2

I was graphing the rule $y = 2x - 3$. Since this rule is of the form $y = mx + c$, we have substituted $m = 2$ and $c = -3$ when $x = 0$. This means the gradient (slope, steepness, etc) of the line $y$ we want to graph will be $2$ since $m$ denotes the gradient. We can mark our first co-ordinate on the graph by knowing that $x = 0$ and $c = -3$, forming the ordered pair $(0, -3)$ which is a point on our graph. Since every point is of the form $(x, y)$ then $c$ is equal to the $y$-intercept (the point where the line crosses the $y$ axis).

Now I have a question. My teacher asked me to let $x = 1$ after I set it equal to $0$, and according to the rule, $y = 2\cdot 1 - 3 = 2 - 3 = -1$ so we form another point $P(1, -1)$ where we write $P$ in front of the ordered pair to make it clear that it is a point on the graph. Since the rule is of the form $y = mx + c$, we are forming a linear graph where our line is always straight (of course since the measured gradient of the line at each of its points must be the same; a constant gradient). So from here, we can just draw a straight line through the two points we have plotted on the graph and Bob's your uncle.

However, I figured out something interesting. We could re-write $y = 2x -3$ as $y + 1 = 2(x - 1)$ and here we have constants $1, -1$ which are exactly the $x$ and $y$ coordinates of the point on our line for which $m = 1$. Is this a coincidence?

I am a little eager and impatient, and I do not want to wait for my next maths class to discuss this.

Thank you in advance.

Edit: I have found my answer here $\longrightarrow$ How to Graph Equations $-$ Linear, Quadratic, Cubic, Radical, & Rational Functions

But thanks for the help anyway :)

Community
  • 1
Mr Pie
  • 9,459
  • 2
    Regarding your re-write at the end, the google search point + slope + form + line will be of interest to you. This way of writing the equation of a line is especially useful in calculus (i.e. linear approximation at a point) and it's also a useful illustrative example to look at when horizontal and vertical shifts are discussed in a precalculus course. – Dave L. Renfro Oct 14 '17 at 12:21
  • @DaveL.Renfro Thanks so much! I know what the shifts are and are pretty useful in quadratic equations. The slope-intercept form I believe is $y = mx + c$ or $y = mx + b$ depending on which one you are taught, but it is the same anyway. The standard form is $ax + by = c$ which is easier because you can find both $x$ and $y$ intercepts and draw a line through the two points. The point-slope form however, I did not know (of) and I believe that probably answered my question, so once again, thanks heaps!! :) – Mr Pie Oct 14 '17 at 12:39
  • 1
    There's also the two-intercept form for the equation of line. These various ways of writing equations of lines can often be found in older algebra textbooks (such as "college algebra" texts from roughly the 1930s to 1950s), and some googling using "internet archive" or "hathitrust" along with "algebra", "line", "equation", etc. should uncover some of these books that are freely available on the internet. – Dave L. Renfro Oct 14 '17 at 12:56
  • Ok. Thank you so much @DaveL.Renfro !! :) – Mr Pie Oct 14 '17 at 22:48

2 Answers2

1

Once you have recognized that the graph will be a straight line, you just need two different points on it to actually draw it with a ruler.

For $y=mx+c$, setting $x=0$ and $x=1$ is a particularly convenient way to get those two points (because those two numbers are both trivially easy to multiply by $m$, so it takes only a single addition to find all the numbers you need). But you don't have to do it that way if you have a different way to choose two points on the line you'd rather use.

If, instead, you have the equation on the form $y-b=m(x-a) $ like you rewrote it to, it makes perfect sense to let $(a, b) $ be one of your two points.

hmakholm left over Monica
  • 286,031
-1

to graph the line $$y=mx+c$$ we will need two different points: if $$x=0$$ we get $$y=c$$ this is the first point: $$P_1(0,c)$$ second: $$y=0$$ we get $$x=-\frac{c}{m}$$ we get $$P_2\left(\frac{-c}{m},0\right)$$

Dr. Sonnhard Graubner
  • 95,283
  • I cannot vote because I reached my daily limit. I can vote in $13$ hours or more. However, this is a really nice and easy way to find another way of plotting a graph!! It's so simple, I didn't know why I didn't think of it before! (By the way, I didn't put the downvote.) – Mr Pie Oct 14 '17 at 10:09
  • And I also think you mean that in the second point we find by letting $y = 0, \ x = -c/m \neq -m/c$ – Mr Pie Oct 14 '17 at 10:14
  • 1
    if $y=0$ then we have $$0=mx+c$$ thus $$x=-\frac{c}{m}$$ – Dr. Sonnhard Graubner Oct 14 '17 at 10:17
  • I know. But in your post, it said that the second point was $$P_2\bigg(\frac{-m}{c}, 0\bigg)$$ which was incor -- oh wait it's changed now. Never mind :) – Mr Pie Oct 14 '17 at 10:19
  • 1
    yes i have made a typo sorry – Dr. Sonnhard Graubner Oct 14 '17 at 10:20
  • 1
    i have written $$-\frac{m}{c}$$ instead $$-\frac{c}{m}$$ – Dr. Sonnhard Graubner Oct 14 '17 at 10:21
  • Haha it's okay. I knew you meant $-c/m$ anyway based on your status, but I just wanted to bring that to your attention since you got a downvote – Mr Pie Oct 14 '17 at 10:22
  • 1
    Your solution works if $c\ne 0.$ But, what if $c=0?$ In such a case $P_1=P_2.$ – mfl Oct 14 '17 at 10:24
  • @mfl if $c = 0$ then we have $y = mx$ and so if $y = 0$, then $x = 0$ assuming we want to make a line that is not coincident to the $x$-axis (or to better put it, does not coincide with the $x$-axis), thus we only have one point $(0, 0)$ which is the origin. So you're right. (See $\rightarrow$ https://math.stackexchange.com/questions/1167754/basic-question-about-y-mx?rq=1) – Mr Pie Oct 14 '17 at 10:26
  • 1
    then we get $$y=mx$$ take $$x=1$$ then is $$y=m$$ and the other point is $$(0;0)$$ – Dr. Sonnhard Graubner Oct 14 '17 at 10:27
  • 1
    @Dr.SonnhardGraubner I think you should include your comment in your answer for completeness. – mfl Oct 14 '17 at 10:29
  • hah I would have upvoted @Dr.SonnhardGraubner's answer anyway :) – Mr Pie Oct 14 '17 at 10:30