- A horizontal asymptote at y=1/3

-A vertical asymptote at x=-4/3

- x-intercept at (-1,0)

-y-intercept at (0,3/4)

-A hole at (1/2, 3/11)
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1 Answer

(1) HA AT y=⅓

The degrees of the numerator and denominator must be the same and the coefficient of the highest degree in the denominator must be 3 times that in the numerator.

(2) VA AT x=-4/3

The factor x+4/3 or 3x+4 must appear in the denominator. 3x+4 would also satisfy (1) in the absence of any other coefficients in the numerator.

(3) x-INTERCEPT (-1,0)

Numerator must contain the factor x+1.

(4) y-INTERCEPT (0,¾)

The factor x+3n (n≠0 is an integer) must appear in the numerator and, because of (2) x+n must be a factor in the denominator. This assumes there are no other constants in the denominator factors apart from 3x+4. If there are other constants, and their product is p then x+n/p replaces x+n. In particular, when p=-1, this changes x+n to x-n.

(5) HOLE AT (½,³⁄₁₁)

The factor x-½ or 2x-1 must appear in both numerator and denominator and when these are removed, the numerical value of the expression must be ³⁄₁₁ when x=½. When constructing the function the common factor can be omitted until the last stage.

y=(x+1)(x+3n)...(2x-1)/((3x+4)(x+n)...(2x-1)).

For all values of x≠½, (2x-1)/(2x-1)=1.

When x has a large magnitude, y≈xⁿ/3xⁿ=⅓;

When x=0, y=3n/(4n)=¾.

When x=-⁴⁄₃, there is a VA.

When x=½, y=³⁄₁₁:

3/11=(½+1)(½+3n).../((³⁄₂+4)(½+n)...,

3/11=3(1+6n).../(11(1+2n)...

For this to happen, (1+6n).../((1+2n)...)=1 when x=½ and n≠0.

But any other factors will produce a different y-intercept (see 4), so it will not be possible to produce a rational function satisfying all the given conditions.

by Top Rated User (1.1m points)

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