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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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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