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Thread: Asymptotes - Help f(x)=x-5x+7/(x-2)

  1. #1
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    Asymptotes - Help f(x)=x-5x+7/(x-2)

    f(x)=x-5x+7/(x-2)

    It asks me to give the vertical, horizontal and slant asymptotes if there are any.

    This is what I have so far:

    vertical asymptote: x cannot equal 2

    I am now trying to find the slant asymptote by synthetic (long division) because x-5x+7 does not factor out.


    I am really stuck so may someone show me how to do this problem? I still need to find the slant and horizontal asymptotes.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by agentlopez View Post
    f(x)=x-5x+7/(x-2)

    It asks me to give the vertical, horizontal and slant asymptotes if there are any.

    This is what I have so far:

    vertical asymptote: x cannot equal 2

    I am now trying to find the slant asymptote by synthetic (long division) because x-5x+7 does not factor out.


    I am really stuck so may someone show me how to do this problem? I still need to find the slant and horizontal asymptotes.
    The vertical asymptote is not "x cannot equal 2" it is the line x = 2.

    $\displaystyle \frac{x^2 - 5x + 7}{x - 2} = x - 3 + \frac{1}{x - 2}$

    So the slant asymptote is $\displaystyle y = x - 3$.

    For horizontal asymptotes we find the limit of the function as x goes to plus or minus infinity. You can see for yourself that this function blows up/down at positive/negative infinity. Or you can simply use the fact that a function with a slant asymptote will never have a horizontal asymptote.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    The vertical asymptote is not "x cannot equal 2" it is the line x = 2.

    $\displaystyle \frac{x^2 - 5x + 7}{x - 2} = x - 3 + \frac{1}{x - 2}$

    So the slant asymptote is $\displaystyle y = x - 3$.

    For horizontal asymptotes we find the limit of the function as x goes to plus or minus infinity. You can see for yourself that this function blows up/down at positive/negative infinity. Or you can simply use the fact that a function with a slant asymptote will never have a horizontal asymptote.

    -Dan
    Can you show me how you got $\displaystyle y = x - 3$ and what would I get as the horizontal asymptote? I know you used synthetic division, but I need help on how to get to $\displaystyle y = x - 3$ and what would be my horizontal asymptote?
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  4. #4
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    You don't need synthetic division:

    $\displaystyle \frac{{x^2 - 5x + 7}}
    {{x - 2}} = \frac{{x^2 - 5x + 6 + 1}}
    {{x - 2}} = \frac{{(x - 2)(x - 3) + 1}}
    {{x - 2}} = x - 3 + \frac{1}
    {{x - 2}}.$

    For horizontal asymptotes, evaluate

    $\displaystyle \lim_{x \to \infty } \frac{{x^2 - 5x + 7}}
    {{x - 2}}$ & $\displaystyle \lim_{x \to -\infty } \frac{{x^2 - 5x + 7}}
    {{x - 2}}.$
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    You don't need synthetic division:

    $\displaystyle \frac{{x^2 - 5x + 7}}
    {{x - 2}} = \frac{{x^2 - 5x + 6 + 1}}
    {{x - 2}} = \frac{{(x - 2)(x - 3) + 1}}
    {{x - 2}} = x - 3 + \frac{1}
    {{x - 2}}.$

    For horizontal asymptotes, evaluate

    $\displaystyle \lim_{x \to \infty } \frac{{x^2 - 5x + 7}}
    {{x - 2}}$ & $\displaystyle \lim_{x \to -\infty } \frac{{x^2 - 5x + 7}}
    {{x - 2}}.$

    Now one has ever showed me that you can just split a number and do that. Something new I learned^^. I still do not understand the horizontal asymptote that you are explaining to me. I need someone to be more detailed with this.
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  6. #6
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    Read carefuly these lines:

    Quote Originally Posted by topsquark View Post
    $\displaystyle \frac{x^2 - 5x + 7}{x - 2} = x - 3 + \frac{1}{x - 2}$

    So the slant asymptote is $\displaystyle y = x - 3$.

    For horizontal asymptotes we find the limit of the function as x goes to plus or minus infinity. You can see for yourself that this function blows up/down at positive/negative infinity. Or you can simply use the fact that a function with a slant asymptote will never have a horizontal asymptote.
    --

    Now, if you evaluate $\displaystyle x - 3 + \frac{1}{x - 2}$ as $\displaystyle x\to\infty$ or $\displaystyle x\to-\infty,$ does that make sense?
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    Read carefuly these lines:



    --

    Now, if you evaluate $\displaystyle x - 3 + \frac{1}{x - 2}$ as $\displaystyle x\to\infty$ or $\displaystyle x\to-\infty,$ does that make sense?

    ^^Not yet^^

    Maybe I was not specific enough when I asked to be more descriptive. I am a beginner at this stuff and I realise that the best way to learn is to figure it out for yourself, but I am really stuck and need this explained 'THOROUGHLY'. Nothing is 'clicking' for me even when you basically just reworded it above. I need this explained as if this is something completely new to me. I really need this so please help me understand.


    What your telling me is all slant asymptotes do not have a horizontal asymptote? I thought you needed the horizontal and vertical asymptotes for when there is a slant asymptote.
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  8. #8
    Senior Member DivideBy0's Avatar
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    Think of it this way...

    As $\displaystyle x \rightarrow \infty$, $\displaystyle f(x) \rightarrow x-3$

    This is because as x goes to infinity, $\displaystyle \frac{1}{x-2}$ becomes smaller and smaller, eventually becoming practically non-existent. But the $\displaystyle x-3$ part keeps rising steadily as $\displaystyle x$ does, so $\displaystyle f(x)$ slowly converges upon $\displaystyle y=x-3$.
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  9. #9
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    You don't need synthetic division:

    How did you know to split the 7 into 6+1?
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  10. #10
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    When I saw Dan's algebra, I was wondering how I could split the original ratio into two sums.

    Then clearly we need to find something on the top which yields $\displaystyle x-2.$ I thought in $\displaystyle (x-2)(x-3)$ 'cause the first two terms are exactly $\displaystyle x^2$ & $\displaystyle -5x,$ so from here, we're done, 'cause the third term is $\displaystyle 6.$ Since on the numerator we have $\displaystyle x^2-5x+7$ the conclusion follows quickly.

    --

    This does not require much effort, it's just a way to avoid the looong division
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  11. #11
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    Quote Originally Posted by DivideBy0 View Post
    Think of it this way...

    As $\displaystyle x \rightarrow \infty$, $\displaystyle f(x) \rightarrow x-3$

    This is because as x goes to infinity, $\displaystyle \frac{1}{x-2}$ becomes smaller and smaller, eventually becoming practically non-existent. But the $\displaystyle x-3$ part keeps rising steadily as $\displaystyle x$ does, so $\displaystyle f(x)$ slowly converges upon $\displaystyle y=x-3$.
    Does that mean the horizontal asymptote is 2?

    x=2
    y=2

    f(x)=x-3

    Is that right?
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  12. #12
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    Quote Originally Posted by agentlopez View Post
    Does that mean the horizontal asymptote is 2?

    x=2
    y=2

    f(x)=x-3

    Is that right?
    Yes, there is a horizontal asymptote at x = 2. Try graphing the function.
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