Question

A linear function is represented in the table below
X|-2|-1|0|1|2|
Y|-3|-1|1|3|5|

Which statement describes why this function is linear?

Answer 1

F.The x values increase by a constant amount.
G.The y values increase by a constant amount.
H.The y values increase by 2 as the x values increase by 1.
J.The y values increase by 2 as the x values decrease by 1.

Answer 2

@whpalmer4

Answer 3

I think it is H but can you just explain it to me why it is H?

Answer 4

Did you make a graph in your investigation?

Answer 5

Well, no. It helps?

Answer 6

If you look at the graph, or calculate the slope between any two points given, you'll see that the slope is indeed \(2\):

\[m = \frac{y_2-y_1}{x_2-x_1} = \frac{5-(-3)}{2-(-2)} = \frac{8}{4} = 2\]

A quick review of slope-intercept form: \(y = mx+b\) shows that we should expect \(y\) to change by \(m\) for every 1 step (positive) of \(x\), because \[y_1 = m(x+1) + b) = mx + m + b\]and\[y = mx+b\]so \[y_1 - y = (mx+m+b) - (mx+b) = m\]

Answer 7

Or if you look at point-slope form, \[y-y_1= m(x-x_1)\]perhaps that makes for a clearer demonstration. The change in \(x\), multiplied by the slope, gives you the change in \(y\).

Answer 8

Ok, lemme look through what you just typed up...

Answer 9

okay, holler if you have more questions...

Answer 10

Whoa... thats a bunch of formulas you keep in your noggin

Answer 11

and that wasn't even all of them ;-)

But a little practice will show you that they really all are the same thing, just different arrangements.

Answer 12

Is what you gave me the definition of a linear function?

Answer 13

no, I wouldn't necessarily say that, although it is consistent with one:

a linear function is a mathematical function in which the variables appear only in the first degree, are multiplied by constants, and are combined only by addition and subtraction

Answer 14

So H is the correct answer right?

Answer 15

I am just trying to get this linear function thing straight :)

Answer 16

Oh, yeah, no question of that! It should be clear from the graph and the theoretical discussion that increasing \(x\) by 1 increases \(y\) by 2, no matter where on the graph you do it, right?

Answer 17

And similarly, decreasing \(x\) by 1 decreases \(y\) by 2.

Answer 18

Wooooooooooohooooooooooooooooo, your like the funnest person on OS that knows his stuff

Answer 19

Thanks @whpalmer4

Answer 20

So, if I got that table of data, and someone said "find the function", the first thing I would do would be to take the first two data points, calculate the slope, then see if the rest of the data points fell on the line that slope and the first two points defined. If they do, great, it's linear, I've already got the equation, and I'm done. If not, then I would probably plot the first three points and start figuring out if it might be a parabola, hyperbola, exponential, log, inverse variation, etc.

Answer 21

If I'm fun, it's probably because I think a lot of this is fun! Of course, it's only fun when you aren't beating your head against your desk because you don't understand :-)

Answer 22

I understood the first part but I lost you at "If not, then I would probably plot the first three points and start figuring out if it might be a parabola, hyperbola, exponential, log, inverse variation, etc."

Answer 23

hahaha thanks

Answer 24

I don't what your background is — have you done formulas like \(y = ax^2+bx+c\) yet? That would be a parabola, such as might describe the flight of a ball through in the air. A hyperbola such as \(xy = k\) would describe indirect variation, such as the relationship between temperature and pressure of an ideal gas in a fixed volume. An exponential such as \(y = 2^x\) could represent bacterial growth, or \(y = 2^{-t/k}\) could represent radioactive decay. There are a handful of forms (the ones that I mentioned) which represent quite a few different physical phenomenon as a decent first approximation (if you get rid of pesky things like air resistance for the ball in flight, etc.)

While it is fun (for me, at least) to tinker with math for tinkering's sake, it's also nice to be able to use it to better understand the real world now and then!

Answer 25

(correction, that should be "the ones that I mentioned among them")