How to find the vertical asymptotes and holes of a rational function (precalculus)

This precalculus tutorial covers finding the vertical asymptotes of a rational function and finding the holes of a rational function. We first set the denominator equal to 0 in order to find the candidates for the vertical asymptotes and the holes. Remember, "nonzero/0" gives you a vertical asymptote, and "0/0 from the original and no nonzero/0 from the reduced" gives you a hole.
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#precalculus
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5:41 0/0 but still gives you a vertical asymptote

Пікірлер: 18

  • @user-sf1gn1iu5j
    @user-sf1gn1iu5j8 ай бұрын

    What a master of exchanging markers....😮

  • @thebingus7243
    @thebingus72439 ай бұрын

    Thank you king!!! Helping me pass my college algebra class !!!

  • @Kai26448
    @Kai264487 ай бұрын

    OMG NOW EVERYTHING IS SEEMS EAZIER THANK YOU ❤

  • @OptimusPhillip
    @OptimusPhillip Жыл бұрын

    First, I factored the denominator to find the poles of this function. I got x = -1 and x = 4. Then, I factored the numerator to see if it shared any zeros with the denominator. I found that both the top and bottom were 0 at x = -1. So I did zero-pole cancellation, and plugged -1 into the cancelled function to see if it approached a finite value. I found that the function had a hole at (-1,0.4). So the function has a vertical asymptote at x = 4, and a hole at x = -1

  • @domosautomotive1929
    @domosautomotive1929 Жыл бұрын

    2 words sum this up.....removable discontinuity

  • @Ninja20704

    @Ninja20704

    Жыл бұрын

    Yeah that’s the more official name

  • @omargoodman2999

    @omargoodman2999

    Жыл бұрын

    Right, but this demonstrates *why,* not what. It's one thing to just say, "this is what it is, this is how to do it, just memorize the process". But by demonstrating the difference, how with one type the equation *would be* continuous if not for the discontinuity, but with the other it's an automatic V.A. in *in spite of* that one exclusion, you understand the reasoning behind the method.

  • @ManjulaMathew-wb3zn
    @ManjulaMathew-wb3zn5 ай бұрын

    The easier way to explain hole at x=a is if direct substitution creates 0/0 and the limit as x->a is finite. If the limit is infinite it becomes an asymptote regardless of the numerator being zero or not.

  • @fredq4332
    @fredq433210 ай бұрын

    yo thank you for this

  • @annfyannfyud2857
    @annfyannfyud28573 күн бұрын

    Thanks!! I was looking for solution when both sides of equation can't be zero. I know this one original is L'Hôpital way

  • @Sg190th
    @Sg190th Жыл бұрын

    well now I know why there's a hole. 0/0

  • @nyandyn
    @nyandyn Жыл бұрын

    Somehow I find L'Hôpital way more straightforward than factoring and reduction 😅

  • @solidpixel

    @solidpixel

    Жыл бұрын

    you do realize you are watching a *PRECALCULUS* tutorial right?

  • @nyandyn

    @nyandyn

    Жыл бұрын

    @@solidpixel Yes. I was pointing out that sometimes the more advanced tools you haven't learnt yet are more convenient.

  • @opus_X
    @opus_X11 ай бұрын

    How does he factor it so fast?

  • @daniaalsadi7208

    @daniaalsadi7208

    11 ай бұрын

    If you practice anything you will ace it and it will be just easy

  • @ethanmatthewapostol881

    @ethanmatthewapostol881

    9 ай бұрын

    you first need to find what makes a number into that from multiplication so like, x^2-x+12 right? then what makes a 12 is 3x4, so it makes sense because if you minus 3-4, you get -1 but in this case, -x. I don't know if I got that correctly nor used the right equation, that's how I was taught

  • @lookingforahookup
    @lookingforahookup7 ай бұрын

    Anything divided into 0 is undefined