Tell me why it took me 2 semesters to figure this out and now i clicked on this video and i FINALLY understand this in 8 mins………. Wow. I’m convinced Professors are gatekeeping. Thanks bro ✊🏻 you the goat fr
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
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.
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.
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
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What a master of exchanging markers....😮
Tell me why it took me 2 semesters to figure this out and now i clicked on this video and i FINALLY understand this in 8 mins………. Wow. I’m convinced Professors are gatekeeping. Thanks bro ✊🏻 you the goat fr
Thank you king!!! Helping me pass my college algebra class !!!
OMG NOW EVERYTHING IS SEEMS EAZIER THANK YOU ❤
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
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.
2 words sum this up.....removable discontinuity
Yeah that’s the more official name
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.
Thanks!! I was looking for solution when both sides of equation can't be zero. I know this one original is L'Hôpital way
yo thank you for this
Somehow I find L'Hôpital way more straightforward than factoring and reduction 😅
you do realize you are watching a *PRECALCULUS* tutorial right?
@@solidpixel Yes. I was pointing out that sometimes the more advanced tools you haven't learnt yet are more convenient.
well now I know why there's a hole. 0/0
How does he factor it so fast?
If you practice anything you will ace it and it will be just easy
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
Anything divided into 0 is undefined