Answer:
Below.
Step-by-step explanation:
f(x)=3x^2-6x-45/x^2-5x
= 3x^2-6x-45 / (x(x - 5))
The denominator is zero when x - 0 and x = 5.
So there will be a vertical asymptote at x = 0 and a hole at x = 5.
If we do the division we get f(x) = 3 remainder 9x - 45
so there will also be a horizontal asymptote where x = 3.
Naw bruv i’m doing the same problems rn and i’m struggling gl tho
