Compassionate
  • Compassionate
Easiest way to graph an equation?
Mathematics
chestercat
  • chestercat
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Compassionate
  • Compassionate
Say I have an equation in y = mx + b ... Lets ay, y = 2x + 4
Compassionate
  • 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
  • Nnesha
slope = rise over run and you can write 2 as 2/1 \[\huge\rm slope =\frac{ rise }{ run }\]

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campbell_st
  • campbell_st
And I thought the easiest way was to use technology can you graph y = x/(x^2 + 1) using this method
Compassionate
  • Compassionate
Right, so would I calculate my slope from my y-intercept.
Compassionate
  • Compassionate
Let's stick to linear equations @campbell_st , all those hyperboles and binomials don't sit well with me
campbell_st
  • campbell_st
well I'd still say technology is the easiest method is if its only straight lines
Compassionate
  • Compassionate
If only I could use graphing technology on standardized testing
Nnesha
  • 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
  • Compassionate
|dw:1443991206348:dw|
Compassionate
  • Compassionate
so it would look something like this?
campbell_st
  • campbell_st
don't forget you should choose 3 independent values.... to avoid errors
campbell_st
  • campbell_st
but what if the line has no slope...
Compassionate
  • 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
  • Nnesha
|dw:1443991283848:dw| looks good to me but to draw a perfect line you can go down down left one as wll
Compassionate
  • 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
  • Compassionate
|dw:1443991488766:dw|
campbell_st
  • 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
  • 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
  • Compassionate
You guys are confusing me with al these different methods.
idku
  • idku
just extract two points (intercepts are easier to find), and connect the dots. This is all.
Compassionate
  • 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
  • idku
and connect intercepts (after you graph them) with a ruler.
Compassionate
  • 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
  • Nnesha
to graph an equation we dont' really need to find intercepts but you CAN if you want
idku
  • idku
well, you can use a graphing calculator, but it is not called you graphing it....
Nnesha
  • Nnesha
yes first you have to rewrite the equation in slope intercept
idku
  • 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
  • Compassionate
y = 2x + 4 6x - 3 = y Now, what if I solve it... Like y = 2x + 4 y = -6x + 4 Is that right?
idku
  • idku
you want to solve the system, right?
Compassionate
  • Compassionate
WEll, I Want to graph it
Nnesha
  • Nnesha
both equations are already in slope intercept form :D
Nnesha
  • Nnesha
i guess there is a typo are you sure it's `6x-3x`=y ?
idku
  • 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
  • idku
I guess it was -4 in eq2?
Compassionate
  • 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
  • idku
sure, good!
idku
  • idku
I for some reason prefered elimination
Compassionate
  • 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
  • campbell_st
perhaps you can graph them to check... that's easy
Compassionate
  • 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
  • campbell_st
well you can always graph them as a valid method, rather than using the algebraic methods
Compassionate
  • 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
  • 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
  • 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
  • Nnesha
right good to have extra information! but i'm saying for graph good luck!

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