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Math Help - How to work it out Graphs of Equations?

  1. #1
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    How to work it out Graphs of Equations?

    I am given;

    y = 1/2x + 3

    I am asked to produce a table of data. I have no problem working this out, however I can't see where the value 1/2 originates from?

    So if I am asked in a different question to work out the equation of the line, and given (-4. -2) and (2, 7) I can put the table together, but y = ? to multiply the values.

    How do I work this out ?

    Please advise.
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  2. #2
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    Quote Originally Posted by David Green View Post
    I am given;

    y = 1/2x + 3

    I am asked to produce a table of data. I have no problem working this out, however I can't see where the value 1/2 originates from?

    So if I am asked in a different question to work out the equation of the line, and given (-4. -2) and (2, 7) I can put the table together, but y = ? to multiply the values.

    How do I work this out ?

    Please advise.
    I'm not sure what the problem is? You say you have the data table. Are you saying you cannot plot the data points for some reason? The 1/2 is just a numerical factor, nothing else. (It's called the "slope" of the line, but it doesn't sound like you are that far yet.)

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    I'm not sure what the problem is? You say you have the data table. Are you saying you cannot plot the data points for some reason? The 1/2 is just a numerical factor, nothing else. (It's called the "slope" of the line, but it doesn't sound like you are that far yet.)

    -Dan
    Thanks for replying Dan, the problem for me at the moment is that I am asked to find the equation of the line, the information I am given is (-4, -2) and (2, 7).

    I have worked out how to find all the corresponding values of y, but I don't know how to work out the value of y before I start multiplying out the y multiplied by x values = new y values?

    Thanks

    David
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  4. #4
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    I am still learning, please advise if what I think I have learned here is right so far.

    In my maths question I am asked to find the equation of the line, the information I have been given is;

    (-4, -2) and (2, 7)

    I have worked out that I am supposed to work out the gradient of the line after plotting the graph, thw gradient I have found to be - 1.71.

    Using this I have worked out that;

    y = -1.71 times (x) + 2

    using the above I can then find my new values of y;

    y = -1.71 x -4 = 8.84 etc

    so the equation of the line would be;

    y - -1.71x + 2

    Please advise if I have worked out the equation of the line correctly and used the right techniques

    Thanks

    David
    Last edited by David Green; May 8th 2011 at 04:48 AM. Reason: made a typo error
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  5. #5
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    Quote Originally Posted by David Green View Post
    I am still learning, please advise if what I think I have learned here is right so far.

    In my maths question I am asked to find the equation of the line, the information I have been given is;

    (-4, -2) and (2, 7)

    I have worked out that I am supposed to work out the gradient of the line after plotting the graph, thw gradient I have found to be - 1.71.

    Using this I have worked out that;

    y = -1.71 times (x) + 2

    using the above I can then find my new values of y;

    y = -1.71 x -4 = 8.84 etc

    so the equation of the line would be;

    y - -1.71x + 2

    Please advise if I have worked out the equation of the line correctly and used the right techniques

    Thanks

    David
    If you plot (-4, -2) and (2, 7) it should be crystal clear that the gradient is not negative. I have no idea how you could get -1.17 (which I also assume is a decimal approximation rather than an exact answer involving a fraction).

    You appear to have many misunderstandings. If you show how you calculated your value for m, some of them can can be pointed out.

    Reviewing your class notes and textbook might help. But from what I have seen I think you will need to see your instructor for extensive one-on-one help.
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  6. #6
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    Any (non-vertical) line in a coordinate system can be written y= ax+ b. Given that the line goes through the two points (-4, -2) and (2, 7) you can put those (x,y) values in the equation to get -2= -4a+ b and 7= 2a+ b which are easy to solve for a and b.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    If you plot (-4, -2) and (2, 7) it should be crystal clear that the gradient is not negative. I have no idea how you could get -1.17 (which I also assume is a decimal approximation rather than an exact answer involving a fraction).

    You appear to have many misunderstandings. If you show how you calculated your value for m, some of them can can be pointed out.

    Reviewing your class notes and textbook might help. But from what I have seen I think you will need to see your instructor for extensive one-on-one help.
    I am glad you said class notes and textbook might help, because it is those that are causing the confusion?

    It appears the books are written in such a style that "key parts" vital to the understanding are "left out" so that misunderstandings become the norm?
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Any (non-vertical) line in a coordinate system can be written y= ax+ b. Given that the line goes through the two points (-4, -2) and (2, 7) you can put those (x,y) values in the equation to get -2= -4a+ b and 7= 2a+ b which are easy to solve for a and b.
    I think what you have said here has just blown all out of proportion?

    Keeping it as simple as possible, I have (-4, -2) and (2, 7). therefore;

    x = -4 when y = -2, and

    x = 2 when y = 7.

    Draw this graph and when x = -4 and y = -2, the slope of the line will CUT the x axis at about -2.6, and when x = 2 and y = 7, the slope of the line will CUT the y axis at y = 4.

    I worked out the gradient of the line, RISE over RUN = 1.54

    My table of data

    x = -4, -3, -2, -1, 0, 1, 2

    y = -4.16, -2.62, -1.08, 0.46, 0, 3.54, 5.08

    I found that y = -4.16 and x = -4

    -4.16 / -4 = 1.04

    y = 1.04 + 2

    gradient = 3

    surely I can't be that far out?
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  9. #9
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    Gradient First:
    RISE = 7 - - 2 = 9
    RUN = 2 - - 4 = 6

    RISE over RUN = 9/6 = 1.5


    Now post your calculation for the intercept
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  10. #10
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    Quote Originally Posted by SpringFan25 View Post
    Gradient First:
    RISE = 7 - - 2 = 9
    RUN = 2 - - 4 = 6

    RISE over RUN = 9/6 = 1.5


    Now post your calculation for the intercept
    Reading through our course book we have not been given any information to calculate the intercept, the book just shows how to read it from a graph.

    In my example the book says; the x intercept is the point where the line crosses the x axis, and the y intercept is the point where the value of y crosses the y axis, so reading from my graph;

    y = 0 then x = -2.6, and when x = 0 then y = 4

    we don't seem to have any other information, and this does'nt sem to answer the original question?

    so the gradient of the line seems to be 1.5, and the equation of the line seems to be;

    y = 1.5x + 2

    Have I got this right?

    Thanks

    David
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  11. #11
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    We have the gradient, so we know that y=1.5 + c

    We need to find c. This can be done using one of our pairs of values, in the above equation. i'll use x=2, y=7

    7 = 1.5*2 + c
    7 = 3 + c
    c = 4

    So the solution is y=1.5x + 4

    I dont see where you got 2 from. It doesn't appear anywhere else in your post. You have correctly identified the y intercept at x=0,y=4.
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  12. #12
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    Quote Originally Posted by SpringFan25 View Post
    We have the gradient, so we know that y=1.5 + c

    We need to find c. This can be done using one of our pairs of values, in the above equation. i'll use x=2, y=7

    7 = 1.5*2 + c
    7 = 3 + c
    c = 4

    So the solution is y=1.5x + 4

    I dont see where you got 2 from. It doesn't appear anywhere else in your post. You have correctly identified the y intercept at x=0,y=4.
    Please advise just so that I am clear, where did you get the +4 from in the equation y = 1.5x + 4

    Thanks

    David
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  13. #13
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    i used one of the points we were given: x=2, y=7.

    The equation for the line is: y = 1.5x + c
    where c is an unknown number to work out.

    substitute in a pair of values we know: x=2, y=7.
    7 = 1.5 * 2 + c
    7 = 3 + c

    Solve that equation in the normal way
    7 - 3 = 0 + c
    4 = c

    Return to the original equation:
    y = 1.5x + c
    y = 1.5x + 4


    This is not the only way of finding the answer. you may have been taught a different one.
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  14. #14
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    Quote Originally Posted by SpringFan25 View Post
    We have the gradient, so we know that y=1.5 + c

    We need to find c. This can be done using one of our pairs of values, in the above equation. i'll use x=2, y=7

    7 = 1.5*2 + c
    7 = 3 + c
    c = 4

    So the solution is y=1.5x + 4

    I dont see where you got 2 from. It doesn't appear anywhere else in your post. You have correctly identified the y intercept at x=0,y=4.
    Just a bit confused how to present the information?

    x = 0 (-2.6, 0) and y = (0, 4)

    does this look OK?
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  15. #15
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    almost, provided rounded answers are acceptable.

    that -2.6 is really -2.666666666.... which rounds to -2.7.
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