Beginner's linear algebra question

[lezardo]

Hi everyone! I recently started taking a basic linear algebra course at my university and I'm already having problems. I used to be pretty good at math in general, but I guess I've unlearned everything over the years. Any help would be appreciated, thanks!

A.
Two points on a plane with the coordinates (x0,y0) and (x1,y1) define a graph with the form:
y = c1x + c0 (polynomial of the 1. degree).

1)Formulate a system of linear equations to find the coefficients of this function.
2) Solve this system of equations to find the function which runs through the points (1;3) and (2;4).


B.
n + 1 points on a plane [(x0, y0), (x1, y1),…; (xn, yn)] define a polynomial of the nth degree:
y = cnx^n + …+ c2x^2 + c1x + c0.

1)Formulate a system of linear equations to find the coefficients of this polynomial.
2) Solve this system to find the polynomial which runs through the points (0;1), (1;10), (2,33) and (3,64).


P.S. Hints or something like that would also be great. It's really frustrating sitting here with nothing but "duhhhhh" in my head.

 

[0rthodontist]

Do you understand the question? It's asking you to find the equation of the polynomial whose graph passes through those points. Here is a hint on part A. The equation you are trying to find is y = c1x + c0 (you are trying to find the c's). You know that the point (x0, y0) satisfies this equation and so does (x1, y1). So you can write, for example
y0 = c1x0 + c0
where y0 and x0 are KNOWN and c1 and c0 are UNKNOWN.
Can you figure it out now?

 

[lezardo]

Do you understand the question? It's asking you to find the equation of the polynomial whose graph passes through those points. Here is a hint on part A. The equation you are trying to find is y = c1x + c0 (you are trying to find the c's). You know that the point (x0, y0) satisfies this equation and so does (x1, y1). So you can write, for example
y0 = c1x0 + c0
where y0 and x0 are KNOWN and c1 and c0 are UNKNOWN.
Can you figure it out now?

Well, i feel stupid. It was staring me right in the face. Thanks a lot!