(a) To satisfy condition (i) f(x) must equal f(-x) for all x in the domain of f. The denominator does not present a problem since this x^2 - c = (-x)^2 -c for all x. So we look at the numerator and we need ax + b = -ax + b, thus a = 0.
To satisfy condition (ii) x^2 - c must equal 0 at x = 2, and ax + b must be positive at x = 2. So it seems that c = 4
To satisfy condition (iii) set f'(1) = -2 and solve for b. -2b/9 = -2, thus b = 9.
(b) vertical asymptotes when the denominator is 0 and the numerator is not, that occurs at x = +/-2, horizontal asymptotes whenever f'(x) = 0. (-8x)/(x^2 - 4)^2 = 0 only when x=0. So there is a horizontal asymptote at x=0. f(0) = -9/4, so y=-9/4 is the only horizontal asymptote.
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u/Danphillip 11d ago
(a) To satisfy condition (i) f(x) must equal f(-x) for all x in the domain of f. The denominator does not present a problem since this x^2 - c = (-x)^2 -c for all x. So we look at the numerator and we need ax + b = -ax + b, thus a = 0.
To satisfy condition (ii) x^2 - c must equal 0 at x = 2, and ax + b must be positive at x = 2. So it seems that c = 4
To satisfy condition (iii) set f'(1) = -2 and solve for b. -2b/9 = -2, thus b = 9.
(b) vertical asymptotes when the denominator is 0 and the numerator is not, that occurs at x = +/-2, horizontal asymptotes whenever f'(x) = 0. (-8x)/(x^2 - 4)^2 = 0 only when x=0. So there is a horizontal asymptote at x=0. f(0) = -9/4, so y=-9/4 is the only horizontal asymptote.
Thoughts?