Find all horizontal asymptotes for f(x) = (4x)/sqrt(x^2+9)

please help :)

Hi,

If you consider the value of 4x/√x² when x is positive, then this approaches a horizontal asymptote at y = 4x/√x² = 4x/x = 4, so y = 4 is the horizontal asymptote when x > 0. <==ANSWER


If you consider the value of 4x/√x² when x is negative, then this approaches a horizontal asymptote at y = 4(-x)/√x² = -4x/x = -4, so y = -4 is the horizontal asymptote when x < 0. <==ANSWER

I hope that helps!! :-)

Write f(x) as

f(x) = (4x)/sqrt(x^2+9)=sqrt(16x^2/x^2+9)

The horizontal asymptotes are when

y= sqrt(16/1)=sqrt(16)= ±4

So y=-4 and y=4