y = kx?
Substitute the x and y value, then take the log of both sides.
log y = k*?*log x
I know that were doing problems were there is a soultion given which is equal to X^? times a constant ...
y = k x^?
So this is problem I have to decide given a probelm x relationship to y (for example iverse square cube and so forth) this unkown variable is denoted witha (?)
So given two points (Q,W) (E,R) how do I determiine the relationsip of x to y... and solve for the (?)
I guess so far in my book the constant is raised to the first power is there any way I could find out if the constant is not raised to first power or not???
Once I come up with the formula I know how to do the rest I just don't know how to obtain the formula and solve for the (?) once you do it's just plug and chug...
so please show me how to solve for (?) given two points
5 answers
can you show me how to do this with the points
(2, 5)
(6, 80)
(2, 5)
(6, 80)
Two points determine a line. Given two points (Q,W) and (E,R), you can find the equation of the unique line, y = m x + b that goes through those two points.
Three points determine a parabola, y = a x^2 + b x + c
Four points determine a cubic, y = a x^3 + b x^2 + c x + d
etc
Now
given (Q,W) and (E,R)
slope = m = (R-W)/(E-Q)
so now you have m
go back and put first point in to get b from b = y - m x = W - m Q
Three points determine a parabola, y = a x^2 + b x + c
Four points determine a cubic, y = a x^3 + b x^2 + c x + d
etc
Now
given (Q,W) and (E,R)
slope = m = (R-W)/(E-Q)
so now you have m
go back and put first point in to get b from b = y - m x = W - m Q
ok
m = (80-5) / (6-2)
m = 75/4
then
b = 5 - (75/4)2
b = 5 - 75/2
b = (10-75)/2
b = -65/2
so
y = (75/4)x -65/2
or
4 y = 75 x - 130
m = (80-5) / (6-2)
m = 75/4
then
b = 5 - (75/4)2
b = 5 - 75/2
b = (10-75)/2
b = -65/2
so
y = (75/4)x -65/2
or
4 y = 75 x - 130
I did not understand that your question related only to direct variation until I saw Dr. Bob's answer.
1 answer