You can put this solution on YOUR website! f(x) = 4x^2/x^2-9
I'm assuming you're asking for Vertical and Horizontal asymptote.
The Vertical Asymptote occurs when the denominator is equal to zero.
x^2-9=0
(x+3)(x-3)=0
x= 3, -3
So the vertical Asymptotes occur at x=3 and x=-3.
To find the asymptote the method I learned was to divide the top and bottom by 1/x^2. You divide top and bottom by 1/x^n(highest power in numerator).This is basically multiplying by 1 but it helps you out.
(4x^2/x^2)/(x^2/x^2 - 9/x^2)
=4/(1-9/x^2)
Now we take the limit of the as x-> infinity and negative infinity. This will mean that 9/x^2 becomes practically zero.
so we get 4/1 or 4. The graph will approach 4 when the x values approach positive or negative infinity.
This means the horizontal Asymptote is at y=4.
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