What is the Cubic Function that Passes Through Given Points?

[Kstan333]

Homework Statement

I have a line of a cubic function that passes through (-10,4), (-2,3), (-1,2), (0,1), (7,0). I must find the function of this line.

Homework Equations

The Attempt at a Solution

I know you can't see this but the line has a "S" curve to it. Making me assume that it is a cubic funtion.

 

[jedishrfu]

have you tried graphing it?

 

[Kstan333]

I need to find the function, using those points, that's all that is given.

 

[jedishrfu]

I don't have an idea of how to solve it but if you look at just the y values what do you see?

Are you sure you aren't simply to find the line that best fits the points?

 

[jedishrfu]

try starting with an equation of the form ax^4 + bx^3 + c^x2 + dx +e = y and generate a set of 5 equations in five unknowns using the the five points one by one.

So start with (0,1) and you get simply e=1 now proceed to find the other constants.

 

[HallsofIvy]

Notice that jedishrfu has a fourth degree polynomial, not a cubic. That is because a cubic, $y= ax^3+ bx^2+ cx+ d$ has four coefficients which take four equations to determine. So there exist a unique cubic through any four given points. You give five points so there may not be such as cubic. Of course, if those points do lie on a cubic, you will find that the coefficient of $x^4$ is 0.

Another way to find the polynomial (though I really prefer jedishrfu's method) is the "Lagrange polynomial":
$$y(x)= 4\frac{(x+2)(x+1)(x)(x- 7)}{(-10+ 2)(-10+ 1)(-10)(-10- 7)}+ 3\frac{(x- 4)(x+ 1)(x)(x- 7)}{(-2+ 10)(-2+ 1)(-2)(-2- 7)}$$
$$+ 2\frac{(x+ 10)(x+ 2)(x)(x- 7)}{(-1+ 10)(-1+2)(-1)(-1-7)}+ 1\frac{(x+10)(x+2)(x+1)(x- 7)}{(0+10)(0+2)(0+1)(0-7)}+ 0\frac{(x+10)(x+2)(x+1)(x)}{(7+10)(7+2)(7+1)(7- 0)}$$

Do you see the idea? Each fraction has factors in the numerator of "x- " each x value except one. And the denominator has factors with that missing x value minus each of the other x values. If x is anyone of the given x values, every fraction except one will be 0 and then the fraction will be 1 so that we just have y value that was in front. And, of course, because there were five points, each fraction has 4 factors involving x and so this will, in general, be a fourth degree polynomial.

 

[symbolipoint]

Ignoring the point, (0,1), the 4 x 5 matrix can be made:
[-1000, 100, -10, 1, 4]
[-8, 4, -2, 1, 3]
[-1, 1, -1, 1, 2]
[343, 49, 7, 1, 0]

Using the online software www.math.purdue.edu/~dvb/matrix.html, the reduced row form is:
1, 0, 0, 0, 0.0106
0, 1, 0, 0, 0.0408
0, 0, 1, 0, -.9518
0, 0, 0, 1, 1.018

The apparent coefficients give an equation that can be run through google and graphed. The result is interesting. Just type in, "graph 0.0106x^3+0.0408x^2-0.9518x+1.018" and see the resulting graph. As you inspect the points, you see a near, not perfect, fit to the five original given points and there are three roots. They are nearly x=-12, x=1, and x=7. My efforts this way may have a mistake somewhere since those roots as binomial factors do not give the same or similar coefficient for the cubic term. Maybe this is off only by a constant value factor?