`f(x) = y= (ax)/(bx+c)`
It is given that at x = 2 we have a vertical asymptote.
`f(2) = (2a)/(2b+c)`
If f(2) is a vertical asymptote then;
`2b+c = 0`
`c = -2b`
`f(x) = y= (ax)/(bx+c)`
`y = (ax)/(bx+c)`
`y(bx+c) = ax`
`x(yb-a) = -yc`
`x = (yc)/(a-yb)`
It is given that at y = 3 we have a vertical asymptote.
`a-3b = 0`
`a = 3b`
So the answers of a and c in terms of b are;
`a = 3b`
`c = -2b`
`y= (ax)/(bx+c)`
`y = (3bx)/(bx-2b)`
`y = (3x)/(x-2)`
The graph will be as follows.
Further Reading
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