Please use paranthesis next time.
Assuming h(x) = a/(x-p) + q
Given,
i) (0,-3) is a point on the curve.
ii) (3,0) is a vertical asymptote
iii) (0,-1) is a horizontal asymptote
We know that a vertical asymptote occurs when denominator = 0
so at x - p = 0 vertical asymptote will occur.
=> 3 - p = 0
=> p = 3
Since in our function, degree of N(x) < D(x) (where N(x) is numerator and D(x) is denominator)
Therefore, the horizontal asymptote should occur at x = 0, but the whole function is shifted down by 1 unit (using ii)
So, q = -1
Now using (i) and values of p and q we can find a,
-3 = a/(0-3) - 1
=> -2 = -a/3
=> a = 6
Graph of the function: