f(x) = p(q)/q(x)
where q(x) = 0 at x=2, so
f(x) = p(x)/(x-2)
horizontal asymptote at y=0 means that the degree of p(x) is less than the degree of q(x). So,
f(x) = 1/(x-2)
This has no x-intercept. There is no x big enough that f(x) = 0.
1/(x-2) = -1/2 at x=2, so we need to scale it by -6 to get y(0) = . So,
f(x) = -6/(x-2)
how do you create a function that has a graph with the given features:
a. A vertical asymptote at x=2; a horizontal asymptote at y=0; no x-intercept; y-intercept is 3
please help.
1 answer
1 answer