Number of parabola types?

Question

Characteristic properties of the graph of a function of the type

f(x) = a·x²+b·x+c

where a≠0, is given by the information about whether each of a, b, c and d is positive, negative or equal to zero. Here d denotes the discriminant d = b²-4·a·c.

Each of the four values has three possible cases, positive, negative or zero, except for the constant a which never equals zero. So we have at most 2·3³ = 54 types of parabolas. But for instance:

a,c<0=b<d is not possible

So the number of possible cases is less than 54. On the other hand any choice of a, b and c will provide a parabola, so we have at least 2·3² = 18 types. How many types are there and why?

Anonymous · Accepted Answer

Define g(n) by:

g(n) = 0 if n=0

g(n) = 1 if n>0

g(n) = -1 if n<0.

Assume we are given g(a), g(b), g(c), g(d).

g(a): determines if parabola points up or down (2 choices)

g(b): given g(a), it determines if max/min point is 0, right of 0, or left of 0. (3 choices)

g(d): determines if there are 0, 1, or 2 roots. (3 choices)

g(c): determines if f(0) is 0, negative, or positive.

The properties specified by g(a),g(b),g(d) are all independent, so given pos/zero/negative information for a,b,d determines 2*3*3 = 18 types.

The information provided by g(c) can SOMETIMES be inferred by g(a),g(b),g(d):

If a>0, no roots, then f(0)>0.

if a>0, 1 root, then f(0) property is determined by g(b).

if a>0, 2 roots, b=0, then f(0)<0.

*if a>0, 2 roots, b nonzero, then f(0) can still be pos/neg/zero.

if a<0, no roots, then f(0)<0

if a<0, 1 root, then f(0) property is determined by g(b)

if a<0, 2 roots, b=0, then f(0)>0.

*if a<0, 2 roots, b nonzero, then f(0) can still be pos/neg/zero.

So the two asterisk cases point out exactly 4 of the 18 original types that need special attention for breaking into sub-cases. So the number of types is:

(18-4)+ 2*3 + 2*3 = 26

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Number of parabola types?

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