Do both x and y values have to have a constant rate of change to be a linear function?
define a linear function, it has different meanings in different contexts
a linear function is either a line equation ... or a function such that f(a+b) = f(a)+f(b), and f(kx) = k f(x) for some constant k
@PeachRings So they both have to have constant rate of change right?
@amistre64 I don't know what you're talking about. lol I mean for linear functions in tables.
a linear function has a common different, a constant slope if we are looking at table values
so yeah, x and y would have a constant rate of change.
0 is still a rate of change :)
But for y values, they have to all have the same amount of change right?
i know how im reading 'amount of change' but im not sure how your reading it. lets say y=1,3,4,7,12 this of course does not appear like a constant rate of change, but that all depends on the x values that relate to them.
if given a table, its points form a line if the slope between all the stated points is the same.
So if the x and y values are data and the y values are not consistent in amounts but they are with a x value it's linear. Sorry for all the questions.
@amistre64
do you have a table to work from? or is this just a mental practice?
lets assume we have a line y=mx+b and see if we cant get a y/x = setup y-b = mx (y-b)/x = m is constant ... if b does not equal zero then we can prolly assess it from table values
\[\frac{y_o-b}{x_o}=\frac{y_1-b}{x_1}\] \[\frac{y_o}{x_o}-\frac{b}{x_o}=\frac{y_1}{x_1}-\frac{b}{x_1}\] \[\frac{y_o}{x_o}-\frac{y_1}{x_1}=\frac{b}{x_o}-\frac{b}{x_1}\] \[\frac{y_o}{x_o}-\frac{y_1}{x_1}=b\left(\frac{1}{x_o}-\frac{1}{x_1}\right)\] \[\frac{y_o~x_1-y_1x_o}{x_o~x_1}=b\left(\frac{x_1-x_o}{x_o~x_1}\right)\] \[\frac{y_o~x_1-y_1x_o}{x_1-x_o}=b\] so, if given a set of points, then we have some work to assess if they form a line if this condition holds for some random points on a line
I got it now. Thanks for all your help and great effort :)
youre welcome
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