Sketching the Graph of a Line (Multiple Strategies)

Video Series: Functions and Their Graphs (Step-by-Step Tutorials for Precalculus)

(D4M) — Here is the video transcript for this video.

00:00
in this episode you’ll learn seven strategies for graphing a line [Music]
okay up first we’re going to be graphing some vertical lines so this will be our
first strategy so if you notice that the equation that you’re given looks like x
equals four x equals five or x equals any real number so i’ll use a constant c
here if x equals a constant then what that means is that we have a vertical
line so here’s an example vertical line right here x equals four
so it’s very easy to graph vertical lines just straight up and down
here would be one say x equals minus two so any real number that you’re given x
equals that real number is vertical line and the vertical lines are straight
forward to sketch and so what about horizontal lines let’s

00:01
look at horizontal lines so horizontal lines look at look like y equals c
again c can be any real any given real number
so to graph a horizontal line we just need to make the horizontal line so for
example this could be y equals five or we could have a negative
say negative one half there’s the line right there
y equals negative one half is just a horizontal line straight across um
and so that’s it that’s all that we need to know for graphing vertical and
horizontal lines um so slanted lines are a lot more fun
to graph so let’s look at some of those so first one is we’ll look at the
slope intercept form of a line so the slope interface intercept form of
a line is very useful for very useful for sketching the graph so for example
if we know the m and the b then we know the y-intercept so the

00:02
y-intercept is at 0 b and then we can use the slope
to go over and do a rise over a run thing
and we can get another point on the line and so we can do that
um however i usually like to graph intercept so i’ll i’ll be using
intercepts to graph the lines but knowing the y intercept already
tells you one point on the line and so then you would just need one
other point on the line to sketch the graph however you want to come up with
that point so in this example here what i’m going to be doing is i know that the
line that i’m looking for has to pass through this point
and so i’m given some other information because one point is not enough to
specify a line so i’m given some information here to find the slope of the line
so this is really two parts here part a and part b so let’s solve part a right
here so for part a we’re going to need the y equals mx plus b

00:03
and we need to find the m and we need to find the b
so first off i’m going to try to find the m so i know that i’m parallel to
this line and what that means is that i have the same slope of this line
so let’s find the slope of this line so this line is 3x plus 4y equals 7
and so i’m going to move the y over and then i’m going to
you know solve this for for y so i’m going to move the 3x over
and i’m going to get 7 minus x and then i’m going to divide by the four
so this will be minus three fourths x and then plus seven fourths
so this will be the m that i choose right here for part a
minus three fourths and that’ll be the m so let’s come up here and use that m
and so now what we need is to find the b so in order to find the b i’m going to

00:04
use this point right here minus two thirds seven eighths so i’m gonna plug
that point in here and see what we get here so i’m going to use seven eighths
equals minus three fourths x and for the x i’m going to use the minus two thirds
so we’re going to solve this for b now the minus threes are going to cancel out
or negative times the negatives will be positive so this would be 7 8 and what
do we get for all this 2 over 4 is left so it’s one half i’m going to move the
one half over and for this one half i’m going to think
about it as 4 over 8 so i can get 7 minus 4 which is 3 8.
are very good so now we have our y intercept form minus 3 4 x
plus our b we just found was 3 over 8 and there’s our y intercept for oops

00:05
forgot my x minus 3 force x plus our b and now we can sketch the graph
again the b gives us the y intercept there so that’s 3 8 right there
and the slope is negative right here negative
three-fourths so we know it’s going to come through here like this
so the question is oh sorry that has positive slope so we have three eights here
and the slope is negative so it’s going to come through right here we can go
find this y intercept right here by setting uh the x-intercept right here by
f setting the y equal to zero so let’s set the y equal to zero and then solve
for the x and you get a one half right here now
um the previous video is we took the time to go through
um and look at how to find the x and y y-intercept so
just just so that you know that that’s there now let’s go to part b here

00:06
so we have the y-intercept form and the in the graph of the line right there
that’s good so for part b now we want to be perpendicular to this
line here so first thing is let’s go find the slope of this line right here
so to find the slope i need to solve this for y so y equals
and this will be seven over two x over three
and so the slope that i’m going to use will be three over two why well
this line right here the slope is minus two thirds that’s the
coefficient of the x that’s the slope of this line but i want
to be perpendicular to this line so that means i need to take the negative
reciprocal and that’s what this is right so this is negative two-thirds so the
negative of the negative two-thirds will be positive and i want to take the
reciprocal of two-thirds so i get three halves here so that’ll be the slope that
i use up here in this problem so i’m gonna start off with y equals m x plus b

00:07
our goal is to find the m and the b we just found the m right here
so this will be y equals three over two x plus the b now to find the b
here i’m gonna use this point here this is the x and the y i’m going to
plug in here and find the b so the y here is seven eighths
we have three over two and the x is minus two thirds plus the b
so here the threes will cancel the twos will cancel this is really a minus one
and so when i move it over to the other side it becomes a positive one
and i’ll write that positive one is eight over eight so b is here is just
fifteen over eight b is just fifteen over eight
so y is equal to three over two x plus the b plus b

00:08
and so now we can go and sketch this this one has positive slope the y
intercept here is fifteen over eight and so we just need one more point
now we know this point right here is zero fifteen over eight
but and we know this point which we can go plot it’s minus two thirds seven
eighths so it’s about right there but i usually like to find the uh the
intercepts to to make my graph look nice so i’m gonna find this intercept real
quick and the the way i do that is i set the y equal to zero and i find the x
when you do that you get minus five fourths and so that would give me the two
intercepts there and the two intercepts of this in this problem here so so part
a and part b is done we have the y intercept forms of the equations and we
have the the graphs right here um and as i said i usually like to find
the intercepts and so we’ll actually do even better than getting to the y equals

00:09
mx plus b there’s even a better form of the line
so but next we’re going to look at the point-slope form of the line
so what does that give us okay so here’s the point slope form of the line
now i kept the example the same here so this is the same point here and we want
to be parallel to the same line and perpendicular to the same line so let’s
recall here that for part a the m that we used was minus three fourths
and the b that we used was three over two now the reason why we use this m is
because that’s the slope of this line and we want to be parallel to it the
reason why we use this m is because we want to be perpendicular to this line
here but the point is is that when we’re looking at the point slope form of the
line what new information do we get here well we get a point

00:10
and we get a slope and so this one is slightly different from the last one the
last one we had y equals mx plus b so we were we were given a slope which
is what we have here and here we’re given the y-intercept the y-intercept was 0b
so um you know this one gives us an arbitrary point x
one y one so this one’s more general however when sketching the graph i don’t
find this one as usual so we can work out part a again so this will be y minus
and this will be y one right here will be the seven eighths
equals the m for part a was minus three fourths and then x minus
and then this will be minus uh x1 here so minus and this already comes with a

00:11
minus so this will be plus two-thirds and so this is the point slope form of
the line but it doesn’t really give us any new
information in terms of trying to sketch the graph
for for part b it would just simply be minus 7 8 and then we’re going to use
the three halves and then x plus two thirds sorry uh yeah x plus two thirds
so it’s a it’s a it’s a minus and then a minus and so it’s positive two thirds
so this would be the point-slope form of the of the line for the equation
but if i was to go sketch these i would actually just solve this for y and solve
this for y and get the y-intercepts and then get
the x-intercept so i would just continue on just like we did on the last example
here so having the point-slope form could be useful trying to find an equation
ultimately when i’m trying to graph it though i usually go back to y equals

00:12
form right there so in terms of graphing this right here doesn’t really give us
much more useful information now let’s look at the two point form for the line
so this one right here also gives us a point x1 y1
but now we have a second point and we’re going to use that second point to find
the slope so this one’s like the last form except the slope isn’t known and so
now we have to go plug it in this right here is just the slope formula right
here so there’s not much more information in terms of this one
except for the fact in the way in which you find the slope given two points
there’s a unique line going through there and so you can find the slope
let’s do that on this and on with these numbers right here so you know we got
two points we got only one line coming through here so this would be the point

00:13
like three minus four and this would be the point here say five six
only one line going through there what is the slope of this line
so we could go find the slope right here it would be using this formula right
here now i can label this was point one or this is point one it doesn’t matter
i will go ahead and label this one right
here just because i see it first so i’ll say this is x one y one
and i’ll label this one right here is point two x two y two
but it really doesn’t matter either way you’re gonna get the same line
so here we go so this will be y minus y one y one is minus four so we’re gonna
get a minus minus and now we’re gonna get the y two minus y one so six
minus a negative four over five minus three and then x minus
x minus and then now what is x one is the three

00:14
and so this right here would be the two point forms uh equation of the line
however you’d probably want to simplify this so this is 10 over
10 over 2 so this is 5 and then x minus 3 here
and so now we can go and sketch this line here
and one way to do that might be to say oh this is 5x and then minus 15
and then move the 4 over you get minus 19
and so we could go sketch this line here we’re down here at -19
the slope is five so it’s coming up through here
and we can go find the x-intercept here so how would we find the x-intercept
right here we would set the y equal to zero and then just solve for the x
so move the nineteen over and then divide by five we see this is nineteen

00:15
over five right here so this would be the sketch of the line right here
and so we’ve we found it using the slope the slope was five right there
all right so very good so now let’s move on to the um
next one which is probably the more standard one of all of them it’s called
the standard form so let’s look at that so the standard form here
we have an ax plus b y equals c so the num the the constants over here
and we’re going to try to make this unique so for example if we had say 10x
minus 14y equals 28 so what would try to do is try to
simplify that and we would try to get rid of the
two and everything so five x minus seven y equals fourteen so this would be a
better they both represent the same line but this would be a better equation for

00:16
the line so we want to specify that a is positive and that we can simplify it
notice this one right here is already in standard form right here five x plus
three y equals two that’s already in standard form so let’s try to graph this
line right here real quick now what i really like about standard
form is it’s very easy to graph it so i’m going to graph it right here
and the way i’m going to do that is i’m going to find the intercepts so here’s
how i can do that when i set y equal to 0 right when y is 0 then that becomes 0
and then solve for x x is two fifths so that’s the intercept here
now when i set x to be zero i can find the y intercept the y intercept is two
thirds so there’s the two thirds right there
so i got the two points that i need to graph a line and then there’s my graph
right coming through there so there’s the line right there

00:17
that’s it that’s all there is to it and so this is my favorite form of the
equation of the line um so you know we could go solve this for y
and get the y-intercept but why do that if all you want to do is
sketch it really quick right there’s the there’s the
intercepts now before i go on to the last form which is actually my favorite
even more than this one so you know let’s get a one here
let’s divide everything by two so if i divide everything by two i get
five over two x plus three over two y equals one
what’s so useful about getting a one here well
i’m actually gonna simplify i’m gonna even rewrite it more what i’m gonna do
is i’m gonna take this five halves and put it in the denominator
so think of it like this five over two x all over one

00:18
plus three over two y all over one equals one
so i’m gonna when i say i’m going to put five halves in the denominator what i
want to do is i’m going to take the reciprocal of it which will be two-fifths
right in other words if you want to move three to the denominator
it’s just one over one third right so the three halves here i want to move
it to the denominator and so it’s going to be just y up here now
and then this is will be two thirds here and as you can see those are the actual
intercepts right there so when i have a x over a plus y over b
and i have it equal to one then those are the actual intercepts right there
and so that’s the actual last strategy here is the two intercept form

00:19
equation for the line so if you have x over a plus y over b equals 1
then you actually know the intercepts right there
so let’s graph this one right here and see how it works
so the x intercept is at one half the y intercept is at four
and so we have our line going right through there
and that’s it that’s all there is to it okay so if you like this video i hope
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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