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March 16th, 2016, 05:05 PM   #1
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graph the line with the given slope m and y-intercept b.

Is there away to find "fractions" or if not called fractions not sure what the correct word is I believe I have found out that if it's 3/2 it's found still between 1 and 2 ? just like if it was 5/2 it would still be the same I think as the bottom number is still 2


If that's wrong sorry.

yet for this problem, and hopefully future ones is there away to know where the "fraction" is found on a graph?
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March 16th, 2016, 06:53 PM   #2
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Points (0, -2) and (3, 2) are on the graph. Can you see why?
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March 16th, 2016, 08:43 PM   #3
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In answer to your question, yes there is a good way. Converting the fractions to mixed fractions will help a lot.

For example, $\dfrac{3}{2} = 1\dfrac{1}{2}$, so you can then infer that it lies halfway between 1 and 2.

Similarly, $\dfrac{5}{2} = 2\dfrac{1}{2}$, so you can infer that it lies halfway between 2 and 3.
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March 17th, 2016, 08:30 AM   #4
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y = 4/3x + c
3y = 4x + 3c
fill in (0,-2)
-6 = 3c
c = -2
y =4/3x - 2
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March 17th, 2016, 10:11 AM   #5
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Quote:
Originally Posted by greg1313 View Post
Points (0, -2) and (3, 2) are on the graph. Can you see why?
I think I see why 0,-2 would be as there is only -2 so I guess that means x is zero.

I don't see how you got (3,2) from the 4/3 I am guessing? unless for some reason we are ignoring the 4?

thanks for the help.
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March 17th, 2016, 10:15 AM   #6
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Quote:
Originally Posted by Azzajazz View Post
In answer to your question, yes there is a good way. Converting the fractions to mixed fractions will help a lot.

For example, $\dfrac{3}{2} = 1\dfrac{1}{2}$, so you can then infer that it lies halfway between 1 and 2.

Similarly, $\dfrac{5}{2} = 2\dfrac{1}{2}$, so you can infer that it lies halfway between 2 and 3.
okay I am thinking for 4/3 I would look for point 1 1/3 not sure how I would find it on this graph though unless I put it on 1 1/2 and the program gives me some budging room.
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March 17th, 2016, 10:17 AM   #7
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Quote:
Originally Posted by manus View Post
y = 4/3x + c
3y = 4x + 3c
fill in (0,-2)
-6 = 3c
c = -2
y =4/3x - 2
I appreciate your help, sadly my understanding in this area is leading me to be confused.
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March 17th, 2016, 10:33 AM   #8
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start at the y-intercept, (0,-2) ... do the rise/run to find at least one more point of the line. Connect.
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March 17th, 2016, 12:37 PM   #9
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start at the y-intercept, (0,-2) ... do the rise/run to find at least one more point of the line. Connect.
Thanks I really am hoping I am getting this, it really has helped to thanks to people's help here. this maybe the wrong way to read it but I think for me I am getting that if it is B this is my y, then if I don't have my x then it is 0.

Then M is where I go from after finding my y.

I know that's not correct but It's how my memory is working. Or Maybe it's better to say My B is my starting point, then M is where I go from finding my starting point?
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March 17th, 2016, 01:33 PM   #10
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you just have to draw a line between the points {x=o and y=-2}
and{x=3 and y=2}
you can do that with a lineal what do you don't understand?
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