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  1. #1
    Junior Member alwaysalillost's Avatar
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    How would I graph...

    How would I graph a line going through the points (3, 2) and (-2, 4).

    How would I graph y = f(x) = x^2

    I get very confused with graphing so if you could please explain how you get the answer I would be very greatful. Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alwaysalillost View Post
    How would I graph a line going through the points (3, 2) and (-2, 4).
    ok, so actually graphing a line through two points is pretty easy. you just draw up your coordinates and then plot the two points and then draw a line thorugh them. however, they usually don't make questions that easy for us. it is usually expected when we draw a graph we should label the x and y-intercapets if any, and so, let's find those first.

    now the equation of a line is of the form y = mx + b where m is the slope of the line and b is the y-intercept. we can find the x-intercept by setting y = 0 and solving for x.

    to find the slope, we do the following:
    given two points, (x1,y1) and (x2,y2) we can find the slope of the line through the points using the formula:

    m = (y2 - y1)/(x2 - x1)

    once we find the m, we can find the equation of the line by the point-slope form, which is:

    y - y1 = m(x - x1)

    we get the equation of the line by solving the above for y, using (x1,y1) as any one of the points we were given. so first let's get the equation of our line.


    we are given (x1,y1) = (3,2) and (x2,y2) = (-2,4), so:

    m = (y2 - y1)/(x2 - x1) = (4 - 2)/(-2 - 3) = 2/-5 = -2/5

    so using m = -2/5 and (x1,y1) = (3,2), we have by the point-slope form:

    y - y1 = m(x - x1)
    => y - 2 = (-2/5)(x - 3)
    => y - 2 = (-2/5)x + 6/5
    => y = (-2/5)x + 16/5

    so now we see that our slope is -2/5 (which means our line is going from the top left to bottom right) and the y-intercept is 16/5 which is just a little bigger than 3. now let's find the x-intercept:

    for x-intercept, set y = 0
    => 0 = (-2/5)x + 16/5
    => (2/5)x = 16/5
    => x = 8

    so our x-intercept is 8.

    now we draw our axis and label the intercepts, and draw a line through them.

    the blue dots are the points that were given, the red dots are the intercepts
    Attached Thumbnails Attached Thumbnails How would I graph...-line.jpg  
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  3. #3
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    Re:

    Quote Originally Posted by alwaysalillost View Post
    How would I graph
    How would I graph y = f(x) = x^2
    RE:
    Attached Thumbnails Attached Thumbnails How would I graph...-graph-8.png  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alwaysalillost View Post
    How would I graph y = f(x) = x^2
    now y = x^2 is a very common graph. if you go any further in math, you'll be seeing this function a LOT. pretty much any math student knows this graph by heart, so i assume this is your first encounter with the graph of this function. basically it looks like a cup, the lowest point passes through the origin, that is, (0,0), and then both ends go up forever and ever. since this is your first time, let's go to the basics. the most basic way to draw a graph is to plot points, and then draw a line through it, you should try to spread the points out so you get a good feel for the graph, try to plot points on both sides of the y-axis if possible. so let's plot some points, five points should be enough for you to get a feel of how the graph is going.

    let x = -2, -1, 0, 1, and 2, and see what happens to y. so we plug each of these x-values into the original function and compute the corresponding values of y, then put them in a table.

    when x = -2
    y = (-2)^2 = 4

    when x = -1
    y = (-1)^2 = 1

    when x = 0
    y = 0^2 = 0

    when x = 1
    y = 1^2 = 1

    when x = 2
    y = 2^2 = 4

    so our table is:

    ....x....|....y
    ---------------
    ...-2...|....4
    ...-1...|....1
    ....0...|....0
    ....1...|....1
    ....2...|....4

    now we will draw our axis up and plot the above points on it. then we draw a smooth curve through the points, and we're done. see the finished product below. i plotted the points on the graph that we got above.
    Attached Thumbnails Attached Thumbnails How would I graph...-xsquared.jpg  
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  5. #5
    Junior Member alwaysalillost's Avatar
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    So what would be the equasion to the first problem? Thanks Jhevon and qbkr21!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alwaysalillost View Post
    So what would be the equasion to the first problem? Thanks!
    i told you. it's y = (-2/5)x + 16/5
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  7. #7
    Junior Member alwaysalillost's Avatar
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    Okay thank you and sorry for being such a pest.
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