anonymous
  • anonymous
What are the vertical asymptotes, horizontal asymptotes, slant asymptotes, and the holes of this equation: (x^3+5x^2)/(x^2-25)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
There is no horizontal asymptotes if numerator's degree greater than denominator. You should make Polynomial Long Division to have slant asymtotes. to find holes and vertical asymtotes make denominator=0
anonymous
  • anonymous
ok. so what are the holes and vertical asymptotes because i am not sure if i did it right
anonymous
  • anonymous
Vertical asymptotes is like Hakanka said, set the denominator equal to zero and solve. Holes come from factoring both the numerator and the denominator and cancelling a common factor...

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anonymous
  • anonymous
ok but im not sure if i did it correctly so can you tell me what the holes and vertical asymptotes are so i can see if i am correct...
anonymous
  • anonymous
\[x^2-25=0\] gives x=-5,5 when you factor everything you get \[x(x+5)(x+5)/((x+5)(x-5))\]
anonymous
  • anonymous
I am sorry for giving wrong info ,Yes, Holes come from factoring both the numerator and the denominator thanks
anonymous
  • anonymous
ok so then what is the slant asymptote and the vertical asymptote?
anonymous
  • anonymous
The vertical asymptote is at x=5, technically it would be at -5 as well, since those are both answers we get when the denominator is set to zero, but there is a hole in the graph there instead. Slant asymptote comes from dividing the denominator into the numerator.
anonymous
  • anonymous
ok, but then what are the holes?
anonymous
  • anonymous
There is a hole at x=-5 because the factor (x+5) appears both in the numerator and the denominator.
anonymous
  • anonymous
ok so tell me if i am right: holes=-5,5 vertical asymptote= -5,5 slant= ?? i dont know. please tell me horizontal asymptote= 0
anonymous
  • anonymous
Ok I want to first say one thing, I factor wrong because I wrote the wrong number down, but it doesn't change any of the answers. Should have factored like \[x^2(x+5)/((x+5)(x-5))\] For some dumb reason I wrote 25 instead of 5 down. But anyway, vertical asymptote is just x=5 you also get x=-5 when you solve, but when you check for holes, you get a hole at -5 which overrides the asymptote. There is no horizontal asymptote.
anonymous
  • anonymous
ok. so what is the slant??
anonymous
  • anonymous
When you take the denominator and divide it into the numerator you get x+5 with a remainder, which doesn't matter so the slant asymptote is x+5. If you were to graph this function, you would see all of these asymptotes in action :)
anonymous
  • anonymous
ok so once again: the vertical asymptote= 5 the holes = -5 is that right?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
thanks man
anonymous
  • anonymous
No problem.
anonymous
  • anonymous
sorry, one more question. is the slant x-5 or x+5 because i think it is x-5 but im not sure
anonymous
  • anonymous
It's x+5, x-5 would go through the graph. x+5 is what you get when you divide correctly :) and doesn't cross the graph.

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