Question

Find Horizontal and Vertical asymptotes for y=(x^2 -9)/(4x^2 +1)

Answer 1

Answer: Horizontal asymptote is $\dfrac{1}{4}$ and vertical asymptotes are $\pm \dfrac{1}{2i}$

Step-by-step explanation:

Since we have given that

$y=\dfrac{x^2-9}{4x^2+1}$

We need to find the horizontal and vertical asymptotes.

Since vertical asymptotes will occur where the denominator becomes zero.

So, here denominator is $4x^2+1$

Now,

$4x^2+1=0\\\\4x^2=-1\\\\x^2=\dfrac{-1}{4}\\\\x=\pm \dfrac{1}{2i}$

And the horizontal asympototes will occur when the coefficient of higher degree of numerator is divided by coefficient of higher degree of denominator.

$y=\dfrac{1}{4}$

Hence, horizontal asymptote is $\dfrac{1}{4}$ and vertical asymptotes are $\pm \dfrac{1}{2i}$