find the horizontal and vertical asymptotes of the curve. you may want to use a graphing calculator (or…

Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero). Set (x^{2}-x^{4}=x^{2}(1 - x^{2})=x^{2}(1 - x)(1 + x)=0). Solving (x^{2}(1 - x)(1 + x)=0), we get (x = 0,x = 1,x=-1). Check numerator at these values: When (x = 0), (y=\frac{6+0}{0 - 0}) (undefined, but since (x^{2}) is a factor of denominator and not of numerator, (x = 0) is not an asymptote). When (x=1), (y=\frac{6 + 1}{1-1}) (undefined and numerator (6 + 1=7\neq0)). When (x=-1), (y=\frac{6+1}{1 - 1}) (undefined and numerator (6 + 1=7\neq0)).

Step2: Find horizontal asymptotes

For horizontal asymptotes, use the limit as (x\rightarrow\pm\infty). Divide numerator and denominator by (x^{4}): (y=\lim_{x\rightarrow\pm\infty}\frac{\frac{6}{x^{4}}+1}{\frac{1}{x^{2}}-1}) As (x\rightarrow\pm\infty), (\lim_{x\rightarrow\pm\infty}\frac{6}{x^{4}} = 0) and (\lim_{x\rightarrow\pm\infty}\frac{1}{x^{2}}=0) So (y=\frac{0 + 1}{0-1}=-1)

Answer:

(x = 1,-1) (y=-1)