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Math Help - Asymptotes and intercepts: help checking work

  1. #1
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    Asymptotes and intercepts: help checking work

    I had to find the vertical and horizontal asymptotes as well as the x an y intercepts..could someone please help me by checking my work...there are only 8 problems...you do not have to check them all, but I wouldreally appreciate the help because I do not know if I am doing these correctly....thank you


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  2. #2
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    Hello, aikenfan!

    You did really good work . . . with a few slips here and there.



    12)\;\;f(x)\:=\:-\frac{x+2}{x+4}
    That minus-sign is in front of the fraction . . .

    For the x-intercept, we have: . -(x +2)\:=\:0 \quad\Rightarrow\quad x + 2 \:=\:0

    . . x \:=\:-2\quad\Rightarrow\quad (-2,0)



    18)\;\;f(x)\:=\:\frac{x^2-25}{x^2+5x}
    VA: . x^2+5x \:=\:0 \quad\Rightarrow\quad x(x+5)\:=\:0\quad\Rightarrow\quad x \:=\:0,-5

    Vertical asymptotes: . x = 0,\;x = -5



    43)\;\;f(x) \:=\:\frac{2x^2+1}{x}
    x-intercept: . 2x^2+1\:=\:0\quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2}\quad\Rightarrow\quad x \:=\:\sqrt{-\frac{1}{2}} . . . not a real number.

    There is no x-intercept.



    44)\;\;g(x) \:=\:\frac{1-x^2}{x}
    x-intercepts: . 1-x^2\:=\:0\quad\Rightarrow\quad x^2 \:=\:1\quad\Rightarrow\quad x \:=\:\pm1\quad\Rightarrow\quad(1,0),\;(-1,0)


    45)\;\;h(x) \:=\:\frac{x^2}{x-1}
    y-intercept: . y \:=\:\frac{0^2}{0-1} \:=\:0\quad\Rightarrow\quad (0,\,0)

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  3. #3
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    Thank you very much for all of your help...I just have one more quick question, if you don't mind...for number 18, how do i figure out the horizontal asymptote? I think it would be 1, but I'm not sure.
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    Hello again, aikenfan!

    For number 18, how do i figure out the horizontal asymptote?
    Yes, the horizontal asymptote is: y = 1


    We want: . \lim_{x\to\infty}\frac{x^2-25}{x^2+5x}

    Divide top and bottom by x^2: . \lim_{x\to\infty}\frac{\frac{x^2}{x^2} - \frac{25}{x^2}}{\frac{x^2}{x^2} + \frac{5x}{x^2}} . **

    Then: . \lim_{x\to\infty}\frac{1 - \frac{25}{x^2}}{1 + \frac{5}{x}} \;=\;\frac{1-0}{1+0} \;=\;1


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Rule: Divide top and bottom by the highest power of x
    . . . . in the denominator.

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    is up to his old tricks again! Jhevon's Avatar
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    Note that we should find \lim_{x \to - \infty} \frac {x^2 - 25}{x^2 + 5} as well. But in this case, it would give the same value, so that's fine
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