totaleclipse2 said:I'm completely lost on how to do this. I don't even have an idea as to where to start. Our teacher decided to not teach us this, and then sadly give it to us for homework. I've looked in my book and on the internet, but I can't seem to find anything that's comprehensible for me. If anyone could help it'd be greatly appreciated.
totaleclipse2 said:Find the roots and set those equal to 0 and solve, place the points on the graph. I think I may have figured it out. Is it that you take the sum of the coefficients, and that is a factor. Then you use that factor to synthetically divide the cubic, to get it into the quadratic. Afterwards, you solve the quadratic by either factoring it our or with the quadratic formula, giving you the remaining two roots? If that is right, how do you determine the shape of the graph with those points? Doesn't the leading coefficient determine the number of turns?
JeffM said:totaleclipse2 said:Find the roots and set those equal to 0 and solve, place the points on the graph. I think I may have figured it out. Is it that you take the sum of the coefficients, and that is a factor. Then you use that factor to synthetically divide the cubic, to get it into the quadratic. Afterwards, you solve the quadratic by either factoring it our or with the quadratic formula, giving you the remaining two roots? If that is right, how do you determine the shape of the graph with those points? Doesn't the leading coefficient determine the number of turns?
Whoa.
f(x) = ax[sup:3p0ib92i]n[/sup:3p0ib92i] + bx[sup:3p0ib92i](n-1)[/sup:3p0ib92i] + ... z.
The leading coefficient of a polynomial in conjunction with the leading power tells you about the sign of the slope and the sign of the value of the polynomial far from zero. The reason is that, when the absolute value of x is large, the term in the largest power of x dominates all the rest of the terms. If |x| >> 0, then [f(x) / ax[sup:3p0ib92i]n[/sup:3p0ib92i]] is approximately 1 so f(x) looks very similar to ax[sup:3p0ib92i]n[/sup:3p0ib92i].
The sum of the coefficients of a cubic is neither a factor nor a root of the cubic.
Example: the roots of x[sup:3p0ib92i]3[/sup:3p0ib92i] - 7x - 6 are - 2, - 1, and 3.
1 - 7 = - 6, which is not a root.
1 - 7 - 6 = - 12, which is not a root.
The MAXIMUM number of turns is (n - 1), 1 less than the highest power in the polynomial. There may be fewer.
If the highest power is odd, 1 <= the number of roots <= n, and 0 <= the number of turns <= (n - 1).
If the highest power is even, 0 <= the number of roots 0 <= n <= n, and 1<= the number of turns <= (n - 1).
Subhotosh Khan said:What level of math you are doing? The answer will depend on that
This equation has one real root - and that too is irrational.
If you know calculus and Newton's method of approximation - then you can use that.
If not - then simply make a table for y values within the domain of -5 to +5 (in x).
Then you refine it as needed.
galactus said:The thing to do is hone in on that real root by trial and error.
You are looking for the values of x that make the cubic equal 0.
So, make some guesses using a calculator.
Try x=-2. This gives -1
Try 0. This gives 5
Since the solution changed from negative to positive, the solution must lie between -2 and 0.
Start honing in on it.
Try x=-1 and get 8. Still positive, so the solution is between -2 and -1.
Keep going like that and see if you can find the root.
This is a rather sloppy way to go about it, but effective.
Try graphing it and checking where it crosses the x-axis. That will give you an idea what you're looking for.
I told you, when the highest power in a polynomial is n, there are at MOST n DISTINCT, REAL roots, but there may be fewer. So a cubic can have at most 3 real roots. You do not have to look all over. Furthermore, if n is odd, there is always at LEAST 1 real root. So, you know you are not looking in vain for a real root to a cubic.totaleclipse2 said:galactus said:The thing to do is hone in on that real root by trial and error.
You are looking for the values of x that make the cubic equal 0.
So, make some guesses using a calculator.
Try x=-2. This gives -1
Try 0. This gives 5
Since the solution changed from negative to positive, the solution must lie between -2 and 0.
Start honing in on it.
Try x=-1 and get 8. Still positive, so the solution is between -2 and -1.
Keep going like that and see if you can find the root.
This is a rather sloppy way to go about it, but effective.
Try graphing it and checking where it crosses the x-axis. That will give you an idea what you're looking for.
So I'm looking for all x values that make y=0, how do I know how many x values are that accomplish that? Obviously I can't look at every number in existence. Also, I can't use a graphing calculator, all has to be by hand. After I find the roots, how can I figure out how it is graphed with only one root?
Once you have roots and where the graph cuts the y axis, you have some concrete points for your graph.totaleclipse2 said:After I find the roots, how can I figure out how it is graphed with only one root?
JeffM said:I told you, when the highest power in a polynomial is n, there are at MOST n DISTINCT, REAL roots, but there may be fewer. So a cubic can have at most 3 real roots. You do not have to look all over. Furthermore, if n is odd, there is always at LEAST 1 real root. So, you know you are not looking in vain for a real root to a cubic.totaleclipse2 said:galactus said:The thing to do is hone in on that real root by trial and error.
You are looking for the values of x that make the cubic equal 0.
So, make some guesses using a calculator.
Try x=-2. This gives -1
Try 0. This gives 5
Since the solution changed from negative to positive, the solution must lie between -2 and 0.
Start honing in on it.
Try x=-1 and get 8. Still positive, so the solution is between -2 and -1.
Keep going like that and see if you can find the root.
This is a rather sloppy way to go about it, but effective.
Try graphing it and checking where it crosses the x-axis. That will give you an idea what you're looking for.
So I'm looking for all x values that make y=0, how do I know how many x values are that accomplish that? Obviously I can't look at every number in existence. Also, I can't use a graphing calculator, all has to be by hand. After I find the roots, how can I figure out how it is graphed with only one root?
Here is a trick to know about polynomials. It is always true that ax[sup:3mow3vmt]n[/sup:3mow3vmt] + bx[sup:3mow3vmt]n-1[/sup:3mow3vmt] + ... z has n roots at r[sub:3mow3vmt]1[/sub:3mow3vmt], r[sub:3mow3vmt]2[/sub:3mow3vmt], ... r[sub:3mow3vmt]n[/sub:3mow3vmt], but some of the roots may not be real and some may be duplicates. And always
ax[sup:3mow3vmt]n[/sup:3mow3vmt] + bx[sup:3mow3vmt]n-1[/sup:3mow3vmt] + ... z = a(x - r[sub:3mow3vmt]1[/sub:3mow3vmt])(x - r[sub:3mow3vmt]2[/sub:3mow3vmt]) ... (x - r[sub:3mow3vmt]n[/sub:3mow3vmt]). With a cubic, if you can find one real root that is APPROXIMATELY q, you can approximate the cubic with (x - q) times a quadratic. Then you can use the quadratic formula to see whether there are any real roots to that quadratic and ,if so, what they are approximately. With a cubic you KNOW that there is at least ONE real root (because 3 is odd). The trick is to find one, at least approximately. Then you can find if any others exist and if so what they are, at least approximately.
totaleclipse2 said:Okay, so for the problem 2x[sup:1wmkuo92]3[/sup:1wmkuo92] - 4x + 5, there is at least one root, maximum of three. Yes, at least 1 real root, at most 3 distinct, real roots.
The graph has opposite end behavior, Not quite sure what you mean. The values of the function at positive x far from zero and at negative x far from zero will have opposite signs, but the signs of the slopes at those distant points will be the same.
and 2 turns because of the leading coefficient, A cubic may have 2 turns but the leading coefficient has nothing to do with it. If n is the highest power to which x is raised, then (n - 1) is the MAXIMUM number of possible turns. There may be fewer.
and it rises right because of a positive leading coefficient. Yes, provided you are far enough from zero.
So now that I know this, I know what the graph looks like, but I'm still at a loss to where is goes. Calculate the value of the function at x = 0, at each approximate point of turn, and at any positive x more than your greatest root and greatest point of turn, and at any negative less than your least root and least point of turn. You know the value of the polynomial at the roots is zero, and you know the roots approximately. You know the approximate value of the points of turn and can calculate y there. You know the sign of the slopes far from zero. So, you have the exact value of y at 3 points and approximate values at one to five other points plus the sign of the slopes far from zero. Draw a smooth curve that takes ALL that info into account. You will have a very decent graph. You can justify your approximations by the method of bisection. I'm sorry I'm not comprehending this, I'm just having a rough time. I tried to sub in values -5 to 5 for x to try to get 0, but it's somewhere in between -1 and -2, do I need to find the exact value, or can I use an approximation, as you stated? Well in this case the root is irrational so an approximation is the best that anyone can do.
JeffM said:totaleclipse2 said:Okay, so for the problem 2x[sup:3u318zii]3[/sup:3u318zii] - 4x + 5, there is at least one root, maximum of three. Yes, at least 1 real root, at most 3 distinct, real roots.
The graph has opposite end behavior, Not quite sure what you mean. The values of the function at positive x far from zero and at negative x far from zero will have opposite signs, but the signs of the slopes at those distant points will be the same.
and 2 turns because of the leading coefficient, A cubic may have 2 turns but the leading coefficient has nothing to do with it. If n is the highest power to which x is raised, then (n - 1) is the MAXIMUM number of possible turns. There may be fewer.
and it rises right because of a positive leading coefficient. Yes, provided you are far enough from zero.
So now that I know this, I know what the graph looks like, but I'm still at a loss to where is goes. Calculate the value of the function at x = 0, at each approximate point of turn, and at any positive x more than your greatest root and greatest point of turn, and at any negative less than your least root and least point of turn. You know the value of the polynomial at the roots is zero, and you know the roots approximately. You know the approximate value of the points of turn and can calculate y there. You know the sign of the slopes far from zero. So, you have the exact value of y at 3 points and approximate values at one to five other points plus the sign of the slopes far from zero. Draw a smooth curve that takes ALL that info into account. You will have a very decent graph. You can justify your approximations by the method of bisection. I'm sorry I'm not comprehending this, I'm just having a rough time. I tried to sub in values -5 to 5 for x to try to get 0, but it's somewhere in between -1 and -2, do I need to find the exact value, or can I use an approximation, as you stated? Well in this case the root is irrational so an approximation is the best that anyone can do.
totaleclipse2 said:I'm completely lost on how to do this. I don't even have an idea as to where to start. Our teacher decided to not teach us this, and then sadly give it to us for homework. I've looked in my book and on the internet, but I can't seem to find anything that's comprehensible for me. If anyone could help it'd be greatly appreciated.