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Math Help - Graphing help.

  1. #1
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    Graphing help.

    Can someone show me how to create a graph where I can plot points for my equation problems.


    I know how to make the graph ( I think) but can't seem to plot the points to show my answer for the equation that required me to show the coordinates on the graph.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Patience View Post
    Can someone show me how to create a graph where I can plot points for my equation problems.


    I know how to make the graph ( I think) but can't seem to plot the points to show my answer for the equation that required me to show the coordinates on the graph.
    Unless you are asking how to plot points on a graph a little more information would be helpful.

    -Dan
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  3. #3
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    Graph

    Here is some math problems that required me to use a graph.
    Solving the problem is not the hard part its the graph I am having a hard time with.


    1. Plot points with coordinates (2, 3), (3, 4), and (4, 5) on the given graph. What do you observe? Can you give the coordinates of another point with the same property?





    2. Graph the equations by first solving for y.

    2x 3y = 12
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    Point plotting

    The graph am looking for is the cartesian coordinates system.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Patience View Post
    1. Plot points with coordinates (2, 3), (3, 4), and (4, 5) on the given graph. What do you observe? Can you give the coordinates of another point with the same property?
    Look at the graph below. It appears that the points fall on a line. Now how can you get other points on this line? Observe that the y coordinate is always 1 more than the x coordinate. So we can guess that
    (5, 6)
    (6, 7)
    or even
    (1, 2)
    (0, 1)
    (-1, 0)
    are all on this same line.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Patience View Post
    2. Graph the equations by first solving for y.

    2x 3y = 12
    Solve for y:

    2x 3y = 12

    -3y = -2x + 12

    y = (2/3)x - 4

    Now set up a chart:
    When x = 0, y = -4
    When x = 3, y = -2
    When x = 6, y = 0
    etc.

    Now graph these points (you should graph more than 3 points, but these are sufficient) and connect them with a line.

    -Dan
    Attached Thumbnails Attached Thumbnails Graphing help.-line2.jpg  
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  7. #7
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    graph

    I understand how to solve the equation I just can't understand how to make a graph.

    Can you send me something in step by step procedure of how to make a graph and how to plot the points also.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Patience View Post
    I understand how to solve the equation I just can't understand how to make a graph.

    Can you send me something in step by step procedure of how to make a graph and how to plot the points also.
    in general, two things you always want to locate when plotting ANY graph are the x and y-intercepts (if they exist).

    the x-intercept is where the graph cuts the x-axis, which is also the line y = 0. the y-intercept is where the graph cuts the y-axis which is the line x = 0. so to find these points, you do the following:

    for x-intercept: set y = 0
    for y-intercept: set x = 0

    so for our function:
    2x 3y = 12 ............we solve for y to get:

    y = (2/3)x - 4

    now for x-intecept: y = 0

    => 0 = (2/3)x - 4
    => 4 = (2/3)x
    => 4(3/2) = x
    => x = 6
    so the line cuts the x-axis at x = 6, as we see in topsquarks graph

    now for y-intercept: x = 0

    => y = (2/3)(0) - 4
    => y = -4
    for straight lines this part is completely unnecessary, the y-interscept always ends up being the lone constant. for other graphs however, this step is crucial.

    now, we're done. this is a straight line, all we need are two points. so we can plot (0, -4) which is the y-intercept and (6,0) which is the x-intercept, and draw a line passing through both, and that's our line.

    for more complicated functions, the procedure of course gets more complicated. for the simplier types of the complicated functions, you can generally get away with it by ploting a few more points and connecting them. we do this as topsquark did, set up a table.

    say, ok, i want to draw this graph in the interval [-3,3].
    then you would take the x-values: -3, -2, -1, 0, 1, 2, 3 and find the corresponding y-values. you do this by plugging them into the function. so for instance, with our line, say you wanted to plot more points, and you wanted to find the point corresponding to x = -3, you do the following:

    y = (2/3)x - 4
    when x = -3

    y = (2/3)(-3) - 4
    => y = -2 - 4 = -6

    so that point is (-3,-6)

    here are a table of values for those points

    ....x.....|......y
    -----------------
    ....3.....|.....-2
    ....2.....|.....-8/3
    ....1.....|.....-10/3
    ....0.....|.....-4
    ....-1...|.....-14/3
    ....-2...|.....-16/3
    ....-3...|.....-6

    so we'd plot these points on a set of axis, draw a line connecting all of them, and we're done.
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