Give an example of a rational function that has vertical asymptote x = 3 and x = -3, horizontal asymptote y = 2 and y-intercept is (0, 4)
2 answers
f(x) = (x-4)/(x^2-9) + 2
Hmmm. f(0) = -4/-9 + 2 ≠ 4
How about
y = -18/(x^2-9) + 2 = 2(x^2-18)/(x^2-9)
That also has two x-intercepts. If you don't want any of those, then you will need to have y>0 for all x.
y = 2*9^2/(x^2-9)^2 + 2
How about
y = -18/(x^2-9) + 2 = 2(x^2-18)/(x^2-9)
That also has two x-intercepts. If you don't want any of those, then you will need to have y>0 for all x.
y = 2*9^2/(x^2-9)^2 + 2
1 answer