Rational functions.

A ratio, or quotient, of two polynomials, of two polynomial functions p(x)/q(x), or call them N(x)/D(x) for numerator and denominator polynomials.
Examples:

A polynomial over a polynomial, denominator can't be zero. cf. rational number: integer over integer, no zero denominator.

Interesting stuff: might have:
domain restrictions,
x intercept(s),
y intercept,
absolute or local extrema,
symmetry,
*vertical asymptote(s),
*a horizontal or oblique asymptote,
*"holes"

Domain of rational function is R minus the values that make q(x) denominator zero (each of these will be either a vertical asymptote or a hole).

The polynomials can simplify by cancelling common factors.
x-intercept(s), if any, are the zeros of the simplified numerator p(x), i.e. p(x)=0 Solve. I.e. what value(s) make the numerator zero (and are in the domain).
The simplified numerator function is a polynomial of degree n → can be as many as n X-intercepts, more precisely n or n-2 or ...or 1 or 0 of them.

y-intercept: set x to 0, if possible, i.e. 0 is in domain (else there is no y-intercept).
The ratio of the two constant terms of p and q.
If numerator has no constant and denominator does, y-intercept is 0
If denominator does not have a constant then 0 is not in domain, so there's no y-intercept.

Asymptote: a line that a curve gets closer and closer to.
Vertical asymptote(s), if any, at x's that make the simplified denominator 0, (i.e. after all cancellations between numerator and denominator, i.e. the rational function is in lowest terms.)
Either by factoring the simplified denominator or by graphing the function, seeing the vertical asymptotes behavior and solving the denominator for =0 to identify the x values.
n-degree simplified denominator → as many as n vertical asymptotes.
Both "sides" at the VA go in same "direction" (i.e. both up or both down) or in opposite "directions" (i.e. one up, the other down).
A discontinuity.
Graph them as dotted orange lines (unless is an axis).

An x value that makes the original unsimplified denominator 0 but not the simplified one is a hole.
The x is not in the domain, so that (x,f(x)) point is a single-point hole.
A discontinuity.
Graph the function. Solve the denominator for =0; spot the non-VA solutions on the graph to see that they are holes ("undefined" in Desmos).

If the denominator q, or D, is a factor of p, or N, this can simplify to a polynomial function. Example:

But any x values that make the q(x) zero are "holes".

"End behavior":
What happens as x approaches ±∞.
Horizontal asymptote:
Dominant/leading terms of numerator and denominator: 3 cases
1. If degrees of N and D are the same:
  axn/bxn --> Horizontal asymptote at y = a/b
E.g.: 2x/3x 2/3    5x2/x2 5    3x3/2x3 3/2 ...
  xn/xn --> y=1 Special case when a=b=1.
E.g.: x/x    x2/x2    x3/x3 ...
2. If degree of N is less than degree of D: (a proper rational function)
  xi/xj   i<j --> Horizontal asymptote at y = 0
E.g.: x/x2    x/x3   ... x2/x3    x2/x4    x3/x4 ...
If have a H.A. (cases 1 or 2), both "ends" approach the same H.A.
3. If degree of N is greater than degree of D:
  xi/xj   i>j --> no horizontal asymptote. Ends go to infinities.
E.g.: x2/x    x3/x   ... x3/x2    x4/x2    x4/x3 ...
  Except see below for oblique asymptote if degree of N is one more than degree of D.
If a rational function has a H.A., both "ends" approach the same H.A.
Graphing the function will show horizontal asymptote.

Graph can cross horizontal or oblique asymptote but not vertical asymptotes. Find any x such that R(x)=horizontal asymptote

Symmetry.
Some are odd, some even, some neither even nor odd but have rotational symmetry around some point, others don't.

Examples:
1/x no intercepts, 0∉domain→VA at x=0, HA y=0, odd, decreasing
1/(x-1) no x-intercepts, y-intercept at -1, 1∉domain→VA at x=1, HA y=0
(x4+1)/(x2+1) no x-intercepts, y-intercept at 1, no asymptotes, even
1/x2 no intercepts, 0∉domain→VA at x=0, HA y=0, even
1/(x2-x) no intercepts, 0,1∉domain→ VAs at x=0 and x=1, HA y=0
1/(x3-x2-x) no intercepts, 3 VA at x=0,φ,1-φ, HA y=0:
(x2+5)/(x2+1) no x-intercepts, y-intercept at 5, HA y=1
(x5+1)/(x2+1) x-intercept at -1, y-intercept at 1
(x4-5x2+4)/(x2-3) 4 x-intercepts at ±1 and ±2, y-intercept at -4/3, ±√3∉domain→VAs at x=±√3, no HA, even
(x3+x2-2x)/(x3-7x+6) 2 x-intercepts at -2 and 0, y-intercept at 0, 2 VA at x=-3 and x=2, hole at x=1, HA at y=1

Examples:
Simplest: y=1/x (reciprocal function. Odd. Hyperbola)
e's geometry
1/x and ln
  

Ratio of linear functions:
  

Various examples:
     
  
                 

Exercises:
           

Oblique/slant asymptote: 0 or 1.
If degree of numerator is one more than degree of denominator:
xi+1/xi
E.g.: x2/x    x3/x2    x4/x3    x5/x4 ...
Only one of horizontal and oblique asymptotes are possible.
Divide N by D to get a linear quotient which is the oblique line, ignore remainder.
N÷D = mx+b + r/D
        


Solution/mixture/alloy problem.
Start with an amount A (e.g. liters, kilograms etc.) at a P% concentration of "stuff"; this contains P%*A of "stuff".
Add amount x of pure (100% concentration) stuff to the A solution/mixture/alloy, resulting in new amount A+x which contains P%*A + x of the "stuff".
Resulting concentration/percentage of pure stuff in the new total amount (A+x):   f(x) = (P%*A + x) / (A + x)
    Graph it. Positive x only. Horizontal asymptote at 1 means 100% of the new amount is pure stuff.

Ex.   Add x amount of chemical X to 6 liters of 40% solution of chemical X.
    f(x) = (0.4*6 + x) / (6 + x)
    Adding 5 liters of pure chemical X brings the concentration up to what?
    How much needs to be added to reach a concentration of 90%?

Ex.   You have 1kg of 12-karat (50%) gold. Add some pure gold (24-karat).
    Function that tells the percentage of gold in the resulting alloy for an amount of pure gold added (x):
    f(x) = (.5*1 + x) / (1 + x)
    Adding 2 kg of pure gold results in what percentage of gold:    f(2) = 0.83333
    To bring the 1 kg of 12-karat gold to 18-karat (75%):    Solve: .75 = (.5*1 + x) / (1 + x)   → x = 1 kg
    To bring the 1 kg of 12-karat gold to 22-karat (91.666%):    Solve: .9166666 = (.5*1 + x) / (1 + x)   → x = 5 kg


Cost of certain task as a function of the percentage completion of the task.
e.g. environmental cleanup, invasive species removal
Let H be the cost to do 50% of the task (e.g. half of the toxins removed).
C(p) = Hp / (100-p), 0≤p<100

H = 1000




Open-top box with square base x, height y of given volume V.
Surface area (of outside or inside, not both) is S(x)=(x3+4V)/x

Cylinder of radius r and volume V.
Surface area: SA(r)=(2πr3+2V)/r