Easiest way to graph an equation?

- Compassionate

Easiest way to graph an equation?

- chestercat

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- Compassionate

Say I have an equation in y = mx + b
... Lets ay, y = 2x + 4

- Compassionate

y-intercept = 4, so I would graph 4 on the y-axis
Now, my slope is 2x, so I could just graph 2? Or would I have to plug in values for 2, like
x = 1
x = 0
x = -1
or will just graphing 2 be enough?

- Nnesha

slope = rise over run
and you can write 2 as 2/1 \[\huge\rm slope =\frac{ rise }{ run }\]

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## More answers

- campbell_st

And I thought the easiest way was to use technology
can you graph y = x/(x^2 + 1) using this method

- Compassionate

Right, so would I calculate my slope from my y-intercept.

- Compassionate

Let's stick to linear equations @campbell_st , all those hyperboles and binomials don't sit well with me

- campbell_st

well I'd still say technology is the easiest method is if its only straight lines

- Compassionate

If only I could use graphing technology on standardized testing

- Nnesha

yes right first draw a point on y-axis (yintercept) and btw slope is just 2 x is a variable not slope
|dw:1443991125916:dw|

- Compassionate

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- Compassionate

so it would look something like this?

- campbell_st

don't forget you should choose 3 independent values.... to avoid errors

- campbell_st

but what if the line has no slope...

- Compassionate

X is independent, right?
y = 2x + 4
y = 2(1) + 4
y = 6
y = 2(2) + 4
y = 8
y = 2(3) + 4
y = 10
Like that?

- Nnesha

|dw:1443991283848:dw|
looks good to me but to draw a perfect line you can go down down left one as wll

- Compassionate

If the line had no slope, say
y = 4
Wouldn't I just graph a horizontal line across the y-axis, since it has no point of intersection on the x-axis? A line with no slope is either parallel to x when y = a number, or parallel to y when x = a number

- Compassionate

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- campbell_st

why not graph the line using intercepts..
y = 2x + 4
x- intercept -2 when y = 0 and y intercept 4 when x = 0
that's an easy method

- idku

y=mx+b
0=mx+b --> -b=mx --> -b/m=x
point: (-b/m, 0)
y=m(0)+b --> y=b
point: (0,b)
Graph the points (0, b) and (-b/m, 0) and connect them with a ruler.

- Compassionate

You guys are confusing me with al these different methods.

- idku

just extract two points (intercepts are easier to find), and connect the dots.
This is all.

- Compassionate

y = 2x + 4
0 = 2x + 4
- 4 = 2x
x = -2
Then y = 0 + 4
y = 4
So I can graph (-2, 0), (0, 4)
Then that will give me bot the intercepts. I get it.

- idku

and connect intercepts (after you graph them) with a ruler.

- Compassionate

What if I'm given two equations to graph now. Say
y = 2x + 4
6x - 3x = y
And one of my equations isn't in y = mx + b, should I convert my second equation to y = mx + b or what?

- Nnesha

to graph an equation we dont' really need to find intercepts but you CAN if you want

- idku

well, you can use a graphing calculator, but it is not called you graphing it....

- Nnesha

yes first you have to rewrite the equation in slope intercept

- idku

the easiest I would do, when I have a y=mx+b to graph,
1) Find y and x intercepts
2) Graph these two points (the y and x intercepts you have found)
3) Connect the intercepts with a rule
Bam!

- Compassionate

y = 2x + 4
6x - 3 = y
Now, what if I solve it... Like
y = 2x + 4
y = -6x + 4
Is that right?

- idku

you want to solve the system, right?

- Compassionate

WEll, I Want to graph it

- Nnesha

both equations are already in slope intercept form :D

- Nnesha

i guess there is a typo are you sure it's `6x-3x`=y ?

- idku

y = 2x + 4
6x - 4 = y
multiply equation#2 times -1.
y = 2x + 4
-6x + 4 = -y
now we will do addition to eliminate the y.
y = 2x + 4
-y = -6x + 4
-------------
0 = -4x + 8

- idku

I guess it was -4 in eq2?

- Compassionate

y = 2x + 4
6x - 4 = y
Why not just
6x - 4 = 2x + 4
4x = 8
x = 2
y = 2(2) + 4
y = 4 + 4
y = 8
So, (2, 8) would be the point of intersection, right? That's where the two lines cross.

- idku

sure, good!

- idku

I for some reason prefered elimination

- Compassionate

If I wanted to graph the lines, would I just put them both in y = mx+ b and just use the aforementioned method to just graph each line individually?

- campbell_st

perhaps you can graph them to check... that's easy

- Compassionate

If I graph them and my point of intersection is (2, 8), then that must mean I graphed them correclly, right? @campbell_st

- campbell_st

well you can always graph them as a valid method, rather than using the algebraic methods

- Compassionate

Since we already graphed the first one, let me try the second one myself.
6x - 4 = y
So: y = 6x - 4
My y intercept is -4
My x intercept is 0 = 6x - 4
4 = 6x
x = 4/6 or 0.67.
I understand on a test you probably wouldn't be asked to graph 0.67, but for now this is just an example, but did I do the process right? These would, in theory, be my y and x-intercepts.

- Nnesha

well if i have to graph that equation
i will draw a point at -4 and then use the fact slope = rise over run
6/1=rise over run
for me this way is easier than the other one
no need to solve for anything :=)

- Compassionate

Ah, right, thank you. But If I was asked to find the x-intercept, I would know how thanks to @campbell_st . Thank you @Nnesha . Wish I could medal you both

- Nnesha

right good to have extra information!
but i'm saying for graph
good luck!

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