What's the horizontal & vertical asymptote of this equation??

Question

I know you have to make the denominator zero to find the vertical asymptote, but I'm unsure as to how to solve it:

y= x^2 (divided by) x(x-1)

Also, in order to find the horizontal asymptote, I know that the degree of the numerator is 1 but what is the degree of the denominator?

Thank you for your help.

Answer 1

NOTE: i know this looks long, but i put in a lot of explanation to help on this problem and any others like it. i think (and hope) that youll find very useful! :)


youre right about the vertical asymptote(VA), you just set the denominator equal to zero:

x(x-1) = 0

you solve it using the zero product theorem. that says that when the product of two things equals zero, at least one of those individual parts must be equal to zero. (because the only way to get zero by multiplying is to multiply by zero. sorry, you probably already knew all that. anyway, you get

x=0
and
x-1=0, which you solve to get x=1

that tells you that the function is undefined at x=0 and 1, but those arent necessarily both VAs. if any part of the denominator can 'cancel' with the a part of the numerator, its considered a 'removable discontinuity' because it cancels out in theory, but you cant really do that because that would totally change the function.

long story short, an x cancels out, so the discontinuity you got from that piece (x=0) is a hole in the graph, not a VA. the x-1 cant cancel, so you know that x=1 is your one and only VA.


about the HA . . .

the degree of the numerator is actually 2. remember that the degree is the highest power of x. since you have x^2 in the numerator, its degree is two.

to get the degree of the denominator, you have to look at what the highest power of x would be if you distributed and FOILed it all the way. if it helps you, go ahead and do that. all you really have to do is count the x's that will be multiplied together though. there are 2, which you know will give you x^2, so the degree of the denominator is also 2.

the rule is that since the degrees of the numerator and denominator are the same, your HA will be y=(leading coeffiecient of the numerator)/(leading coeff. of the denominator)

remember that the leading coeff. is the coeff of the x with the highest power.

for the numerator thats easy, its 1

for the denominator, you have you use the distributed version, which is also 1.

follow the rule, and y=1/1, or y=1



there you have it! final answer . . .

VA: x=1
HA: y=1


of course feel free to send me a message if you have a question :)

Answer 2

A. vertical asymptot

The vertical asymptot will be at point where
x(x-1) = 0
i.e x=0 or x=1

Since I can't have a zero in the denominator, then I can't have x = 0 or x = 1 in the domain. This tells me that the vertical asymptotes (which tell me where the graph can't go) will be at x = 0 or x = 1.

B. Horizontal asymptote

To find it we consider
lim x---> infty (y)
= lim x---> infty [x^2 / (x(x-1)) ] = lim x---> infty [x^2 / (x^2-x)]
= lim x---> infty [2x / (2x-1))]
= lim x---> infty [2 / (2]
=1
[This is because, if you want to avoid 0/0 or any any indefinate form as a result of limits, you take derivatives on numerator and denominator, according to a well known theorem]

So, the horizontal aymptot is at y = 1.
That also justified by the graph.

I hope, i answere your question.

All the best.





And that what the graph suggest.

Answer 3

one million. Vertical asymptotes ensue for values of the variable that makes the denominator 0, yet, would not make the numerator 0 ... on the grounds that x= -one million makes the denominator 0 and the numerator not 0, x= -one million is the vertical asymptote. 2. Horizontal asymptotes ensue if the decrease as x is going to inf or -inf exist. because of the fact the decrease of (3x-2) / (x+one million) as x is going to inf is 3, y=3 is a horizontal asymptote. And, because of the fact the decrease of (3x-2) / (x+one million) as x is going to -inf is 3, y=3 is, returned, a horizontal asymptote. ?????????? answer x = -one million ... vertical asymptote y = 3 .... horizontal asymptote .

Answer 4

For the horizontal asymptote, you should look to see what the function approaches as x approaches infinity or negative infinity. Put a large number in, like 100,000, and you will be very close to your asymptote. For the vertical asymptote, look for a value of x that causes the equation to "blow up". That is, is there a number you can put into the equation that causes the denominator to be zero? (There is more than one for this equation.)

Answer 5

A better way is to find the limits

For horizontal, x must approach infinity

1/x approaches zero

This gives y = 1/(1-1/x)

Thus the horizontal asymptote is y=1.

Answer 6

y = x^2/ x(x-1) = x/(x-1)

If x goes to +- infinity the expression will goes to 1 so y=1 is horizontal asymptote

If x goes to 1 form right the expression will goes to + infinity
if x goes to 1 from left the expression will goes to - infinity
so x=1 is vertical asymptote