Question

(I NEED URGENT HELP FOR THIS) Given the info. that a rational function has a vertical asymptote of x=3 and x=-3, horizontal asymptote at y=-4, no x-intercepts, and y-intercept: (0,2/3), no removable discontinuity, and no end behavior, what would the function be exactly? I know the denominator would be (x^2)-9, but I'm having trouble figuring out what the numerator would be since there are no x-intercepts.

Answer 1

Can someone help???

Answer 2

The graph of -1/(x^2-9) has the vertical asymptotes with y-intercept at (0,1/9)
To get that to be 2/3, we need
y = -6/(x^2-9)
That still has a horizontal asymptote at y=0, so we need to shift that down. But just subtracting 4 would give us x-intercepts.
So, let's try
y = -1/(x^2-9) - 1/18
That gives us a y-intercept of 1/18, and a horizontal asymptote of -1/18.
Now, to get that to y = -4, multiply by 72
y = 72(-1/(x^2-9) - 1/18) does that.
But now we have a y-intercept of (0,4)
So, we need to rethink the shift and scale.
y = a(-1/(x^2-9) - b)
such that
a(1/9 - b) = 2/3
a(0-b) = -4
a = 3-√3
b = -2(3+√3)/3
y = (3-√3)(-1/(x^2-9) + 2(3+√3)/3)

see the graph at
https://www.wolframalpha.com/input/?i=%283-%E2%88%9A3%29%28-1%2F%28x%5E2-9%29+%2B+2%283%2B%E2%88%9A3%29%2F3%29