1.) f(x) = 4x^2+4x+1/x^2+3x+2 2.) f(x) = 2x^2-5/3x^2+x-7 3.) f(x) = 3x+4/2x^2+3x+1
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1 Answer

1) f(x) = 4x^2+4x+1/x^2+3x+2=(2x+1)^2/((x+1)(x+2)).

Asymptotes occur when the denominator is zero: x=-1 and x=-2, but these are vertical asymptotes. The horizontal asymptotes are found by making x very large or very small (large and negative). Under these conditions, the numbers 1 and 2 become insignificant, so f(x) tends to 4x^2/x^2=4. So f(x)=4 is the asymptote.

2) f(x) = 2x^2-5/3x^2+x-7. Like (1) the horizontal asymptotes occur for large x: 2x^2/3x^2=2/3, so f(x)=2/3.

3.) f(x) = 3x+4/2x^2+3x+1. Unlike (1) and (2) the denominator is "heavier" than the numerator by a power of x, which means that as x gets larger f(x) approaches zero, the horizontal axis.

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