# graphWatch

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Thread starter 3 weeks ago
#1
I have a graph (going to try insert a picture) can someone tell me if these answers are correct?

(a) write down the equations of the two line shown in the diagram - y=mx+c (I feel like this is wrong...)
(b) What can we say about the relationship between these two lines - They are linear, Don't pass through origin but both cross x+y axes, so there is an intercept.
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Thread starter 3 weeks ago
#2 0
3 weeks ago
#3
For the equation of the lines, you'd be expected to work out what "m" and "c" for each line.
If they meet (they do), put the coordinates of the point where they meet.
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Thread starter 3 weeks ago
#4
(Original post by mqb2766)
For the equation of the lines, you'd be expected to work out what "m" and "c" for each line.
If they meet (they do), put the coordinates of the point where they meet.
How do I work out the gradient?
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3 weeks ago
#5
(Original post by jqhsehs)
How do I work out the gradient?
(change in y) / (change in x)

If you move along 4 (x) and up one (y), the gradient is 1/4. Try for these two lines?
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Thread starter 3 weeks ago
#6
(Original post by mqb2766)
(change in y) / (change in x)

If you move along 4 (x) and up one (y), the gradient is 1/4. Try for these two lines?
Does it have to be only up one for y?
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3 weeks ago
#7
(Original post by jqhsehs)
Does it have to be only up one for y?
You must have a book with this in, or just a quick google for the equation of a line.
Pick a point on the line, then move to another point on the line. Count how many you've gone along (x) and how many you've gone up (y).
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Thread starter 3 weeks ago
#8
(Original post by mqb2766)
You must have a book with this in, or just a quick google for the equation of a line.
Pick a point on the line, then move to another point on the line. Count how many you've gone along (x) and how many you've gone up (y).
But, there's so many different points you can pick won't that mean the gradient could be wrong or something...
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3 weeks ago
#9
(Original post by jqhsehs)
But, there's so many different points you can pick won't that mean the gradient could be wrong or something...
Whichever two points you pick, "m" is a constant so it does not change. That is what a line is.
Pick two points which make counting the changes in x and y as easy as possible.
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Thread starter 3 weeks ago
#10
(Original post by mqb2766)
Whichever two points you pick, "m" is a constant so it does not change. That is what a line is.
Pick two points which make counting the changes in x and y as easy as possible.
would this be correct - y=2.5x+-1
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3 weeks ago
#11
(Original post by jqhsehs)
would this be correct - y=2.5x+-1
Neither line is of that form. If you want to check yourself, put a few values of x into the equation and see what they evaluate to (do they lie on the line)?
Pls show how you came up with those numbers?
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Thread starter 3 weeks ago
#12
(Original post by mqb2766)
Neither line is of that form. If you want to check yourself, put a few values of x into the equation and see what they evaluate to (do they lie on the line)?
Pls show how you came up with those numbers?
I went along 2 on the x axes and up 5 on the y axes then divided them
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3 weeks ago
#13
(Original post by jqhsehs)
I went along 2 on the x axes and up 5 on the y axes then divided them
Look at the scales on the x and y axes, don't just count boxes.
Also, did you go up 4 rather than 5?
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Thread starter 3 weeks ago
#14
(Original post by mqb2766)
Look at the scales on the x and y axes, don't just count boxes.
Also, did you go up 4 rather than 5?
would the answer be
y=2x+-1
y= -0.5x+-1 ?
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3 weeks ago
#15
(Original post by jqhsehs)
would the answer be
y=2x+-1
y= -0.5x+-1 ?
Gradients look good this time. Not sure about the constants though, what is +-1?
If you're unsure, put x = 0. What are the values of y, are they the same on the graph?
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Thread starter 3 weeks ago
#16
(Original post by mqb2766)
Gradients look good this time. Not sure about the constants though, what is +-1?
If you're unsure, put x = 0. What are the values of y, are they the same on the graph?
I looked online and it said the 'c' in the equation means the y intercept so I presumed it was just -1
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3 weeks ago
#17
(Original post by jqhsehs)
I looked online and it said the 'c' in the equation means the y intercept so I presumed it was just -1
Definition of c is fine.
The lines do not cross the y axis when y = -1? Close, but not right.
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Thread starter 3 weeks ago
#18
(Original post by mqb2766)
Definition of c is fine.
The lines do not cross the y axis when y = -1? Close, but not right.
1?
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3 weeks ago
#19
(Original post by jqhsehs)
1?
So the equations of the two lines are ....
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Thread starter 3 weeks ago
#20
(Original post by mqb2766)
So the equations of the two lines are ....
y= 2x+1
y= -0.5x+1 ?
0
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