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Math Help - Advanced functions

  1. #1
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    Advanced functions

    Hello again people. I need a tad of help with my homework. Here are the questions I am having trouble with.

    f(x) =  \frac {7x - m}{2 - nx}

    So, I need to find the values of m and n such that the vertical asymptote is x = 6 and the x intercept is 5. Im guessing for the value to be an assymptote, we have to look at just the bottom and that has to = to 0. So I get
    2 - n(6) = 0
    -2 = -n(6)
    -2/6 = -n = -1/3
    or
    n = 1/3
    but im not sure if thats right.

    Ok now the other question.
    Find the value if p for which f(x) would have a removable discontinuity and then define f(p) so that f(x) = \frac {x^2 - 6x +9}{x-p}

    I have no clue where I am even supposed to start with this question. Someone explain this to me because wouldn't there always be a value that would give you 0 on the bottom?

    Thanks
    Chris
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    Quote Originally Posted by ffezz View Post
    Hello again people. I need a tad of help with my homework. Here are the questions I am having trouble with.

    f(x) =  \frac {7x - m}{2 - nx}

    So, I need to find the values of m and n such that the vertical asymptote is x = 6 and the x intercept is 5. Im guessing for the value to be an assymptote, we have to look at just the bottom and that has to = to 0. So I get
    2 - n(6) = 0
    -2 = -n(6)
    -2/6 = -n = -1/3
    or
    n = 1/3
    but im not sure if thats right.
    That is correct. Now, what condition do you have to make happen to make the x-intercept be 5?

    Quote Originally Posted by ffezz View Post
    Ok now the other question.
    Find the value if p for which f(x) would have a removable discontinuity and then define f(p) so that f(x) = \frac {x^2 - 6x +9}{x-p}
    To have a "removable" discontinuity means that we would have to cancel the factor in the denominator by a factor from the numerator. So what are the factors of the numerator?

    -Dan
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  3. #3
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    ok so umm for the first question because its an x intercept, the y value has to = 0 so i get 7(5) - n which gives me n as -35. So the value of n is -35 and m is 1/3?

    and for the 2nd question

    (x-3)(x-3) so we can factor out and x-3 which means the value of p is 3

    how would i find a value which would make the functions continuous for all values?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ffezz View Post
    ok so umm for the first question because its an x intercept, the y value has to = 0 so i get 7(5) - n which gives me n as -35. So the value of n is -35 and m is 1/3?
    Yes.

    Quote Originally Posted by ffezz View Post
    and for the 2nd question

    (x-3)(x-3) so we can factor out and x-3 which means the value of p is 3

    how would i find a value which would make the functions continuous for all values?
    The given function cannot be made continuous for all values of x. The function \frac{x^2 - 6x + 9}{x - 3} is not defined at x = 3, and thus the function is not continuous there. The best you can do is the function with p = 3.

    -Dan
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  5. #5
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    hehe alright =) i missed a questions though somehow =\

    Find constants a and b such that the function is continuous for all of x
    f(x) = ax+3, if x>5
    8 if x=5
    x^2 + bx + a if x<5
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  6. #6
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    sorry about this but a bump a day keeps page 59 away
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  7. #7
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    Quote Originally Posted by ffezz View Post
    hehe alright =) i missed a questions though somehow =\

    Find constants a and b such that the function is continuous for all of x
    f(x) = ax+3, if x>5
    8 if x=5
    x^2 + bx + a if x<5
    New questions should go in new threads.

    It looks to me like you simply need to find an a and b such that
    ax + 3 = x^2 + bx + a = 8 when x = 5.

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ffezz View Post
    sorry about this but a bump a day keeps page 59 away
    (sigh) You waited a less than two hours to "bump?" Sheesh, that's impatient!

    Bumping is rude. More than that, please see rule #10 here.

    -Dan
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