Giving Lessons: Finding Slope From An Equation
If you don’t already know… for the past few days I have decided to start Giving Lessons. If you want to see the rest of them just go on my profile, click on my questions, and click on the ones that start with: “Giving Lessons”. I feel that this is definitely helping some of you, so I decided that this is a great way to help you guys out! If you already know this, then tag someone who doesn’t, but if you don’t already know this, then hopefully this will help you out! Let’s Get Started! First I would like to show you guys the template of a slope formula. It is: \[y = mx + b\] In this equation m is the slope and b is the y-intercept. Since in this lesson we are finding the SLOPE from an equation we will be focusing on the variable, m. Let’s continue. The first sample problem we are going to be working with is: \[y = -\frac{ 5 }{ 2 }x - 5\] In this equation, all we did is simply plug in a few numbers for the variables that were in the original equation. So we simply need to find out, what is in place of the variable m. We can see that \[-\frac{ 5 }{ 2 }\] is in place of m! So the slope of this equation is: \[-\frac{ 5 }{ 2 }\] DONE! GREAT JOB! :) ————————— Still Confused? Don’t worry! We will be doing another sample problem! ————————— The next sample problem we will be working with is: \[y = -\frac{ 4 }{ 3 }x - 1\] Now, in this sample problem it is the same thing. What is substituting the variable m? It is: \[-\frac{ 4 }{ 3 }\] So the answer is: \[-\frac{ 4 }{ 3 }\] DONE! GREAT JOB! :) ————————— Still Confused? Don’t worry! We will be doing one last sample problem! ————————— Our last sample problem is: \[y = -x + 3\] This is a little different from our other problems, because we don’t really SEE what is substituting for the variable m. But if there is a negative sign with no number next to it, we must assume that there is a 1. This is because, in this equation, if you simply put a negative sign, you don’t need to put the number 1. So the slope in this equation is: \[-1\] DONE! GREAT JOB! :) ————————— Still Confused? Don’t Worry! If you have any questions just comment them down below! I will try my best to answer them as soon as possible! ————————— I hope this helped you in some way! Once again, If you already knew this, then tag someone who doesn’t. But if you didn’t and this helped you… then I am very glad that I helped you out! Alright, bye now! :)
xD cool, i thought you gonna put some interesting thing in the end, o_O
What do you mean by interesting? xD @rishabh.mission
like a puzzle for others or something else...
btw, thats not a Question in any way.
Oh haha. Well they are welcome to post any questions in the comments lol :) I have done many "Giving Lessons" so maybe in the next one I could do something like that! :) @rishabh.mission
Cool, Good lucky!
luck* -_-
Thank you! :) @rishabh.mission
perhaps you should have used the phrase Linear Equations. Just a thought.
Oh okay. Thanks for that info! :) Will definitely think about that next time! :) @campbell_st
well linear equations are the only equations that have a slope....
\(\color{blue}{\text{Originally Posted by}}\) @AihberKhan First I would like to show you guys the template of a slope formula. It is: y=mx+b \(\color{blue}{\text{End of Quote}}\) y = mx + b is actually called the slope-intercept form. Also, you might want to use `\( \)` instead of `\[ \]` when using \(\LaTeX\). The one I suggest will help your text be inside the line instead of forming a different line.
Oh wow thanks! That will be very helpful info! :) @TheSmartOne
See the difference? The slope is \(\Large -\frac{5}{2}\) `The slope is \(\Large -\frac{5}{2}\)` The slope is \[\Large -\frac{5}{2}\] `The slope is \[\Large -\frac{5}{2}\]` And, I'm glad to be of some help! :)
Oh yes! Thank you! :)
Something I find lacking in this tutorial, and also in some textbooks, is the theoric deduction or at least, a theoric structurization of the slope formula to then generalize a point a find then, the y=mx+c form. Of course, let's take into consideration a referential plane XoY and an arbitrary line, the slope will be defined as the angle of the line respective to the x-axis, in this case, \(\alpha\). |dw:1451255824180:dw| Since the known modules are indeed \(\left| \left| x_2 - x_1 \right| \right|\) and \(\left| \left| y_2-y_1 \right| \right|\) we can then define the slope "m" as the tangent on the angle \(\alpha\). \[\alpha=\tan (\alpha)\] And tangent is the division of the catets, these being the opposite by the adyacent: \[\tan (\alpha)=\frac{ y_2-y_1 }{ x_2-x_1 }\] Therefore: \[m=\tan (\alpha) = \frac{ y_2-y_1 }{ x_2-x_1 } \iff m=\frac{ y_2-y_1}{ x_2-x_1 }\] And that defines "slope of a line".
Yes, I completely agree. However, this is simply a basic version just to start some people off to get more into and in depth for this concept. :) @Owlcoffee
Nice!!
Thank you! :) @noahbred
@Owlcoffee u tryhard
Great Job! This tutorial clearly shows on what m means in the slope-intercept formula :D
Thank you! I appreciate your nice and well thought out comment! :) @563blackghost
Np ^^
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