Question

What are the vertical asymptotes, horizontal asymptotes, slant asymptotes, and the holes of this equation:

(x^3+5x^2)/(x^2-25)

Answer 1

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

Answer 2

ok. so what are the holes and vertical asymptotes because i am not sure if i did it right

Answer 3

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...

Answer 4

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...

Answer 5

\[x^2-25=0\] gives x=-5,5 when you factor everything you get \[x(x+5)(x+5)/((x+5)(x-5))\]

Answer 6

I am sorry for giving wrong info ,Yes, Holes come from factoring both the numerator and the denominator thanks

Answer 7

ok so then what is the slant asymptote and the vertical asymptote?

Answer 8

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.

Answer 9

ok, but then what are the holes?

Answer 10

There is a hole at x=-5 because the factor (x+5) appears both in the numerator and the denominator.

Answer 11

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

Answer 12

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.

Answer 13

ok. so what is the slant??

Answer 14

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 :)

Answer 15

ok so once again:

the vertical asymptote= 5
the holes = -5

is that right?

Answer 16

Yes.

Answer 17

thanks man

Answer 18

No problem.

Answer 19

sorry, one more question. is the slant x-5 or x+5 because i think it is x-5 but im not sure

Answer 20

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.