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Math Help - Find pattern, equation, then graph...

  1. #1
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    Find pattern, equation, then graph...

    Directions: Find the pattern in each problem. Find the equation. Graph the line on a coordinate plane.

    ********
    i'm in a class that mixes algebra 2 w/pre-cal and i have absolutely no idea on how to proceed on finding the equation. on #2 i noticed a +2 pattern on the second row. thanks in advance
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    Quote Originally Posted by trancefanatic View Post
    Directions: Find the pattern in each problem. Find the equation. Graph the line on a coordinate plane.

    ********
    i'm in a class that mixes algebra 2 w/pre-cal and i have absolutely no idea on how to proceed on finding the equation. on #2 i noticed a +2 pattern on the second row. thanks in advance
    here's a hint for the first, the rest seem to be similar.

    note that since for each new term we are adding or subtracting a constant number, we can describe the pattern by an arithmetic sequence.

    recall that an arithmetic sequence is one of the form: a_n = a_1 + (n - 1)d

    where a_n is the nth term, a_1 is the first term, n is the current number of the term, and d is the common difference (that is, the constant we add to one term to get the next term).

    fill in the corresponding value in the formula and simplify. the result will be in the form of the equation of a line


    an alternative is to use the point-slope form of the equation of a line to find the answer.

    the point slope form is: y - y_1 = m(x - x_1), where m is the slope, which we can find from any two points, and (x_1,y_1) is a point the line passes through (in this case, of course, n is the input value, so n would replace x in the formulas)

    got it?
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  3. #3
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    for #2, i got 2n-7. but when i tried to create an equation, i got m=-8 and on the second attempt with another pair of coordinates, i got m=2. then i thought it'd be something like y=2x-7... i think i have an idea.

    but what about #4? ...is it a quadratic?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by trancefanatic View Post
    for #2, i got 2n-7. but when i tried to create an equation, i got m=-8 and on the second attempt with another pair of coordinates, i got m=2. then i thought it'd be something like y=2x-7... i think i have an idea.
    wait, i'll see what i get

    but what about #4? ...is it a quadratic?
    i doubt it. i'd try to fit it to a cubic. do you know how? Hint: we can use simultaneous equations



    EDIT: ok, i got (i wrote them as the lines you will plot, so i replaced n with x and a_n with y):

    #1: y = 2 - 3x

    #2: y = 2x - 7

    #3: y = 2 - 2x
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    i hated simultaneous equations and never did understand it... you can graph those??
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    Quote Originally Posted by trancefanatic View Post
    i hated simultaneous equations and never did understand it... you can graph those??
    yes, they are straight lines. they are relatively easy to graph, especially in comparison to a cubic, which is what i believe #4 is.

    to graph the lines, find the x- and y-intercepts, plot them on the axis, and draw a straight line through them, that's it. you do know how to find the intercepts, right?
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  7. #7
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    Quote Originally Posted by Jhevon View Post

    EDIT: ok, i got (i wrote them as the lines you will plot, so i replaced n with x and a_n with y):

    #1: y = 2 - 3x

    #2: y = 2x - 7

    #3: y = 2 - 2x
    i got
    y=2x-7
    y=-2x+2
    y=-3x+2

    which i think are the same as yours. on #4 i tried to see if there were any patterns and ended up factoring 18. but that's pretty much it...
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    yes, they are straight lines. they are relatively easy to graph, especially in comparison to a cubic, which is what i believe #4 is.

    to graph the lines, find the x- and y-intercepts, plot them on the axis, and draw a straight line through them, that's it. you do know how to find the intercepts, right?
    yes, i realized that now that i thought about what simultaneous equations looks like.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by trancefanatic View Post
    yes, i realized that now that i thought about what simultaneous equations looks like.
    a hint for #4:

    Let the cubic be of the form: y = ax^3 + bx^2 + cx + d

    Note that: f(1) = 6, f(2) = 2, f(3) = 40, and f(4) = 60

    do you think you can take it from here? to set up the simultaneous equations and solve them?
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    a hint for #4:

    Let the cubic be of the form: y = ax^3 + bx^2 + cx + d

    Note that: f(1) = 6, f(2) = 2, f(3) = 40, and f(4) = 60

    do you think you can take it from here? to set up the simultaneous equations and solve them?
    i'm definitely lost...
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    Quote Originally Posted by trancefanatic View Post
    i'm definitely lost...
    well, f(x) = y, so we have f(x) = ax^3 + bx^2 + cx + d

    " f(1) = 6" means, when x = 1, y= 6

    so since f(1) = 6, we have:

    6 = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d

    since f(2) = 2, we have:

    2 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d

    since f(3) = 40, we have:

    40 = a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d

    since f(4) = 60, we have:

    60 = a(4)^3 + b(4)^2 + c(4) + d = 64a + 16b + 4c + d


    Thus, to find the coefficients in our cubic, we need to solve the system:

    6 = a + b + c + d ......................(1)

    2 = 8a + 4b + 2c + d .................(2)

    40 = 27a + 9b + 3c + d ..............(3)

    60 = 64a + 16b + 4c + d ............(4)


    Hopefully you can do that
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  12. #12
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    OH ok now i get it. you didn't have to do them all, but i really appreciate it! thanks. i'll probably redo them just for practice and comprehension check.
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