageable pieces and working on one piece at a time.
If you don't read anything else I've written read this: Talk to your teacher. If your teacher cannot explain it to your satisfaction then find a tutor. Trying to read a math text book is generally an exercise in futility. I am saying this as a math teacher. If you can learn to read them, that's wonderful and you will go far. But I've never read one that was better than mediocre at explaining things to a beginner. For whatever reason people who are both good at math and good at explaining things just do not write text books. Find someone who knows the material and can explain things to you in terms you can understand.
0431-6094-6446-7088
It's the other way around. A function f(x) only has one value for a given x, but different values of x can yield the same value of f(x).
The problem you have initially posted has nothing to do with complex factoring. It's just plain ol' real factoring.
What major are you? What level math did you go up to in high school and what level math are you in now? Maybe you placed higher than you could handle to start at? Generally courses in college don't hold your hand through every little step of progress, especially in math and science, with a fair amount of reading on your own and extrapolation of current knowledge to solve new problems.
Side note: It's not a failure of the American public math curriculum as a whole, though I can't speak for the particular individual you had as an instructor back then.
Fizban, I realize you're trying really really hard to get this but you post a lot of questions about math that show you might have a poor understanding of basic operations, like factoring
In the grand scheme of things, no algebra is not hard. You in particular may have a learning style that isn't compatible with how your teacher presents things, or how quickly he/she moves through material. As I've suggested to you before, you should probably look into paying for a tutor to help you identify why exactly you're having such difficulty with the material, and how you can learn new techniques to get through the class.
H/A is all well and good if you're stuck on a problem once in a while, but this seems to be a regularly frustrating thing for you
Consider the equation
X^2 = 1
In words it means "x times itself equals 1"
We can then ask what x is. Because the only constraint is that "x times itself equals 1" x can be equal to any number that multiplies by itself to equal 1.
In fact there are two numbers that x can equal. They are 1 and -1. Both 1*1 = 1 and -1*-1 = 1. We have no way to prefer one over the other and they both fulfill the requirements we set for x.
The thing to keep in mind about math is that it is a system of rules. As long as you follow the extremely general rules everything you do is valid.
For the question you were given the equation was rearranged into the form 0= x * (2*x -1) cause that's the easiest way to find the solution to polynomial equations's (i.e. solve for x when some x's are put to powers).If x is anything but 0 or 1/2 the equation is false. We can know this because if two numbers are multiplied together, and neither are 0, then the result cannot be zero. Let's consider A*B = 0 as having 'a' groups of 'b' objects (e.g. 3 baskets or 4 apples, 3 debts of -15 dollars). You can't have 'a' groups of 'b' objects and then multiply to have 0 objects. At the same time if you either you have 0 groups or 0 objects in each group you necessarily have 0 objects.
Thus for 0 = x * (2*x-1) to be true you know that either "a" (if we say a is x) = 0 or "b" (if we say b is 2*x - 1) = 0
And because we know 0 = x * (2*x-1) is true
then x = 0 or x = 2*x-1
Ya, factoring takes time to get used to. It sounds like you kinda got fucked over by your highschool. If you multiply out (x^2 - 1) times itself you'll get their equation. If you do not know how to do that google multiplying polynomials. Factoring (which is that process in reverse) is something you really need to practice a lot before you can do it easily. You just need to get used to the rules of the game.
There are people here saying algebra isn't hard. That's half true. It isn't hard once you know how to do it. But it's a bitch to learn. You remember how hard it was to learn to read? Algebra is kind of like that.