WHY!! asmyptotes -_- (1 Viewer)

[ajyt]

It might be easier remember that if Deg(numerator) >= Deg(denomenator), then there will be an oblique asymptote. To find its equation, do long division with the fraction (eg, x divided by x+1). Then the oblique asymptote has equation: y=quotient

 

[Seraph]

hmm people isnt it possible to find all asymptotes by dividing the numerator by the denominator

how is this done???

 

[Eeko]

where the denom cant = 0 hehehehe

 

[shkspeare]

yup then make like p(x)/a(x) = q(x) + r(x)/a(x)

then q(x) will be yer non vertical asymptote (usin limits)

this is when the degree of num > degree of denom

 

[CM_Tutor]

For y = x / (x + 1), the vertical asymptote (x = -1) is easily found - just look when the denominator is zero. For the horizontal asymptote, I use limits. To do this, divide every term by the highest power of x in the denominator. In this case it's x, so we get:

y = x / (x + 1) = (x / x) / (x / x + 1 / x) = 1 / (1 + 1 / x)

As x increases (or decreases) without bound - ie goes to infinity - the 1 / x term goes to zero (as would any term of the form 1 / x^n for n a positive integer). Thus,

y ---> 1 / (1 + 0) = 1. Thus, y = 1 is a horizontal asymptote.

The long division method also works, but its useful to remember the easy way to do this - rewrite the numerator until its the same as the denominator (+ extra terms) and then separate. In this case:

y = x / (x + 1) = (x + 1 - 1) / (x+1) = (x + 1) / (x + 1) - 1 / (x + 1) = 1 - 1 / (x + 1)

Now 1 / (x + 1) can be very close to zero, but it can't ever equal zero. Thus, y can be close to (but never reach) 1. Thus, y = 1 is a horizontal asymptote.

By the way, anyone who prefers the 'make x the subject' approach should be familiar with one of the other methods as well. For example, limits will show that y = (3x^2 - 2) / (x^2 + 5x - 7) has a horizontal asymptote at y = 3 quite easily, but you wouldn't want to have to try to make x the subject!