How can i find what "c" is in one of these equations if i know the points on the graph?

the "c" equations (just need to know one "c" for any one of the equations, doens't matter):





y=cx


y=cx^2


y=c square root of absolute x


y= c/x





the points are (-4,-1) (-1, -.25) (0,0) (1, .25) (4,1) ???

How can i find what "c" is in one of these equations if i know the points on the graph?
e.g.





points (x, y) = (-4 ,-1) so first equation





-1 =c*(-4) = -4c





c = 1/4





2) y = cx^2 implies





-25 = c* (-1)^2 = c*1 = c





so c = 25


--------------------------------------...





P.S.





ok, i get your point, if all these points are in a graph, then we can find the 'c' which is the slope (gradient of that graph)





i.e y = mx + 'k' , where 'm = c' in this case is usually the slope and 'k' is a constant





so pick any two points (x1, y1) and (x2, y2)





c = (y2 - y1 )/ (x2 - x1)





so if we pick the first two points,





c = (-25 - (-1) )/ (-1 - (-4))





= (-25 + 1) / (-1 +4)





= -24/3 = -8





so your 'c' = -8 iff these two particular points are in a graph.





eqn of the line is





y - y1 = c (x - x1), for give points (x1, y1)





pick any of the pair of points we have used to get 'c'





say (-4, -1) = (x1, y1)





y = cx + k





to find 'k' take (-4 , -1) = (x , y)





so (-1) = (-8)*(-4) + k





hence 'k' = -33





thus





equation of the graph passing through these two points is





y = - 8*x - 33





Clearly not one of the equations you have there ... so try another paring, let me think of a shorter way to get around the problem, mean while i hope you get the idea ...
Reply:c = (-3,7)