Can someone explain this graph?

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King Friday

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How we get from the expression in the screenshot (I don't know yet how to use LaTex) To the graph shown?

to this graph:
 

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King Friday said:
How do we get from: -4/3 x

to this graph: View attachment 40015
Click to expand...
The graph represents the equation [imath]y=-\frac{4}{3}x[/imath]. For any x, you can calculate y, and plot the point [imath](x,y)[/imath].

For example, if [imath]x=0[/imath], then [imath]y=-\frac{4}{3}\cdot0=0[/imath], so you plot [imath](0,0)[/imath].

If [imath]x=3[/imath], then [imath]y=-\frac{4}{3}\cdot3=-4[/imath], so you plot [imath](3,-4)[/imath].

Then draw a straight line through those points; or plot more points to convince yourself they all lie on a line.
 
Dr.Peterson said:
The graph represents the equation [imath]y=-\frac{4}{3}x[/imath]. For any x, you can calculate y, and plot the point [imath](x,y)[/imath].

For example, if [imath]x=0[/imath], then [imath]y=-\frac{4}{3}\cdot0=0[/imath], so you plot [imath](0,0)[/imath].

If [imath]x=3[/imath], then [imath]y=-\frac{4}{3}\cdot3=-4[/imath], so you plot [imath](3,-4)[/imath].

Then draw a straight line through those points; or plot more points to convince yourself they all lie on a line.
Click to expand...
How do I assume that x = 0? I'm not seeing the rise/run in this.
 
blamocur said:
You don't. But you can see in the graph that [imath]y(0) = 0[/imath], and you can safely assume that the graph is a straight line. Finding another point (y(3) in the @Dr.Peterson's example) gives you enough info to figure out the equation.
Click to expand...
Oh. So if the b expression is absent, that is the same as it equaling Zero?
And that is where we plot our next two points from.
 
King Friday said:
Oh. So if the b expression is absent, that is the same as it equaling Zero?
Click to expand...
Not sure how to interpret this statement. But general equation for non-vertical lines is [imath]y=ax+b[/imath]. What do you get for [imath]y(0)[/imath] in the general case?
 
blamocur said:
Not sure how to interpret this statement. But general equation for non-vertical lines is [imath]y=ax+b[/imath]. What do you get for [imath]y(0)[/imath] in the general case?
Click to expand...
How do I know the b expression is Zero? I have no starting point to plot points from if all I have is [math]-\dfrac{4}{3}[/math]
 
King Friday said:
Oh. So if the b expression is absent, that is the same as it equaling Zero?
And that is where we plot our next two points from.
Click to expand...
King Friday said:
How do I know the b expression is Zero? I have no starting point to plot points from if all I have is [math]-\dfrac{4}{3}[/math]
Click to expand...
It would have been helpful if your initial question focused on your specific thoughts about the problem. It appears that you are not just asking about the equation in general, but about its relation to the "slope-intercept form", which in America is written as y = mx + b. Since your equation is not exactly written in this form, you wonder how to make it fit. Right?

To put [imath]y=-\frac{4}{3}x[/imath] in the form [imath]y=mx+b[/imath], we can observe that the one term you have, containing x, corresponds to mx (slope times x), and the b term (y-intercept) is absent. So we can write your equation as [math]y=-\frac{4}{3}x+0[/math]
This makes it clear that the slope (rise/run) is -4/3, so you can use a rise of -4 (fall of 4) with a run of 3. And the y-intercept is 0, meaning that the graph starts at (0,0).
 
Dr.Peterson said:
It would have been helpful if your initial question focused on your specific thoughts about the problem. It appears that you are not just asking about the equation in general, but about its relation to the "slope-intercept form", which in America is written as y = mx + b. Since your equation is not exactly written in this form, you wonder how to make it fit. Right?

To put [imath]y=-\frac{4}{3}x[/imath] in the form [imath]y=mx+b[/imath], we can observe that the one term you have, containing x, corresponds to mx (slope times x), and the b term (y-intercept) is absent. So we can write your equation as [math]y=-\frac{4}{3}x+0[/math]
This makes it clear that the slope (rise/run) is -4/3, so you can use a rise of -4 (fall of 4) with a run of 3. And the y-intercept is 0, meaning that the graph starts at (0,0).
Click to expand...
So, the absence of the y intercept is to be interpreted as "0" To state it would be redundant?
 
King Friday said:
So, the absence of the y intercept is to be interpreted as "0" To state it would be redundant?
Click to expand...
It's not that there isn't a y-intercept, but that there is no need to write a 0 term in an equation.

What we are doing here is rewriting any equation of a line in a particular form, from which we can infer the slope and intercept. If the equation as given has only an "mx" term, then you can write "+0" to make it explicitly look like "mx+b". Or, as we do with experience, just imagine it being there. The key idea is that the equation you're given will not always directly show these numbers; you have to make them visible by the way you rewrite it.
King Friday said:
My question exactly. It's needlessly confusing to a someone early in the path.
Click to expand...
Maybe this course is not as detailed as you need. Different people need different amounts of support, based on their level of experience.

Or maybe it was mentioned somewhere, but you didn't notice it! Sometimes there is a separate section on special cases (line through the origin, horizontal line, vertical line), and this would be emphasized there; sometimes it is just mentioned in one of several examples., and you need to have read closely to see it.
 
Dr.Peterson said:
It's not that there isn't a y-intercept, but that there is no need to write a 0 term in an equation.

What we are doing here is rewriting any equation of a line in a particular form, from which we can infer the slope and intercept. If the equation as given has only an "mx" term, then you can write "+0" to make it explicitly look like "mx+b". Or, as we do with experience, just imagine it being there. The key idea is that the equation you're given will not always directly show these numbers; you have to make them visible by the way you rewrite it.

Maybe this course is not as detailed as you need. Different people need different amounts of support, based on their level of experience.
Click to expand...
It is rather detailed and I'm making good progress. I'm just noticing some minor shortcomings that matter. Mostly in the chapter tests.
Dr.Peterson said:
Or maybe it was mentioned somewhere, but you didn't notice it! Sometimes there is a separate section on special cases (line through the origin, horizontal line, vertical line), and this would be emphasized there; sometimes it is just mentioned in one of several examples., and you need to have read closely to see it.
Click to expand...
I did go through the course description and it is for Beginners. I'm 68 years old and this is something I should have learned in school, had I been paying attention. :)
And thank you all for your time here helping a beginner!
 
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