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Math Help - Graphs and asymptotes

  1. #1
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    Graphs and asymptotes

    State the conditions on k such that the graph of  y=\frac{1}{x^2+kx+16} has:

    a) two vertical asympototes
    b) only one vertical asymptote
    c) no vertical asymptotes.

    Answers:
    a) (-\infty ,-8)\cup(8,\infty)
    b) k=\pm8
    c) (-8,8)

    However if the equation above were to have two vertical asymptotes wouldn't the factorised denominator of the equation above be:

     \frac{1}{(x+8)(x+2)} or \frac{1}{(x-8)(x-2)} or \frac{1}{(x+1)(x+16)} or \frac{1}{(x-1)(x-16)}

    So wouldn't k=\pm10 or k=\pm17? If not, why not?

    And if it were to have only one vertical asymptote, the equation with the factorised denominator would be \frac{1}{(x+4)^2} or \frac{1}{(x-4)^2}

    Hence k=\pm8 makes sense.

    If the equation were to have no vertical asymptote, wouldn't x simply equal x=0? If x did equal zero, how would you find the range of values for k with no vertical asymptote?

    Any help regarding questions a and c above and how to get the answer would be appreciated!
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  2. #2
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    Quote Originally Posted by Joker37 View Post
    However if the equation above were to have two vertical asymptotes wouldn't the factorised denominator of the equation above be:

     \frac{1}{(x+8)(x+2)} or \frac{1}{(x-8)(x-2)} or \frac{1}{(x+1)(x+16)} or \frac{1}{(x-1)(x-16)}

    So wouldn't k=\pm10 or k=\pm17? If not, why not?
    k=\pm10 or k=\pm17 certainly work. Note that these values belong to (-\infty ,-8)\cup(8,\infty). \pm10 and \pm17 are not the whole answer because when the denominator factors into (x-a)(x-b), a and b don't have to be integer.

    If the equation were to have no vertical asymptote, wouldn't x simply equal x=0?
    I am not sure I understand. Under what conditions would x be 0? The question was about the possible values of k.

    The graph of \frac{1}{x^2+kx+16} has n vertical asymptotes, where n=0,1,2, iff x^2+kx+16=0 has n roots, so all you need is to examine the discriminant of this quadratic equation.
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  3. #3
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    Quote Originally Posted by emakarov View Post
    k=\pm10 or k=\pm17 certainly work. Note that these values belong to (-\infty ,-8)\cup(8,\infty).
    But how do you know that the values belong to (-\infty, -8)\cup(8,\infty) based on the information given? What are the steps to figuring this out?


    Quote Originally Posted by emakarov View Post
    The graph of \frac{1}{x^2+kx+16} has n vertical asymptotes, where n=0,1,2, iff x^2+kx+16=0 has n roots, so all you need is to examine the discriminant of this quadratic equation.
    But if I did this wouldn't that mean that there would be two unknown variables in the one equation. I'm sorry, I'm still a bit confused. I'm not very good at maths and need further assistance.
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  4. #4
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    The discriminant of x^2+kx+16 is k^2-4\cdot16=k^2-64. There are two roots iff k^2-64>0, i.e., k<-8 or k>8. There is one root iff k^2-64=0, i.e., k=\pm8. There are no roots iff k^2-64<0, i.e., -8<k<8.
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  5. #5
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    Quote Originally Posted by Joker37 View Post
    State the conditions on k such that the graph of  y=\frac{1}{x^2+kx+16} has:

    a) two vertical asympototes
    b) only one vertical asymptote
    c) no vertical asymptotes.

    Answers:
    a) (-\infty ,-8)\cup(8,\infty)
    b) k=\pm8
    c) (-8,8)

    However if the equation above were to have two vertical asymptotes wouldn't the factorised denominator of the equation above be:

     \frac{1}{(x+8)(x+2)} or \frac{1}{(x-8)(x-2)} or \frac{1}{(x+1)(x+16)} or \frac{1}{(x-1)(x-16)}

    So wouldn't k=\pm10 or k=\pm17? If not, why not?
    In order to have two asyptotes, the denominator must factor into two distinct factors, but not necessarily those two!
    [tex]x^2+ kt+ 16= (x+ a)(x+ b)= x+ (a+b)x+ ab. Take a and b to be any two numbers, not necessarily integers, calculate k from that. Take a to be any number at all, and calculate b= 16/a. The only values of a that won't work is a= 4 or a= -4 because then b= 16/4= 4 or b= 16/-4= -4 so a and b are not different. If a and b are both 4, or a and b are both -4, then k= 8. For any other k, there will be two factors.

    And if it were to have only one vertical asymptote, the equation with the factorised denominator would be \frac{1}{(x+4)^2} or \frac{1}{(x-4)^2}

    Hence k=\pm8 makes sense.
    Yes, good!

    If the equation were to have no vertical asymptote, wouldn't x simply equal x=0? If x did equal zero, how would you find the range of values for k with no vertical asymptote?
    What? No, x is the independent variable. It can be any number (when there are no asymptotes). And there will be no asymptotes if and only the denominator is never 0. The values of x that make a quadratic, ax^2+ bx+ c, 0 are given by the quadratic formula: \frac{-b\pm\sqrt{b^2- 4ac}}{2a}. There will be NO real number x that makes that 0 if the "discriminant", b^2- 4ac is negative. Here, your quadratic is x^2+ kx+16 so a= 1, b= k, c= 16. That means that the discriminant is k^2- 4(1)(16)= k^2- 64. There will be no real x that makes the quadrtic in the denominator 0, and so no vertical asymptote as long as k^2- 64< 0 which is the same as -8< k< 8.

    Any help regarding questions a and c above and how to get the answer would be appreciated!
    Thanks to Emakarov for pointing out a typo.
    Last edited by HallsofIvy; January 27th 2011 at 05:08 PM.
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  6. #6
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    Quote Originally Posted by Joker37 View Post
    State the conditions on k such that the graph of

     y=\frac{1}{x^2+kx+16}

    has:

    a) two vertical asympototes
    b) only one vertical asymptote
    c) no vertical asymptotes.


    Answers:

    a) (-\infty ,-8)\cup(8,\infty)

    b) k=\pm8

    c) (-8,8)

    However if the equation above were to have two vertical asymptotes wouldn't the factorised denominator of the equation above be:

     \frac{1}{(x+8)(x+2)},\;\; or\;\;\frac{1}{(x-8)(x-2)},\;\; or \;\;\frac{1}{(x+1)(x+16)},\;\; or\;\; \frac{1}{(x-1)(x-16)}

    So wouldn't

    k=\pm10,\;\; or \;\;k=\pm17\;\;?

    If not, why not ?
    This problem is similar to your recent one containing roots.

    x^2+bx+c=(x-\alpha)(x-\beta)

    You have found the "integer roots".
    However, 16 factorises in numerous ways... 32 and 0.5 being another example.

    (x-32)(x-0.5)=x^2-0.5x-32x+16=x^2-32.5x+16

    So you see, there are endless values of k possible.

    You can also write the roots of the denominator being zero using the quadratic formula.

    \displaystyle\ x^2+kx+16=0\Rightarrow\alpha,\;\beta=\frac{-k\pm\sqrt{k^2-(4)16}}{2}=\frac{-k\pm\sqrt{k^2-64}}{2}

    -\infty<k<-8 gives a positive value under the square root.

    8<k<\infty also gives a positive value under the square root.

    These values of "k" cause \alpha,\;\beta to be distinct.

    We cannot have a negative value under the square root.
    That accounts for the solution for (a).


    And if it were to have only one vertical asymptote, the equation with the factorised denominator would be

    \frac{1}{(x+4)^2}\;\;or \;\;\frac{1}{(x-4)^2}

    Hence

    k=\pm8

    makes sense.
    Good!

    If the equation were to have no vertical asymptote, wouldn't x simply equal x=0? If x did equal zero, how would you find the range of values for k with no vertical asymptote?

    Any help regarding questions a and c above and how to get the answer would be appreciated!
    Yes, x=0 works. However, the question refers to the possible values of "k".

    However, if the value of k lies in between -8 and 8, then x^2+kx+16 is never zero,

    and so the denominator will not be zero and cause the function to have an asymptote.

    If a quadratic equation has only complex roots (no real roots), then it never crosses
    the x-axis, so it is never zero.
    You have all of that to study.
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  7. #7
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    Thanks for the help everybody!
    I understand now.
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