Understanding the Concept of "y=mx+b

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The discussion centers on understanding the equation y = mx + b, which represents a linear relationship where m is the slope (or steepness) of the line and b is the y-intercept, indicating where the line crosses the y-axis. Participants explain that the equation defines a straight line, with variations leading to different shapes like parabolas when additional terms are included. The reasoning behind the equation is that it captures the relationship between x and y in a linear form, where any straight line can be expressed as such. Additionally, the conversation touches on solving systems of equations and the implications of different outcomes, such as "no solution" or "infinite solutions," which relate to the slopes and y-intercepts of the lines involved. Overall, the thread emphasizes the foundational concepts of linear equations and their graphical interpretations.
  • #31
Okay...

What I meant was, the line the slope was describing cannot go into the 2nd quadrant.
true?
 
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  • #32
Sure it can.

x = -5, y = 21.4

(-5,21.4) is in the second quadrant.

Here's a question, though. Can this line ever go into the third quadrant?

cookiemonster
 
  • #33
okay, I got the graph right. I sort of cheated though.
Thank you cookiemonster.
 
  • #34
cookiemonster said:
Sure it can.

x = -5, y = 21.4

(-5,21.4) is in the second quadrant.

Here's a question, though. Can this line ever go into the third quadrant?

cookiemonster

according to my graph, it can only go into the 4th quad. Did you graph it? I could be wrong u know.
 
  • #35
Now I'm doing the addition method. (The last part of the poster)

x+y=26.4
5y=x

Should you isolate the y intercept in 5y=x into the form y-mx=b?
 
  • #36
how do you edit posts? When I press "edit" it asks me why I want to delete the post.

About my recently aforementioned inquiry (on my last post), if you did infact convert 5y=x into the form y-mx=b, you'd get 5y-x=0. When you add it with x + y=26.4,

x+y=26.4
+ -x+5y=0 the x's cancel out. so I guess it okay to convert it.
-------------
0 + 6y=26.4
y= 4.4

YES! It correct right? (the way I did it?)
 
  • #37
Correct.

And the line y = -x + 26.4 only passes through the first, second, and fourth quadrants. This brings up an interesting question. What is the maximum and minimum number of quadrants a line can pass through?

And if you want to edit, just hit the edit button and type the changes into the box below the Delete this Message box.

cookiemonster
 
  • #38
oh, you come up with marvelous questions.

The maximum number (I would guess) is 3. This can only be true for a straight line.
The minimum number, (I would guess) is 1. This can only be true for all forms of lines, not neccesarily linear.

Oh boy, was that conjecture even close to being right?
 
  • #39
Well, once you bring non-straight lines into it, everything changes and you can get functions that are in no quadrants and functions that are in all quadrants!

But what about for a straight line of the form y = mx + b? Don't consider the lines y = 0 and x = 0 (the coordinate axes), they just complicate things.

cookiemonster
 
  • #40
I still say 3.

Um, I'm doing some algebra homework and I need help:

"Sunshine Trucks Rentals rents a truck at a daily rate of 57.99 plus $.48 per mile. City Rentals rent the same size truck at $58.95 plus $.46 per mile. For what mileage is the cost the same?"

I know I have to set them to be equal, so here's what I've got:

57.99 + .48m = 58.95 + 46m


Is that the write equation?
 
Last edited:
  • #41
It's almost the "write" (oops!) equation. You're missing the decimal on the .46m.

And yes, a straight line can be in at most three quadrants. How about the minimum number?

cookiemonster
 
  • #42
a minimum number of what?
 
  • #43
What is the minimum number of quadrants of the plane a line can be in?
 
  • #44
3? :confused:
 
  • #45
Try a simpler line. =] It can be in fewer than three.

cookiemonster
 
  • #46
1?

:rolleyes:
 
  • #47
Now you're just picking numbers out of a hat. =\

What's the equation of a line that is in only one quadrant?

cookiemonster
 
  • #48
Lol!

dude!

All I know for a line is y=mx+b.
 
  • #49
C'mon, c'mon, let's think about what that means a bit!

Let's start with x. We know that x can (for a non-vertical line) take on any value between negative infinity and infinity, right? So that gives us three possible answers right there:

2 quadrants, the positive and negative x versions of either a positive y or a negative y. That means it can be in quadrants I and II or III and IV. Can you think of a line that's like this?

3 quadrants, like quadrants I, II, and III or quadrants I, II, and IV, etc. You already have an example of a line like this, right?

4 quadrants. We already know this can't happen.

So it's either 2 quadrants or 3. So can you find a line that's only in two quadrants?

Here's a hint, find a line that's in only quadrants I and II. What would the slope of such a line have to be?

cookiemonster
 
  • #50
OMG!

Doesn't the slope vary?

THE ANSWER IS 2!

hey cookiemonster, I'd like to hear (read) your comments on my stem cell research essay. I posted it in the thread "Morals" in General Philosophy.
 
  • #51
2's the right answer, whether by elimination or thinking it out, nobody cares? ;)

For reference, a vertical or horizontal line will be in only two quadrants. For example, if m = 0 and b = 1, we have the line

y = 1

which is only in the first and second quadrants.

Stem cell research, eh? I guess I can have a look...

cookiemonster
 
  • #52
I noticed that this thread has changed subject a couple times. There is no hard policy on this, but information will later be easier to find if we use new threads for new subjects.

My point is that, if you have some other, new question, feel free to open a new thread (unless, of course, it is closely related to this one).
 
  • #53
cookiemonster said:
For reference, a vertical or horizontal line will be in only two quadrants.

cookiemonster

But you never said a vertical/horizontal line; you said non-vertical and non horizontal. Lines can be diagonal too. But I guess it was my folly here; I didn't ask specifically.

thanks so much!

Sorry about that arhkron.
 
  • #54
Ah, sorry. I remember saying not to consider vertical lines because the equation to describe them doesn't work like you normally expect y = mx + b to.

Anyway, I mean that only vertical or horizontal lines will be in two quadrants. If a line is diagonal in the slightest, it will end up in three quadrants. Except for maybe lines that pass through the origin...

cookiemonster
 

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