Curve Sketching using derivatives

[preet]

When you're looking for vertical asymptotes in the graph of a rational function, you check to see whether or not the denominator has any values of x that could make it zero. I don't have any problems understanding this, or visualizing it.

However, when you check for horizontal or oblique asymptotes, I don't understand the logic of either method. This is the way I've learned to do it...

For a horizontal asymptote [first check to see if the function has one, the denominator must have a greater to or equal order exponent from the numerator] divide by the highest order term and evaluate the limit as x -> infinity.

For an oblique asymptote [first check to see if the function has one, the numerator must be higher order than the denominator] divide the numerator by the denominator and evaluate the lim as x -> infinity.

For horizontal asymptotes, I understand the part about evaluating the limit... the expected value for the limit will be what y approaches but never touches. But I don't understand how the 'check' works or why you divide by the highest order term before evaluating the limit as x-> infinity.

The same goes for oblique asymptotes. I would think that when you just evaluate the rational function for x-> infinity, you would get a linear equation that tells you what y is approaching... again, I don't understand how the check works, or why you divide the numerator by the denominator... can someone explain?

Thanks in advance
Preet

 

[HallsofIvy]

For horizontal asymptotes, I understand the part about evaluating the limit... the expected value for the limit will be what y approaches but never touches. But I don't understand how the 'check' works or why you divide by the highest order term before evaluating the limit as x-> infinity.

xⁿ, as x goes to infinity, for n> 0, is undefined. On the other hand, x^-n, as n goes to infinity, is 0 and a lot easier to handle. That's why you divide each term, in both numerator and denominator, by the highest power of x: to get as many negative powers as you can.

The same goes for oblique asymptotes. I would think that when you just evaluate the rational function for x-> infinity, you would get a linear equation that tells you what y is approaching... again, I don't understand how the check works, or why you divide the numerator by the denominator... can someone explain?

A fractions is a division! P(x)/Q(x) means "P(x) divided by Q(x)". Obviously, if the degree of the denominator, Q, is greater than that of P, then the fraction goes to 0 as x goes to infinity and so y= 0 is a horizontal asymptote. If degree of Q is greater than or equal to degree of P, then P(x)/Q(x)= A(x)+ B(x)/C(x) where, again, degree of C is greater than degree of B (A(x) is the "quotient", B(x) is the "remainder" when you divide). Now, as x goes to infinity, B(x)/C(x) goes to 0 leaving A(x) as the asymptote. By the way, that is NOT necessarily linear. If degree of Q is equal to the degree of P, then A(x) is a constant and there is a horizontal, y not equal to 0, asymptote. If the degree of Q is exactly one more than the degree of P, then A(x) is linear and there is a linear, oblique, asymptote. If degree of the Q is 2 or more greater than the degree of P, then A(x) is a polynomial of degree 2 or more and there is a "curved" asymptote. You've probably only seen examples of straight line asymptotes.

 

[benorin]

If y=h(x) is a slant asymptote of f(x), then we must have

$$\lim_{x\rightarrow \infty} (f(x)-h(x))=0$$

suppose that f(x) is a rational function such that the degree of the numerator is higher than that of the denominator [i.e. degree(P)>=degree(Q) ], then we have

$$f(x):=\frac{P(x)}{Q(x)}= g(x)+\frac{R(x)}{Q(x)}$$

where the latter representation is obtained by long division, g(x) denotes the quotient and R(x) denotes the remainder which is such that degree(R)<degree(Q), so that

$$\lim_{x\rightarrow \infty} \frac{R(x)}{Q(x)}=0$$

now we have

$$\lim_{x\rightarrow \infty} (f(x)-h(x)) = \lim_{x\rightarrow \infty}\left( g(x)+\frac{R(x)}{Q(x)} -h(x)\right) = \lim_{x\rightarrow \infty}\left( g(x)-h(x)\right) = 0$$

so clearly if if we take the quotient g(x) to be the slant asymptote h(x) it will work, that is the above limit will indeed be zero.