Could someone summary how to find the restrictions for problems like the ones below?

Question

3x-27/19x+7

3x+27/18x

3x^2/4x^2+9x+8

3x/9x^3

sry if these don't work out perfectly as i'm making them up in my head. so pretty much, the question is how do you find the domain, range, and vertical/& horizontal asymptotes?

all i know is that if the degree is same in the numerator and the denominator then to find the horizontal asymptote just divide the leading coefficient in the numerator by the leading coefficient in the denominator. to find the vertical asymptote, set teh denominator to zero. but these only works if the degree is the same in both the numerator and denominator. what about if the degrees are different?


b) how would you also find the domain and range of exponential problems? like (1/5)(5^x)=y

Answer 1

Good questions!

The easiest one is domain. The only real numbers *not* in the domain of a rational function are the ones that make you divide by 0.

You will find a vertical asymptote only at these numbers, but there isn't always one. For example, the function y=(x-2)/(x-2) is a horizontal line with a hole at (2,1). When a factor appears upstairs and downstairs, as in this example, there's a little tug-of-war. The easiest way to explain it is this: cancel as many of that factor as you can and graph the function accordingly--but at the very end, if there isn't a vertical asymptote at that x-value, put a hole there. For the example I gave, I would cancel both factors of (x-2) and be left with y=1. I would graph that line, and then make a hole in the graph at x=2. Another example (the reciprocal of one of yours): y = (9x^3)/(3x). We cancel as much as possible, getting y = 3x^2. Nice, a parabola. We graph it, and then put a hole in it at x=0 (because our original function is undefined at x=0).

Next, we have the non-vertical asymptote. You mentioned horizontal asymptotes, and most books on the topic mention slant asymptotes, but the truth is more elaborate: ANY polynomial can serve as an asymptote for a rational function. Let's take these cases one at a time.

Case 1: Denominator is higher degree than numerator.
When downstairs wins, the rational function decays over time. The x-axis is the horizontal asymptote.

Example: y=(2x+2)(x+3)/[(x-1)^2(x+1)]
Multiply out: y = (2x^2+8x+6)/(x^3-x^2-x+1)
Notice how the numerator is degree 2, while the denominator is degree 3? Let's reduce both by the smaller number. That is, divide top and bottom by x^2.
y = (2 + 8/x + 6/x^2) / (x - 1 - 1/x + 1/x^2)
For large x, the numerator gets close to 2, while the denominator gets big. So, the whole souffle collapses.

Case 2: Numerator and Denominator have same degree.
In this case, they strike a balance at some horizontal asymptote *other* than the x-axis. You know the shortcut: compare leading coefficients. That's actually a fast version of the trick I did a few lines ago. A slow-mo example:

y = 3x(x+2)/(2x+1)^2 = (3x^2+6x) / (4x^2+4x+1)
Degree 2, top and bottom. Divide both by x^2:
y = (3 + 6/x) / (4 + 4/x + 1/x^2)
For large x, the fractional terms vanish, leaving us with 3/4. One thing I want to point out is that the leading coefficient shortcut works only when top and bottom are multiplied out. Don't let the original form of this function fool you into saying the horizontal asymptote is at 3/2.

Case 3: Numerator is higher degree than denominator.
Rational functions in this category do not have horizontal asymptotes at all. If you're willing to blur your vision a little, you could just compare the leading terms to get a sense of the function's end behavior. Example:

y= (x^3+x^2-6x) / (x^2 - x - 2)

If we just look at x^3 / x^2, we get y~x. What we learn from this:
(1) As x gets large (positive), so does y.
(2) As x gets large (negative), so does y.
(3) This function has a slant asymptote of slope 1.

However, y=x is NOT the slant asymptote. I know--cruel trick, huh. In fact, our usual "divide top and bottom by some power of x" routine is also no good.

Good news: there is a single catch-all method to handle all of these Case 3 rational functions. Polynomial Long Division is its name. (Those three words are good for searching with, if you want instructions.) When I used polynomial long division, I found that (x^3+x^2-6x)/(x^2-x-2) = x + 2 + (-2x+4)/(x^2-x-2).

That last cluttery bit is the leftovers; the polynomial part (x+2) is the asymptote. To be clear, it's the slant asymptote y=x+2.

As an alternative to polynomial long division, we can look for common factors:

(x^3+x^2-6x)/(x^2-x-2) = [x(x^2+x-6)]/[(x-2)(x+1)]
= [x(x+3)(x-2)]/[(x-2)(x+1)]
= x(x+3)/(x+1) (with a hole at x=2)
= (x^2+3x) / (x+1)

This gives us a smaller polynomial to work with; polynomial long division would be easier now. Or, we could use Synthetic Division.

You also asked about the range of a rational function. If you're lucky, it has a strand that reaches from bottom to top, so the range is all reals. That doesn't always happen, though; you frequently need calculus (or mad algebra skills) in order to figure out exactly what the range is.

Answer 2

Domain - whenever numerator and denominator are both defined, and denominator is not zero. In these examples, numerator and denominator are always defined, so the domain is all real numbers except where the denominator is 0.

Horizontal asymptote - as you said for equal degree; if the denominator has higher degree, horizontal asymptote is y = 0; if the numerator has higher degree, there are no horizontal asymptotes.

Vertical asymptote - set denominator to 0, for any degree. Check that the numerator is not 0 at this point. If both are 0, simplify the expression and repeat. If denominator is never 0 there are no vertical asymptotes.

Range is tricky. If numerator and denominator have equal degree, but do not share any linear factors, then the range will be all of R except the horizontal asymptote. Otherwise it's hard to describe generally. For example, in your last case the function simplifies to 1 / 3x^2 which has a range of (0, ∞), but if it was 1 / 3x^3 it would have a range of all R.

b) a^b is defined for every b if a>0, and for b > 0 if a = 0. Also, if a>0 then a^b > 0 for every b. So for y = 5^x the domain is all of R and the range is (0, ∞). (Hence the same for y = (1/5) 5^x.)