Find the vertical and horizontal asymptotes of the function

Question

F(x)=x=(√9x^2)/(2x+1)

Answer 1

Not sure what the extra "x=" means, so I'll ignore it. And, assuming the usual carelssness with parentheses,
√(9x^2) = |3x|
so, f(x) = |3x|/(2x+1)

the vertical asymptotes occur when the denominator is zero. so where is
2x+1 = 0 ?

the horizontal asymptotes are at x = ±3/2

Answer 2

thanks